Complex number study notes tute0066
update 2016-05-24




<a name=tute0066_index> cplxdraw Image List
■ Complex number drawing board cplxdraw.htm Cauchy-Riemann Equations, begin 
■ CR is necessary condition
■ CR talk about smoothness
■ d[g(x)]/dx change g(x) physics meaning
■ derive The Cauchy–Riemann Equations
■ eqn.BW125 is valid for real and imag
■ apply eqn.BW125 to real axis 
■ real axis CR result
■ apply eqn.BW125 to imag axis 
■ imag axis CR result
■ Cauchy-Riemann Eq. in Cartesian form 
■ Analytic function section 
■ Analytic = CR + single-valued 
■ Analytic = CR + continuous
■ Analytic = CR + neighbor CR too
■ CR in polar math.msu.edu//shapiro
■ CR polar r vary, θ constant begin 
■ CR polar r vary, θ constant result 
■ CR polar r constant, θ vary begin 
■ multiply by one is a key step 
■ coefficient limit value 
■ multiply by one more (red) 'i'  
■ CR polar r constant, θ vary result
■ Cauchy-Riemann Eq. in polar form (r∂θ), ∂r OK ; (∂θ), ∂r ERROR 
■ puzzle, CR is NOT a water mellon complex draw Image List
■ derive f(z)=sin(z*z)+cos(z*z)
■ cplxdra1.jpg text 
■ cplxdra2.jpg text 
■ cplxdra3.jpg text 
■ cplxdra4.jpg text 
■ cplxdra5.jpg text 
■ cplxdra6.jpg text 
■ cplxdra7.jpg text 
■ cplxdra8.jpg text 
■ cplxdra9.jpg text 
■ cplxdr10.jpg text 
■ cplxdr11.jpg text 
■ cplxdr12.jpg text 
■ cplxdr13.jpg text 
■ cf=exp(-1/z/z/z/z), ux,uy,vx,vy
■ cplxdr14.jpg text 

<a name="docA001">
2016-02-11-14-27 start 
On 2016-01-29 Liu,Hsinhan begin write a page 
named "Complex number drawing board" 
http://freeman2.com/cplxdraw.htm
In input type method, listed 
Unit circle; Broken line; Parametric; Limit; 

<a name="docA002">
Unit circle section help user build a circle 
locus points in complex form '1+2i' or in 
(x,y) coordinate form '1,2'. For example 
circle center﹕0+0i   <== use digits and 'i'
circle radius﹕ 1     positive number
step length ﹕  6     3=3 degree, 360=full circle
start angle ﹕  0     0=start on +x axis, 45=start on y=x line
number steps ﹕ 60    6*60=360=go 360 steps, full circle

<a name="docA003">
Broken line section allow user input broken 
line corner (x,y) values. Program build points 
sequence for this broken line. For example 
1,0;0.866,0.5;0.5,0.866;0,1;-0.4999,0.866;-0.8659,0.5;-0.9999,0;
is input corner (x,y) values for upper broken 
circle

2016-02-11-14-38 call from 310-617-6444

<a name="docA004">
Parametric section let user input equation in 
parametric form, for example 
parabola
x(t)=t
y(t)=t*t
Hypotrochoid 
x(t)=1.4*cos(t)+.3*cos(14*t/2)
y(t)=1.4*sin(t)-.3*sin(14*t/2)

<a name="docA005">
2016-02-09 LiuHH shift attention to 
Limit section 
Target at Cauchy-Riemann Equations, 
abbreviated as "CR". 
Liu,Hsinhan study notes may 
contain error! Please be alert!
Let 
i=√(-1) ---eqn.BW101
be a imaginary unit. Equivalent form is 
i*i=-1 ---eqn.BW102
Assume z is a complex number 
z = x+i*y ---eqn.BW103
Both x and y are real numbers.

<a name="docA006">
Assume f(z) is a complex function, which 
take one input complex number and output 
another complex number u+i*v. Both u and v 
are real numbers. u be function of x and y 
u=u(x,y), similarly v=v(x,y). Complex 
function equation has the following form. 
f(z)=f(x+i*y)=u(x,y)+i*v(x,y) ---eqn.BW104
Equality in eqn.BW104 is NOT curve1 cross 
curve2, meet point equality.
Equality in eqn.BW104 is "re-write as"
<a name="docA007">
For example, let
f(z)=z*z=(x+i*y)*(x+i*y) ---eqn.BW105
f(z)=(x+i*y)*x + i*y*(x+i*y)
f(z)=x*x+i*y*x + i*y*x + i*y*i*y
f(z)=x*x +(i*y*x + i*y*x) + i*i*y*y
f(z)=x*x +i*(y*x + y*x) -y*y
f(z)=x*x-y*y + 2i*x*y ---eqn.BW106
<a name="docA008">
Input z, split to real x and imag i*y. 
Output f(z) do the same thing, re-write 
eqn.BW106 as 
f(z)=f(x,y)=u(x,y) + i*v(x,y) ---eqn.BW107
where 
u(x,y)=x*x-y*y ---eqn.BW108
v(x,y)=2*x*y   ---eqn.BW109
Above eqn.BW105 to eqn.BW109 emphasize 
equality in eqn.BW104 is "re-write as"
not two curve meet equality.

<a name="docA009">
Complex function differentiability is 
different from real function differentiability.
In one variable real function g(x) variable x 
approach target point x0 from greater than x0 
side approach (reduce value) to x0. Another 
approach is variable x approach target point 
x0 from smaller than x0 side approach (increase 
value) to x0. Total two side approach. variable 
x move on one dimensional line. 

<a name="docA010">
Complex function differentiability is another 
story. Complex function variable z approach 
target point z0 from horizontal path where 
z = x+i*y ---eqn.BW103
y is a constant y0, variable z is then 
z = x+i*y0 (only x vary) ---eqn.BW110

Another possible, variable z approach target 
point z0 from vertical path, in eqn.BW101
x is a constant x0, variable z is then 
z = x0+i*y (only y vary) ---eqn.BW111

<a name="docA011">
Third possible, variable z approach target 
point z0 from third path. In z=x+i*y if 
relate x and y by a constraint,
m*(x-x0)=(y-y0) ---eqn.BW112
here, m,x0,y0 all be constants. x and y are 
two variables. Since two variables related 
by eqn.BW112, one is independent variable, 
the other is dependent variable. In eqn.BW112 
solve for y, get 
y=m*(x-x0)+y0 ---eqn.BW113
<a name="docA012">
Substitute eqn.BW113 to eqn.BW103, replace y.
Complex variable z (only x vary)
z = x+i*[m*(x-x0)+y0] ---eqn.BW114
Here m is the slope of approaching path.
If m=0, eqn.BW114 change to eqn.BW110
If m→∞, exam eqn.BW112 ∞*(x-x0)=(y-y0)
∞*0=any value
This is true only if (x-x0)=0, which 
require x=x0 a constant value. y is free.
In eqn.BW114 x=x0 and use y=m*(x-x0)+y0
If m→∞, eqn.BW114 approach to eqn.BW111

<a name="docA013">
Change m value, change approach direction.
There are infinity many m, there are 
infinity many different path approach to 
target point z0=x0+i*y0 (constant point)
Complex function, infinity many path compare 
with real function, only two path. This is 
very different.
Cauchy-Riemann Equations are necessary equation 
for complex function differentiability. There 
is NO similar equations in one variable real 
function analysis.
2016-02-11-15-46 stop

<a name="docA014">
2016-02-11-20-15 start
However, there are similar requirement 
between real function analysis and 
complex function analysis. That is, for 
differentiability be true at target point, 
first, moving point must cross target point
continuously. This is asking no crack, 
one side function value = 
other side function value 
<a name="docA014a">
second, moving point must cross target point
smoothly. This is asking no wrinkle, 
one side function change rate value = 
other side function change rate value 
Cauchy-Riemann Equations are wrinkle detector.
CR are necessary, not sufficient, because CR 
cannot detect crack. 2016-02-14-11-49 
ALERT! Liu,Hsinhan words may be wrong!!
<a name="docA015">
For example, if input domain is discontinuous 
require moving point far jump from one circle 
to another circle, that is not differentiable. 
Both real analysis and complex analysis require 
domain be simple connected. 
On the other hand, if input domain is 
continuous, but output range has sudden 
jump like stair way, first step sudden jump 
to second step. Both real analysis and 
complex analysis consider sudden jump to 
be not differentiable. 
<a name="docA016">
If input domain is continuous, and output 
range is also continuous, but output range 
is not smooth. The real function 
g(x)=|x| ---eqn.BW115
has sharp corner at x=0. Both real analysis 
and complex analysis treat this sharp corner 
(wrinkle) as not differentiable point. 
<a name="docA017">
Cauchy-Riemann Equations is used 
in complex analysis to guarantee 
output range smoothness. Because 
CR equation involve slope only. 
ux, uy, vx, vy are all slope 
quantity and care smoothness.
2016-02-11-20-46 here

<a name="docA018">
In real function analysis we express curve 
slope by 
[g(x+h)-g(x)]/[(x+h)-(x)] ---eqn.BW116
In which x is an arbitrary point in domain. 
h is a tiny quantity. Domain items are all 
input values. Range items are all output.
Input (x+h), output g(x+h).
Input (x),   output g(x).
<a name="docA019">
eqn.BW116 say output difference divide by 
input difference. (x+h)-(x) can be simplify 
to h. eqn.BW116 is a slope expression. Use 
g' to represent slope of function g(x) at x.
Rewrite eqn.BW116 as next 
g'(x)=[g(x+h)-g(x)]/h ---eqn.BW117
In g'(x) prime represent gradient if x,h 
are length (space).
In g'(x) prime represent time change rate 
if x,h are time. 

<a name="docA020">
Both space change rate and time change rate 
change original function g(x) physics meaning.
Assume g(x) represent temperature in air. 
Then g'(x) represent temperature space 
change rate in air, temperature gradient.

<a name="docA021">
Assume g(t) represent distance traveled. 
Then g'(t) represent distance traveled in 
one second. That is speed. 
We cannot add length with speed.
We cannot add density with gradient.

Just like we cannot add one hour with two 
miles, we cannot add g(x) with g'(x). But, 
multiplication and division are allowed. 
Divide g(x) by distance, get gradient.
Divide g(t) by time, get speed.
Divide speed by time, get acceleration.
Multiply evaporation rate with seconds 
get evaporated water mass in that time 
period.

<a name="docA022">
Slope, or change rate, or gradient are 
expressed by //next is  ---eqn.BW118
[secondOutput - firstOutput]/[secondInput - firstInput]
Earlier 
g'(x)=[g(x+h)-g(x)]/h ---eqn.BW117
satisfy the Slope definition eqn.BW118.
This page use g(x) for real analysis. 
f(x+i*y) for complex analysis.

<a name="docA023">
Next derive The Cauchy–Riemann Equations.
Start from OutputChange/InputChange and 
take limit for very short InputChange.
 
f '(z)
limit
h→0
f(z+h) - f(z)

(z+h) - (z)
limit
h→0
f(z+h) - f(z)

h
---eqn.BW119
width of above equation
Why take limit for very short InputChange?
For long InputChange, we measure "from one 
o'clock to three o'clock traveled 130 miles". 
We say "The average speed in two hours is  
65 miles per hour." But this two hour average 
speed is not interesting, because police send 
ticket base on instant speed. To measure instant 
speed we take limit for very short InputChange. 

<a name="docA024">
2016-02-11-21-38 here
In eqn.BW119 both z and h are complex number, 
and both are input. z is an arbitrary point. 
h is a small quantity approach zero. If given 
target point z0, in eqn.BW119 change z to z0.

z is input, f(z) is output, still a complex 
number. For general representation, write 
f(z) as real part plus imaginary part, next
f(z) = u(x,y) + i*v(x,y) ---eqn.BW120 
<a name="docA025">
Earlier example, if 
f(z)=z*z=(x+i*y)*(x+i*y) ---eqn.BW105
then 
u(x,y)=x*x-y*y ---eqn.BW108
v(x,y)=2*x*y   ---eqn.BW109
Another example, if 
f(z)=complexConjugate(z)=x-i*y ---eqn.BW121
then 
u(x,y)= x ---eqn.BW122
v(x,y)=-y ---eqn.BW123
Recall 
z = x+i*y ---eqn.BW103

<a name="docA026">
Because f(z) has many different form. Use 
u(x,y) and v(x,y) is a better choice.

Re-write eqn.BW119 in u(x,y) and v(x,y) 
form as next. 
 
f'(z) |
|z=z0
limit
h→0
f(z0+h) - f(z0)

(z0+h) - (z0)
limit
h→0
[u(z0+h)+i*v(z0+h)] - [u(z0)+i*v(z0)]

h
---eqn.BW124 ; z0 is a fixed point, only h move.
width of above equation
First equality is from eqn.BW118 and take limit.
Second equality apply eqn.BW120 twice.
<a name="docA027">
2016-02-11-22-11 here
In eqn.BW124, z0 is a given point, not moving 
and 'moving' responsibility fall to h shoulder.
Symbol "h→0" say h approach to zero. Next 
split limit(u+i*v) to limit(u) + i*limit(v) 
Re-write eqn.BW124 right hand side as next 
f'(z0) = limit[h→0]{[u(z0+h)-u(z0)]/h}
      +i*limit[h→0]{[v(z0+h)-v(z0)]/h} ---eqn.BW125
eqn.BW125 is valid for both real axis approach 
and imaginary axis approach. 

<a name="docA028">
Approach target point z0, complex number 
analysis has two choices. One is along real 
axis. Second is along imaginary axis. 
When move along real axis, real component x 
change and imaginary component y is constant.
When move along imaginary axis, real component x 
is constant and imaginary component y change.

Because complex analysis input plane is 
a two dimensional space, therefore, 
move along real axis and 
move along imaginary axis 
take care of ALL possibilities.

<a name="docA029">
First move along real axis, in eqn.BW125 
require h be real number, no imaginary. 
In eqn.BW125 write fixed point z0 as x0+i*y0
In eqn.BW125 write moving component h as 
∆x+i*0 Drop zero "+i*0", get 
h=∆x (move on real axis only) ---eqn.BW126 
eqn.BW125 become 
<a name="docA030">
 
f '(z)
limit
∆x→0
u(x0+i*y0+∆x)+i*v(x0+i*y0+∆x) - u(x0+i*y0)-i*v(x0+i*y0)

∆x
---eqn.BW127
REAL axis
width of above equation
<a name="docA031">
2016-02-11-22-45 
Our goal is to write f'(z) as sum of ∂u(x,y)/∂x 
and ∂v(x,y)/∂x . It is better write u(x,y) and 
v(x,y) in two separate terms. Change eqn.BW127 
to next.
f '(z)=
 
limit
∆x→0
u(x0+∆x +i*y0) - u(x0+i*y0)

∆x
+i*
limit
∆x→0
v(x0+∆x +i*y0)-v(x0+i*y0)

∆x
---eqn.BW128
REAL axis
width of above equation
<a name="docA032">
2016-02-11-23-03
In eqn.BW128, left limit blue terms are 
imaginary component and not change. But 
red terms changed from x0+∆x to x0
eqn.BW128 left limit term is ∂u(x,y)/∂x 
Similarly, 
eqn.BW128 right limit term is ∂v(x,y)/∂x 
Abbreviate ∂u(x,y)/∂x as ux, 
abbreviate ∂v(x,y)/∂x as vx
eqn.BW128 has the final form 
f'(z)= ux +i*vx REAL axis ---eqn.BW129 
2016-02-11-23-12 stop

<a name="docA033">
2016-02-12-09-48 start
eqn.BW129 start from eqn.BW125 and set 
h=∆x ---eqn.BW126 
This requirement let point move on real 
axis, the result is eqn.BW129.

Since complex analysis has both real and 
imaginary to deal with. 
<a name="docA034">
Next start from eqn.BW125 and let 
h=∆x+i*∆y ---eqn.BW130 
move on imaginary axis, it is necessary 
that ∆x=0, x coordinate not change. Get 
h=i*∆y ---eqn.BW131
y move on imaginary axis. y itself is a 
real number, i*y make y an imaginary.
∆y in eqn.BW131 is small change of y . 
Now substitute eqn.BW131 to eqn.BW125
<a name="docA035">
 
f '(z)
limit
∆y→0
u(x0+i*y0+i*∆y)+i*v(x0+i*y0+i*∆y) - u(x0+i*y0)-i*v(x0+i*y0)

i*∆y
---eqn.BW132
IMAG axis
width of above equation
<a name="docA036">
2016-02-12-10-06
Our goal is to write f'(z) as sum of ∂u(x,y)/∂y 
and ∂v(x,y)/∂y . It is better write u(x,y) and 
v(x,y) in two separate terms. Change eqn.BW132 
to next.
f '(z)=
 
limit
∆y→0
u(x0 +i*y0+i*∆y) - u(x0+i*y0)

i*∆y
+i*
limit
∆y→0
v(x0 +i*y0+i*∆y)-v(x0+i*y0)

i*∆y
---eqn.BW133
IMAG axis
width of above equation
<a name="docA037">
2016-02-12-10-14
In eqn.BW133, left limit blue terms are 
imaginary component and change from y0+∆y 
to y0 due to ∆y. Red terms NOT change
eqn.BW133 left limit term is ∂u(x,y)/∂y 
Similarly, 
eqn.BW133 right limit term is ∂v(x,y)/∂y 
Abbreviate ∂u(x,y)/∂y as uy, 
abbreviate ∂v(x,y)/∂y as vy
<a name="docA038">
eqn.BW133 has the form 
f'(z)= uy/i +vy IMAG axis ---eqn.BW134
where eqn.BW133 two purple i* cancel, and 
carry red i from eqn.BW133 to eqn.BW134
In eqn.BW134, move red i from denominator 
to numerator, get final form

<a name="docA039"> f'(z)= -i*uy +vy IMAG axis ---eqn.BW135 Recall real axis result f'(z)= ux +i*vx REAL axis ---eqn.BW129 To keep derivative continuity, we insist REAL axis f'(z) and IMAG axis f'(z) be equal. In eqn.BW129 and eqn.BW135 equate real term get ux = vy ---eqn.BW136 In eqn.BW129 and eqn.BW135 equate imaginary term get -uy = vx ---eqn.BW137

<a name="docA040">
Cauchy-Riemann Equations
in Cartesian form
Let input be
z = x+i*y ---eqn.BW103
Write complex function result in real and imaginary
f(z) = u(x,y) + i*v(x,y) ---eqn.BW120
If f(z) is differentiable at point z=x0+i*y0, it is necessary that
next two relations hold.
 
∂u(x+iy)

∂x
|
|
|


x=x0
y=y0
∂v(x+iy)

∂y
|
|
|


x=x0
y=y0
---eqn.BW138
 
∂u(x+iy)

∂y
|
|
|


x=x0
y=y0
=-
∂v(x+iy)

∂x
|
|
|


x=x0
y=y0
---eqn.BW139
width of above equation
<a name=a502141446>
How to remember eqn.BW138 and eqn.BW139? //a502141446
Output real/imag: u(z)/v(z) ; Input real/imag: x/y ;
real partial real = + imag partial imag : ux = vy ---eqn.BW136
imag partial real = - real partial imag : vx =-uy ---eqn.BW137
∂u/∂x=∂v/∂y , ∂u/∂y=-∂v/∂x //a502191232add
2016-02-14-14-52

2016-02-12-11-23 stop

<a name="docA041"> Analytic function section 2016-02-12-13-15 What is the difference between analytic complex function f(z) and differentiable complex function f(z) ? 2016-02-10-17-38 Liu,Hsinhan download http://physics.oregonstate.edu/~wwarren/COURSES/ph461/505/h7.pdf physics.oregonstate.edu_~wwarren_h7.pdf say <a name=a502121315> [[ The function f(z) is analytic in a domain R in the complex plane if it is differentiable and also single-valued within R. ]] 2016-02-12-13-27 here <a name="docA042"> 2016-02-10-17-31 Liu,Hsinhan download http://sym.lboro.ac.uk/resources/Handout_Analytic.pdf sym.lboro.ac.uk-Handout_Analytic.pdf say [[ 3 Key points  Always split the function into real and imaginary parts, identify these as functions u(x; y) and v(x; y) respectively.  Use partial differentiation to find the terms in the Cauchy-Riemann equations, the function is analytic only where the Cauchy-Riemann equations are satisfied and continuous. ]] <a name=a502132224> 2016-02-13-22-24 open faculty.uaeu.ac.ae_jaelee_LN.pdf [[ §2.13 Analytic Functions. Definition 2.13.1. Let f be a complex function. 1. If f has a derivative at each point in some neighborhood of z, then f is analytic (or regular, holomorphic) at z. 2. If f is analytic at each point in a domain D or region R, then f is analytic in D or R. (Recall: a domain D means a connected open set and a region R means a domain with some, none, or all of its boundary points.) 2 3. If f is analytic in the entire complex plane, then f is an entire function. 4. If f is not analytic at z0 but it is analytic at some point in every neighborhood of z0, then z0 is a singular point or singularity of f. ]]

<a name="docA043"> 2016-02-10-17-43 Liu,Hsinhan access http://users.math.msu.edu/users/shapiro/teaching/classes/425/crpolar.pdf users.math.msu.edu_crpolar.pdf 2016-02-13-11-00 start The following is Cauchy-Riemann Equations in polar form. It is a study notes from reading users.math.msu.edu_crpolar.pdf <a name="docA044"> Complex number z in polar form is z=r*exp(i*θ) ---eqn.BW140 where r is distance from target point z=r*exp(i*θ) to complex zero point 0+0*i z is always greater than or equal to zero. z never negative. θ is angle between radial line r and real x axis line. Positive θ start from x axis counter clockwise rotation. Negative θ start from x axis clockwise rotation. If rotation < 90 degree, first quadrant angle has positive θ value. fourth quadrant angle has negative θ value. <a name="docA045"> Let f(z) be a complex function transform from z=r*exp(i*θ) to another complex number f(z)=f(r*exp(i*θ))=u(r,θ)+i*v(r,θ) ---eqn.BW141 u(r,θ) is a real number, no i in u(r,θ). v(r,θ) is a real number, no i in v(r,θ). i*v(r,θ) is a pure imaginary number. Differentiation in polar form, take r and θ as independent variables. f'(z)[z at z0]=limit[z→z0]{[f(z)-f(z0)]/[z-z0]} ---eqn.BW142 First vary r and hold θ be constant θ0, this requirement let z→z0 become r→r0
<a name="docA046">
f '(z0)=
 
limit
r→r0
f(re0) - [f(r0e0)]

re0 - r0e0
1

e0
limit
r→r0
u(r0) +i*v(r0) - [u(r00) +i*v(r00)]

r - r0
---eqn.BW143
θ=θ0=const
red r vary
width of above equation
2016-02-13-11-49 here 
eqn.BW143 left limit is re-write eqn.BW142 for 
θ=θ0=constant
eqn.BW143 left limit to right limit 
apply eqn.BW141 to eqn.BW143 numerator and move 
out denominator common factor e0 

<a name="docA047">
Next in eqn.BW143 regroup u() together and 
regroup v() together. It is easier to see the 
differentiation u'() and v'() in new grouped 
expression. 
Next also re-write 1/exp(iθ) as exp(-iθ).
 
f '(z0)
e-iθ0
limit
r→r0
u(r0) - u(r00) +i*[v(r0) - v(r00)]

r - r0
---eqn.BW144
θ=θ0=const
red r vary
width of above equation
<a name="docA048">
Next write u() with a limit and write v() with a limit .
f '(z0)=
 
e-iθ0
limit
r→r0
u(r0) - u(r00)

r - r0
e-iθ0
limit
r→r0
i*[v(r0) - v(r00)]

r - r0
---eqn.BW145
width of above equation
<a name="docA049">
2016-02-13-12-30
eqn.BW145 left limit is [e-iθ0]*ur 
eqn.BW145 right limit is i*[e-iθ0]*vr 
eqn.BW145 has a form indicate that f'(z0) is a 
sum of ur and vr 
 
f '(z0)=
e-iθ0
[
∂u(r,θ)

∂r
|
 
 
θ=θ0
r→r0
i*
∂v(r,θ)

∂r
|
 
 
θ=θ0
r→r0
]
---eqn.BW146
width of above equation compare with eqn.BW156

eqn.BW146 is differential equation along a 
radial path, keep angle θ be constant θ0.
2016-02-13-12-54

<a name="docA050">
2016-02-13-14-10 
Above differential equation along a radial 
path, angle θ be constant.
Below differential equation along a circular 
path, radius r be constant.

Differentiation in polar form, take r and θ 
as independent variables.
f'(z)[z at z0]=limit[z→z0]{[f(z)-f(z0)]/[z-z0]} ---eqn.BW142
Now vary θ and hold r be constant r0, this 
requirement let z→z0 become θ→θ0
<a name="docA051">
f '(z0)=
 
limit
θ→θ0
f(r0 e) - [f(r0 e0 )]

r0 e - r0 e0
1

r0
limit
θ→θ0
u(r0,θ) +i*v(r0,θ) - [u(r0,θ0) +i*v(r0,θ0) ]

e - e0
---eqn.BW147
r=r0=const
red θ vary
width of above equation
<a name="docA052">
2016-02-13-14-31
In eqn.BW147, moved 1/r0 out of limit expression.
Next, in eqn.BW147 re-group, put u() together 
and put v() together. 
 
f '(z0)=
1

r0
limit
θ→θ0
u(r0,θ) - u(r0,θ0) +i*v(r0,θ) -i*v(r0,θ0)

e - e0
θ-θ0

θ-θ0
---eqn.BW148
width of above equation
<a name="docA053">
2016-02-13-14-52
For differentiation consideration, see eqn.BW149 
θ is a variable, (θ-θ0) should be denominator, 
but in eqn.BW148 (e - e0) is denominator 
term. Modify equation to better shape as next. 
In eqn.BW148, multiply whole equation by one.
See blue (θ-θ0)/(θ-θ0) This multiplication do 
not change equation, but make derivation easier. 
Now write eqn.BW148 in two limit form and 
switch denominator get next equation.
<a name="docA054">
f '(z0)=
 
1

r0
limit
θ→θ0
{
[
u(r0,θ) - u(r0,θ0)

θ-θ0
i*
v(r0,θ) -v(r0,θ0)

θ-θ0
]
θ-θ0

e - e0
}
---eqn.BW149
r=r0=const
red θ vary
width of above equation
<a name="docA055">
2016-02-13-15-17
In eqn.BW149 
limit[θ→θ0]{u(r0,θ)-u(r00)}/{θ-θ0} is very 
close to ∂u(r,θ)/∂θ and 
limit[θ→θ0]{v(r0,θ)-v(r00)}/{θ-θ0} is very 
close to ∂v(r,θ)/∂θ , IF (θ-θ0)/(e - e0) 
has a limit. Now study the reciprocal of 
the fraction term. Next equation is ---eqn.BW150
<a name="docA056">
Left equality use Euler's formula
 
e - e0

θ-θ0
cos(θ)+i*sin(θ) - [cos(θ0)+i*sin(θ0)]

θ-θ0
cos(θ)-cos(θ0) +i*[sin(θ)-sin(θ0)]

θ-θ0
width of above equation Above equation ---eqn.BW150
<a name="docA057">
2016-02-13-15-39
In eqn.BW149, there is a limit symbol. Apply 
limit[θ→θ0] to eqn.BW150
limit[θ→θ0]{[cos(θ)-cos(θ0)]/[θ-θ0]}=-sin(θ0) ---eqn.BW151
limit[θ→θ0]{[sin(θ)-sin(θ0)]/[θ-θ0]}=+cos(θ0) ---eqn.BW152 
eqn.BW151 is d[cos(θ)]/dθ at θ=θ0.
eqn.BW152 is d[sin(θ)]/dθ at θ=θ0.
eqn.BW150 has a limit value -sin(θ0)+i*cos(θ0).

<a name="docA058">
-sin(θ0)+i*cos(θ0)
=i*i*sin(θ0)+i*cos(θ0)
=i*[i*sin(θ0)+cos(θ0)] //use Euler's formula get next 
=i*e0 ---eqn.BW153 
Because (θ-θ0)/(e - e0) has a reciprocal limit 
value i*e0, substitute this value back to eqn.BW149
<a name="docA059">
f '(z0)=
 
1

i*e0 r0
limit
θ→θ0
{
u(r0,θ) - u(r0,θ0)

θ-θ0
i*
v(r0,θ) -v(r0,θ0)

θ-θ0
}
---eqn.BW154
width of above equation
<a name="docA060">
2016-02-13-16-04
eqn.BW154 purple term come from eqn.BW150 to 
eqn.BW153. 
In eqn.BW154 write purple denominator 1/e0 as 
a numerator term and change power sign e-iθ0
In eqn.BW154 both numerator and denominator multiply 
by 'i*' (i/i is one) eqn.BW154 become next form.
<a name="docA061">
f '(z0)=
 
e-iθ0

i*i*r0
limit
θ→θ0
{
i*
u(r0,θ) - u(r00)

θ-θ0
i*i*
v(r0,θ) -v(r00)

θ-θ0
}
---eqn.BW155
width of above equation
<a name="docA062">
2016-02-13-16-22
In eqn.BW155
limit{u(r0,θ)-u(r00)]/[θ-θ0]} is uθ(r,θ)
limit{v(r0,θ)-v(r00)]/[θ-θ0]} is vθ(r,θ)
In eqn.BW155 i*i=-1 we get next 
 
f '(z0)=
e-iθ0

r0
*
[
-i*
∂u(r,θ)

∂θ
|
 
 
θ→θ0
r=r0
∂v(r,θ)

∂θ
|
 
 
θ→θ0
r=r0
]
---eqn.BW156
width of above equation compare with eqn.BW146
<a name="docA063">
2016-02-13-16-33
Compare eqn.BW156 with eqn.BW146, we demand 
f'(z0) from "θ constant and r vary" formula 
f'(z0) from "θ vary and r constant" formula 
both derivative be the same. Get 
Cauchy-Riemann Equations in polar form 

<a name="docA064">
Cauchy-Riemann Equations
in polar form
Let input be
z=r*exp(i*θ) ---eqn.BW140
Write complex function result in real and imaginary
f(z)=f(r*exp(i*θ))=u(r,θ)+i*v(r,θ) ---eqn.BW141
If f(z) is differentiable at point z=r*exp(i*θ), it is necessary that
next two relations hold.
<a name="docA065">
 
∂u(r,θ)

∂r
|
|
|


r=r0
θ=θ0
1

r
∂v(r,θ)

∂θ
|
|
|


r=r0
θ=θ0
---eqn.BW157
 
∂v(r,θ)

∂r
|
|
|


r=r0
θ=θ0
= -
1

r
∂u(r,θ)

∂θ
|
|
|


r=r0
θ=θ0
---eqn.BW158
width of above equation
eqn.BW157 and eqn.BW158 are equation (3) page 2/2 in
http://users.math.msu.edu/users/shapiro/teaching/classes/425/crpolar.pdf

eqn.BW157 and eqn.BW158 blue r come from 
eqn.BW156 blue r.
2016-02-13-16-50 stop

<a name="docA066">
2016-02-13-18-37 start 
Cauchy-Riemann Equations in Cartesian form
∂u/∂x = ∂v/∂y ---eqn.BW159
∂u/∂y =-∂v/∂x ---eqn.BW160 
Cauchy-Riemann Equations in Polar form
∂u/∂r = ∂v/(r∂θ)  ---eqn.BW161 
∂u/(r∂θ) =-∂v/∂r  ---eqn.BW162 

<a name="docA067">
Cartesian form denominator has ∂x and ∂y
Polar form denominator has (r∂θ) and ∂r
∂x and ∂y no extra variable attach to either 
one. But why (r∂θ) has an 'r' attach to ∂θ?
'r' in (r∂θ) is necessary. If in eqn.BW161 
and eqn.BW161 drop 'r', write 
∂u/∂r = ∂v/(∂θ)  ---eqn.ER161 
∂u/(∂θ) =-∂v/∂r  ---eqn.ER162 
that is absolutely error ! 
<a name="docA068">
Start from 
z=r*exp(i*θ) ---eqn.BW140
dz=d[r*exp(i*θ)]
dz=∂r*[exp(i*θ)]+r*∂[exp(i*θ)]
dz=∂r*[exp(i*θ)]+r*[exp(i*θ)]∂(i*θ) 
dz=∂r*[exp(i*θ)]+r*[exp(i*θ)]*i*∂θ ---eqn.BW163
In eqn.BW163, r*∂θ show up as ONE PIECE.
<a name="docA069">
Another view point.
θ is an angle, arc length divide by radius.
θ is length divide by length, a pure number.
∂r has physics dimension length, meter, inch etc.
∂θ is a pure number, length/length.
r*∂θ is length*pure number = length. 
r*∂θ has physics dimension length, meter, inch etc.
Use ∂r with r*∂θ is consistent.
2016-02-13-18-56

<a name="docA070">
2016-02-13-19-27
A puzzle. 
Complex function one variable input is in a 
real axis and imaginary axis 2-Dimensional 
plane. 
2016-02-11-23-20 Liu,Hsinhan had a question 
Cauchy–Riemann Equations
make sure 
f'(z0) on REAL axis and 
f'(z0) on IMAG axis 
both equal. 
<a name="docA071">
But !! Is it necessary?
Put a water mellon on table, take a cutting 
board lean on water mellon. Board tangent to 
water mellon and touch at one point P. 
Draw a in-board tangent parallel to table 
and pass point P. 
Draw another in-board tangent perpendicular 
to first tangent and pass point P . In this 
case first tangent has slope zero and second 
tangent has slope one! Why 
Cauchy and Riemann demand two tangent have 
same slope?!

<a name="docA072">
In few second, LiuHH got answer.
water mellon is in three dimensional space. 
Complex function input is in a 2-Dimensional 
space and output to another 2-Dimensional 
space. Total four Dimensional space. 
water mellon comparison is inadequate. 

<a name="docA073">
Draw a parabola and a tangent to parabola. 
A point approach to tangent point from upper 
side or from lower side, must have same 
slope at tangent point.
Complex function input/output problem should 
compare with parabola and tangent problem. 
From either real or imaginary axis approach 
a target point, we should get same derivative. 
Although we view x+i*y has two dimension. But 
cplxFunc(x+i*y) view x+i*y as one dimension.

<a name="docA074">
2014-05-21-23-40 access 
http://www.math.columbia.edu/~rf/complex2.pdf
www.math.columbia.edu_~rf_complex2.pdf
This page say 
[[
2.3 Complex derivatives
Having discussed some of the basic 
properties of functions, we ask now
what it means for a function to have 
a complex derivative. Here we will
see something quite new: 
<a name="docA075">
this is very 
different from asking that its real and
imaginary parts have partial derivatives 
with respect to x and y. We will
not worry about the meaning of the 
derivative in terms of slope, but only
ask that the usual difference quotient exists.
]]
water mellon comparison use tangent line slope 
which is a "not worry" topic. 

<a name="docA076">
2016-02-12-17-16
http://people.reed.edu/~mayer/math112.html/html2/node24.html
people.reed.edu_~mayer_node24.html
[[
10.1 Derivatives of Complex Functions
You are familiar with derivatives of
 functions from  $\mbox{{\bf R}}$ to  $
\mbox{{\bf R}}$, and with the motivation
 of the definition of derivative as the
 slope of the tangent to a curve. 

<a name="docA077">
For complex functions, the geometrical
 motivation is missing, but the definition
 is formally the same as the definition
 for derivatives of real functions.
]]
water mellon comparison use tangent line slope 
which is a "motivation is missing" topic. 
2016-02-13-20-13 stop
2016-02-14-15-49 done proofread.

<a name="docA078">
2016-02-19-11-51 start 
Liu,Hsinhan write 
Complex number drawing board
Cauchy-Riemann Equations drawing board
http://freeman2.com/cplxdraw.htm
need build one example. The following 
is derivation draft work. 
<a name="docA079">
Define 
f(z)=sin(z*z)+cos(z*z) ---eqn.BW171
Given //see freeman2.com/tute0006.htm#csin01
z=x+i*y ---eqn.BW172
sin(x+i*y)= sin(x)*(exp(-y)+exp(+y))/2
         +i*cos(x)*(exp(+y)-exp(-y))/2 ---eqn.BW173
Change x to q, change y to r get 
sin(q+i*r)= sin(q)*(exp(-r)+exp(+r))/2
         +i*cos(q)*(exp(+r)-exp(-r))/2 ---eqn.BW174
<a name="docA080">
Next consider sin(z*z) 
sin(z*z)=sin((x+i*y)*(x+i*y))
sin(z*z)=sin(x*x-y*y+i*2*x*y) ---eqn.BW175
Compare eqn.BW175 with eqn.BW174
Let 
q=(x*x-y*y) ---eqn.BW176
r=(2*x*y)   ---eqn.BW177
<a name="docA081">
Re-write eqn.BW174 as next
sin(q+i*r)= 
sin(x*x-y*y+i*2*x*y)= ---eqn.BW178
   sin(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2
+i*cos(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2

<a name="docA082">
Next see 
cos(z*z)=cos(x*x-y*y+i*2*x*y) ---eqn.BW179
and 
cos(x+iy)=cos(x)*(exp(-y)+exp(+y))/2
     +i*(-sin(x)*(exp(+y)-exp(-y))/2) ---eqn.BW180
Re-write cos(x+iy) as cos(q+ir)
cos(q+ir)=cos(q)*(exp(-r)+exp(+r))/2
     +i*(-sin(q)*(exp(+r)-exp(-r))/2) ---eqn.BW180
<a name="docA083">
cos(q+ir)=
cos(x*x-y*y+i*2*x*y)= ---eqn.BW181
     cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2
+i*(-sin(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2)
Original problem 
f(z)=sin(z*z)+cos(z*z) ---eqn.BW171
become eqn.BW178 + eqn.BW181
That is 
<a name="docA084">
sin(z*z)+cos(z*z)= ---eqn.BW182
   sin(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2
+i*cos(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2
     cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2
+i*(-sin(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2)

<a name="docA085">
Next write 
sin(z*z)+cos(z*z) = u+i*v ---eqn.BW183
Define 
u(x,y)=  ---eqn.BW184
 sin(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2
+cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2
Define 
v(x,y)=  ---eqn.BW185
 cos(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2
-sin(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2

<a name="docA086">
Cauchy-Riemann Equations are next 
∂u/∂x=∂v/∂y  ---eqn.BW186
∂u/∂y=-∂v/∂x ---eqn.BW187
Need find ∂u(x,y)/∂x , ∂v(x,y)/∂y and 
∂u(x,y)/∂y , ∂v(x,y)/∂x as next.

<a name="docA087">
∂u(x,y)/∂x =  ---eqn.BW188
∂[ sin(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2
  +cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2]/∂x 

=∂[sin(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2]/∂x 
+∂[cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2]/∂x 

<a name="docA088">
  =[∂sin(x*x-y*y)/∂x]*(exp(-2*x*y)+exp(+2*x*y))/2
  +sin(x*x-y*y)*∂[(exp(-2*x*y)+exp(+2*x*y))/2]/∂x 
+[∂cos(x*x-y*y)]/∂x*(exp(-2*x*y)+exp(+2*x*y))/2 
+cos(x*x-y*y)*∂[(exp(-2*x*y)+exp(+2*x*y))/2]/∂x 

  =[cos(x*x-y*y)*2*x]*(exp(-2*x*y)+exp(+2*x*y))/2
  +sin(x*x-y*y)*[(-2*y*exp(-2*x*y)+2*y*exp(+2*x*y))/2]
+[-sin(x*x-y*y)*2*x]*(exp(-2*x*y)+exp(+2*x*y))/2 
+cos(x*x-y*y)*[(-2*y*exp(-2*x*y)+2*y*exp(+2*x*y))/2]

<a name="docA089">
Next cancel 2/2 
∂u(x,y)/∂x =  ---eqn.BW189
  [cos(x*x-y*y)*x]*(exp(-2*x*y)+exp(+2*x*y))
  +sin(x*x-y*y)*(-y*exp(-2*x*y)+y*exp(+2*x*y))
+[-sin(x*x-y*y)*x]*(exp(-2*x*y)+exp(+2*x*y))
+cos(x*x-y*y)*(-y*exp(-2*x*y)+y*exp(+2*x*y))

<a name="docA090">
∂u(x,y)/∂x =  ---eqn.BW190
   x*cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))
  +y*sin(x*x-y*y)*(-exp(-2*x*y)+exp(+2*x*y))
-x*sin(x*x-y*y)*( exp(-2*x*y)+exp(+2*x*y))
+y*cos(x*x-y*y)*(-exp(-2*x*y)+exp(+2*x*y))

<a name="docA091">
Next find ∂v(x,y)/∂y
v(x,y)=  ---eqn.BW185
 cos(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2
-sin(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2
∂v(x,y)/∂y= ---eqn.BW191
∂[cos(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2
 -sin(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2]/∂y

=∂[cos(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2]/∂y
+∂[-sin(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2]/∂y

<a name="docA092">
  =[∂cos(x*x-y*y)/∂y]*(exp(+2*x*y)-exp(-2*x*y))/2
  +cos(x*x-y*y)*∂[(exp(+2*x*y)-exp(-2*x*y))/2]/∂y
+[-∂sin(x*x-y*y)/∂y]*(exp(+2*x*y)-exp(-2*x*y))/2
-sin(x*x-y*y)*∂[(exp(+2*x*y)-exp(-2*x*y))/2]/∂y

  =[-sin(x*x-y*y)*(-2*y)]*(exp(+2*x*y)-exp(-2*x*y))/2
  +cos(x*x-y*y)*[(2*x*exp(+2*x*y)+2*x*exp(-2*x*y))/2]
+[-cos(x*x-y*y)*(-2*y)]*(exp(+2*x*y)-exp(-2*x*y))/2
-sin(x*x-y*y)*[(2*x*exp(+2*x*y)+2*x*exp(-2*x*y))/2]

<a name="docA093">
Next cancel 2/2 
∂v(x,y)/∂y= ---eqn.BW192
  [-sin(x*x-y*y)*(-y)]*(exp(+2*x*y)-exp(-2*x*y))
  +cos(x*x-y*y)*[(x*exp(+2*x*y)+x*exp(-2*x*y))]
+[-cos(x*x-y*y)*(-y)]*(exp(+2*x*y)-exp(-2*x*y))
-sin(x*x-y*y)*[(x*exp(+2*x*y)+x*exp(-2*x*y))]

<a name="docA094">
∂v(x,y)/∂y= ---eqn.BW193 //a502191306
  =y*sin(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))
  +x*cos(x*x-y*y)*(exp(+2*x*y)+exp(-2*x*y))
+y*cos(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))
-x*sin(x*x-y*y)*(exp(+2*x*y)+exp(-2*x*y))

<a name="docA095">
compare with
∂u(x,y)/∂x =  ---eqn.BW190
   x*cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))
  +y*sin(x*x-y*y)*(-exp(-2*x*y)+exp(+2*x*y))
-x*sin(x*x-y*y)*( exp(-2*x*y)+exp(+2*x*y))
+y*cos(x*x-y*y)*(-exp(-2*x*y)+exp(+2*x*y))

∂v(x,y)/∂y=∂u(x,y)/∂x  ---eqn.BW186
is true a502191309
2016-02-19-13-09 stop 

<a name="docA096">
2016-02-19-14-05 start 
Cauchy-Riemann Equations are next 
∂u/∂x=∂v/∂y  ---eqn.BW186
∂u/∂y=-∂v/∂x ---eqn.BW187
Need find ∂u(x,y)/∂x , ∂v(x,y)/∂y and 
∂u(x,y)/∂y , ∂v(x,y)/∂x as next.

Next find 
∂v(x,y)/∂x=-∂u(x,y)/∂y ---eqn.BW187

<a name="docA097">
u(x,y)=  ---eqn.BW184
 sin(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2
+cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2
and 
v(x,y)=  ---eqn.BW185
 cos(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2
-sin(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2

<a name="docA098">
∂u(x,y)/∂y= ---eqn.BW194
∂[sin(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2
+cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2]/∂y
=∂[sin(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2]/∂y
+∂[cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2]/∂y

  =[∂sin(x*x-y*y)/∂y]*(exp(-2*x*y)+exp(+2*x*y))/2
  +sin(x*x-y*y)*∂[(exp(-2*x*y)+exp(+2*x*y))/2]/∂y
+[∂cos(x*x-y*y)/∂y]*(exp(-2*x*y)+exp(+2*x*y))/2
+cos(x*x-y*y)*∂[(exp(-2*x*y)+exp(+2*x*y))/2]/∂y

<a name="docA099">
  =[cos(x*x-y*y)*(-2*y)]*(exp(-2*x*y)+exp(+2*x*y))/2
  +sin(x*x-y*y)*[(-2*x*exp(-2*x*y)+2*x*exp(+2*x*y))/2]
+[-sin(x*x-y*y)*(-2*y)]*(exp(-2*x*y)+exp(+2*x*y))/2
+cos(x*x-y*y)*[(-2*x*exp(-2*x*y)+2*x*exp(+2*x*y))/2]

Next cancel 2/2
∂u(x,y)/∂y= ---eqn.BW195
  =cos(x*x-y*y)*(-y)*(exp(-2*x*y)+exp(+2*x*y))
  +sin(x*x-y*y)*(-x*exp(-2*x*y)+x*exp(+2*x*y))
-sin(x*x-y*y)*(-y)*(exp(-2*x*y)+exp(+2*x*y))
+cos(x*x-y*y)*(-x*exp(-2*x*y)+x*exp(+2*x*y))

<a name="docA100">
Final ∂u/∂y
∂u(x,y)/∂y= ---eqn.BW196
  =(-y)*cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))
    +x*sin(x*x-y*y)*(-exp(-2*x*y)+exp(+2*x*y))
+y*sin(x*x-y*y)*( exp(-2*x*y)+exp(+2*x*y))
+x*cos(x*x-y*y)*(-exp(-2*x*y)+exp(+2*x*y))

<a name="docA101">
Next find 
∂v(x,y)/∂x= ---eqn.BW197
∂[cos(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2
-sin(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2]/∂x
=∂[cos(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2]/∂x
-∂[sin(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2]/∂x

  =[∂cos(x*x-y*y)/∂x]*(exp(+2*x*y)-exp(-2*x*y))/2
  +cos(x*x-y*y)*∂[(exp(+2*x*y)-exp(-2*x*y))/2]/∂x
-[∂sin(x*x-y*y)/∂x]*(exp(+2*x*y)-exp(-2*x*y))/2
-sin(x*x-y*y)*∂[(exp(+2*x*y)-exp(-2*x*y))/2]/∂x

<a name="docA102">
  =[-sin(x*x-y*y)*2*x]*(exp(+2*x*y)-exp(-2*x*y))/2
  +cos(x*x-y*y)*(2*y*exp(+2*x*y)+2*y*exp(-2*x*y))/2
-[cos(x*x-y*y)*2*x]*(exp(+2*x*y)-exp(-2*x*y))/2
-sin(x*x-y*y)*(2*y*exp(+2*x*y)+2*y*exp(-2*x*y))/2

Next cancel 2/2
∂v(x,y)/∂x= ---eqn.BW198
  -sin(x*x-y*y)*x*(exp(+2*x*y)-exp(-2*x*y))
  +cos(x*x-y*y)*(y*exp(+2*x*y)+y*exp(-2*x*y))
-cos(x*x-y*y)*x*(exp(+2*x*y)-exp(-2*x*y))
-sin(x*x-y*y)*(y*exp(+2*x*y)+y*exp(-2*x*y))

<a name="docA103">
Next final ∂v/∂x
∂v(x,y)/∂x= ---eqn.BW199
  -x*sin(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))
  +y*cos(x*x-y*y)*(exp(+2*x*y)+exp(-2*x*y))
-x*cos(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))
-y*sin(x*x-y*y)*(exp(+2*x*y)+exp(-2*x*y))

<a name="docA104">
compare ∂v/∂x with  ∂u/∂y
∂u(x,y)/∂y= ---eqn.BW196
  -y*cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))
  +x*sin(x*x-y*y)*(-exp(-2*x*y)+exp(+2*x*y))
+y*sin(x*x-y*y)*( exp(-2*x*y)+exp(+2*x*y))
+x*cos(x*x-y*y)*(-exp(-2*x*y)+exp(+2*x*y))

confirmed 
∂v(x,y)/∂x=-∂u(x,y)/∂y ---eqn.BW187
2016-02-19-14-29 

<a name="docA105">
Summerize six equations 
u(x,y)=  ---eqn.BW184
 sin(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2
+cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))/2

v(x,y)=  ---eqn.BW185
 cos(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2
-sin(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))/2

<a name="docA106">
∂v(x,y)/∂x= ---eqn.BW199
  -x*sin(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))
  +y*cos(x*x-y*y)*(exp(+2*x*y)+exp(-2*x*y))
-x*cos(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))
-y*sin(x*x-y*y)*(exp(+2*x*y)+exp(-2*x*y))

<a name="docA107">
compare ∂v/∂x with  ∂u/∂y
∂u(x,y)/∂y= ---eqn.BW196
  -y*cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))
  +x*sin(x*x-y*y)*(-exp(-2*x*y)+exp(+2*x*y))
+y*sin(x*x-y*y)*( exp(-2*x*y)+exp(+2*x*y))
+x*cos(x*x-y*y)*(-exp(-2*x*y)+exp(+2*x*y))

<a name="docA108">
∂v(x,y)/∂y= ---eqn.BW193 //a502191306
  =y*sin(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))
  +x*cos(x*x-y*y)*(exp(+2*x*y)+exp(-2*x*y))
+y*cos(x*x-y*y)*(exp(+2*x*y)-exp(-2*x*y))
-x*sin(x*x-y*y)*(exp(+2*x*y)+exp(-2*x*y))

<a name="docA109">
compare ∂u/∂x with ∂v/∂y
∂u(x,y)/∂x =  ---eqn.BW190
   x*cos(x*x-y*y)*(exp(-2*x*y)+exp(+2*x*y))
  +y*sin(x*x-y*y)*(-exp(-2*x*y)+exp(+2*x*y))
-x*sin(x*x-y*y)*( exp(-2*x*y)+exp(+2*x*y))
+y*cos(x*x-y*y)*(-exp(-2*x*y)+exp(+2*x*y))
Equation first two lines come from sin(z*z)
second two lines come from cos(z*z).

<a name="docA110">
f(z)=sin(z*z)+cos(z*z)
u(x,y) , v(x,y) and ∂u/∂x=∂v/∂y , ∂u/∂y=-∂v/∂x 
are used in 
http://freeman2.com/cplxdraw.htm#CRboard3
Click [a2] button to write f(z)=sin(z*z)+cos(z*z)
to Box41, then click CRRUN⇒[CRdraw] button 
to draw graph. Possibly most complicate one 
in sixteen examples. 
2016-02-19-14-35 

<a name=cplxdrawImageList>
Complex number curve image files
http://freeman2.com/complex2.gif ; complex2.jpg
sin(complex), cos(complex), input unit circle.
http://freeman2.com/complex4.gif ; complex4.jpg
exp(complex), tan(complex), input unit circle.
http://freeman2.com/complex6.gif
sin(complex), cos(complex), input unit SQUARE.
http://freeman2.com/cplxdraw.jpg  text
sin(complex), cos(complex), input unit circle.
http://freeman2.com/cplxdra1.jpg  text
Epitrochoid, Hypotrochoid. ISBN 0-486-60288-5 J. Dennis Lawrence
http://freeman2.com/cplxdra2.jpg  text
Involute of a circle ; Huygens, 1693
http://freeman2.com/cplxdra3.jpg  text
sin(z*z)+cos(z*z) where z=x+i*y input is a ╋ path.
http://freeman2.com/cplxdra4.jpg  text
sin(z*z)+cos(z*z) where z=x+i*y input is unit circle.
2016-02-20-14-26 record
http://freeman2.com/cplxdra5.jpg  text
cos(sin(z)) and sin(z)*sin(z)+cos(z)*cos(z)≡1 . a502241937
http://freeman2.com/cplxdra6.jpg  text
gear example.
Input unit circle with sine. Output f(z)=sin(z)^cos(z) . a502261653

<a name=cplxdra7open>

http://freeman2.com/cplxdra7.jpg  text
Circular mesh complex function mapping graphs. a503010923

<a name=cplxdra8open>

http://freeman2.com/cplxdra8.jpg  text
http://freeman2.com/cplxdra9.jpg  text
Square mesh cexpf(z) equal z blue lines. a503152213
Square mesh complex function mapping graphs. a503010925
2016-03-01-09-40 record cplxdra5.jpg to cplxdra8.jpg

http://freeman2.com/cplxdr10.jpg  text
Square mesh 1/z confirmed from other source. a504101702
http://i.stack.imgur.com/kPZVZ.png
http://freeman2.com/cplxdr11.jpg  text
Move on critical line x=0.5,y=t (input) zeta function czeta(z) graph
Online has many other graph confirm cplxdr11.jpg
http://freeman2.com/cplxdr12.jpg  text
zeta function czeta(z) root 1 to root 10 output mesh sit on (0,0)
Online may not have other graph confirm cplxdr12.jpg
http://freeman2.com/cplxdr13.jpg  text
zeta function czeta(z) input path from z=0.5-15i to z=0.5+15i
Parameter t, tstep=0.01; tbgn0=-15; tend0=15. root 1 = 0.5+i*14.134725
w=czeta(0.5+t*i) . cplxdr13.jpg draw [t,real(w)] and [t,imag(w)]
https://www.wolframalpha.com/ plot zeta confirm cplxdr13.jpg
http://freeman2.com/cplxdr14.jpg  text
f(z)=exp(-1/z/z/z/z) approach to zero and approach to infinity at same time

<a name=cplxdrawjpg>
2016-02-20-12-17 
Following is text printed on complex graph 

cplxdraw.jpg
http://freeman2.com/cplxdraw.jpg default graph
Liu,HsinHan off line Acer1 computer. a501312340

<a name=cplxdra1jpg>
cplxdra1.jpg
ISBN 0-486-60288-5 page 160 Epitrochoid, third set [18*]
ISBN 0-486-60288-5 page 165 Hypotrochoid, second set [14*]
Epitrochoid Dürer, 1525; ISBN 0-486-60288-5 J. Dennis Lawrence
also see http://freeman2.com/graph09e.htm draw 607

ISBN 0-486-60288-5 J. Dennis Lawrence 
p160-164 Epitrochoid, blue; Dürer, 1525
page 165-167 Hypotrochoid, red;
x(t)=1.6*cos(t); y(t)=1.6*sin(t)
above purple circle.
x(t)=1.4*cos(t)+.3*cos(14*t/2)
y(t)=1.4*sin(t)-.3*sin(14*t/2)
above Hypotrochoid; below Epitrochoid
x(t)=1.8*cos(t)-.3*cos(18*(t+0.34)/2)
y(t)=1.8*sin(t)-.3*sin(18*(t+0.34)/2)
3rd quadrant triangle,z= 32 equal z pts
4th quadrant triangle,z=142 equal z pts
2nd quadrant triangle,z=272 equal z pts
graph http://freeman2.com/cplxdra1.jpg
tools http://freeman2.com/cplxdraw.htm
2016-02-08-21-37 Liu,Hsinhan 劉鑫漢 

<a name=cplxdra2jpg>
cplxdra2.jpg
A Catalog of Special Plane Curves 
J. Dennis Lawrence , ISBN 0-486-60288-5
page 190, 191 Involute of a circle 
Huygens, 1693
x(t)=1*(cos(t)+t*sin(t));
y(t)=1*(sin(t)+t*cos(t));
parameter t bgn=-30;t end=30; t step=0.1
graph http://freeman2.com/cplxdra2.jpg
tools http://freeman2.com/cplxdraw.htm
2016-02-08-22-29 Liu,Hsinhan 劉鑫漢 

<a name=cplxdra3jpg>
cplxdra3.jpg
2016-02-19-16-14
This graph is drawing of sin(z*z)+cos(z*z) where z=x+i*y
Thin cross is input. ╋ represent approaching to target from
four directions. User define target point x0+i*y0 and define 
start point x1+i*y1. From xy1 to xy0 is purple line. Rotate 
90 deg. draw red, one more draw blue. third time black. After 
build four path, evaluate function value f(z)=u+i*v. Lower 
left is first draw u+i*v. Upper left 2nd draw ux+i*uy,vx+i*vy
here ux=∂u/∂x. Upper right 3rd draw ux+i*vy,uy+i*vx this is 
"✕" shape output. Lower right 4th draw, (ux,step),(uy,step) 
(vx,step),(vy,step) Cauchy-Riemann Equations show up in U.R. 
and L.R. draws. ∂u/∂y=-∂v/∂x made (uy,step),(vx,step) up/down
reflect. ∂u/∂x=∂v/∂y made (ux,step),(vy,step) red on yellow.
sin(z*z)+cos(z*z) http://freeman2.com/tute0066.htm#docA078
Cauchy-Riemann Eq. drawing freeman2.com/cplxdraw.htm#CRboard3
This graph http://freeman2.com/cplxdra3.jpg
Curve no confirmation! 2016-02-19-16-48 Liu,Hsinhan 劉鑫漢 
2016-02-20-12-24 done record

<a name=cplxdra4jpg>
cplxdra4.jpg
We know (sin z)squared+(cos z)squared≡one
Whoknow sin(z squared)+cos(z squared)≡what
This graph, input purple unit circle. 
Red sin(z squared), blue cos(z squared)
Black curve sin(z squared)+cos(z squared)
purple one turn;red/black two turn;blue 4
This draw http://freeman2.com/cplxdra4.jpg
ComplexDraw page freeman2.com/cplxdraw.htm
2016-02-20-14-18 Liu,Hsinhan 劉鑫漢 
2016-02-20-14-26 done record

<a name=cplxdra5jpg>
cplxdra5.jpg
2016-03-01-09-31 start 
update 2016-03-01 add cplxdra5.jpg, cplxdra6.jpg
cplxdra7.jpg and cplxdra8.jpg graph text

cplxdra5.jpg graph text is next 

Input purple circle radius=1.2 Output two composite 
functions. 1. cos(sin(z)) red curve , z is points 
on purple circle. 2. sin(z)*sin(z)+cos(z)*cos(z)≡1 
This 'curve' is invisible. Because they all stay on 
one point 1+0*i But, three triangles. All have one 
corner stay on 1+0*i. This concurrent phenomena say
the existence of sin(z)*sin(z)+cos(z)*cos(z) ≡ 1 
Composite function capability added to cplxdraw.htm 
Allow more function to be studied. Complex number 
drawing board and Cauchy-Riemann Equations 
drawing board  http://freeman2.com/cplxdraw.htm
This graph URL http://freeman2.com/cplxdra5.jpg
2016-02-24-19-16 Liu,Hsinhan 劉鑫漢

<a name=cplxdra6jpg>
cplxdra6.jpg
Input unit circle with sine. Parametric function 
x(t)=(1+0.05*sin(24*t))*cos(t); 
y(t)=(1+0.05*sin(24*t))*sin(t);        劉鑫漢
Input teeth are even spaced. Output curve also 
has teeth not even spaced. Aid with check points
(three blue lines) We know which input arc is 
squeezed and which is relaxed. Output curve is 
f(z)=sin(z)^cos(z) or cpowf(csinf(z),ccosf(z))
Begin not meet end on purpose. 2016-02-26-16-35
http://freeman2.com/cplxdra6.jpg Liu,Hsinhan
http://freeman2.com/cplxdraw.htm#a502251031 

<a name=cplxdra7jpg>
cplxdra7.jpg
Complex function mapping graph
Purple curcles input. Red curves output. 
csinf(z)  ccosf(z)
csinh(z)  ccosh(z)
cexpf(z)  cmulf(z,z)
Graph program  http://freeman2.com/cplxdraw.htm
These graph need to be confirmed. Reader must 
verify first before accept these graphs.
Graph page URL http://freeman2.com/cplxdra7.jpg
2016-02-29-16-17 Liu,Hsinhan 劉鑫漢 

<a name=cplxdra8jpg>
cplxdra8.jpg
Complex function mapping graph
Purple square input. Red curves output
cexpf(z)  cmulf(z,z)
csinf(z)  ccosf(z)
csinh(z)  ccosh(z)
These graph need to be confirmed. Reader must verify first
before accept these graphs. Blue lines link equal z points
Graph program  http://freeman2.com/cplxdraw.htm
Graph page URL http://freeman2.com/cplxdra8.jpg
2016-03-01-08-17 Liu,Hsinhan 劉鑫漢 
2016-03-01-09-56 record stop

<a name=cplxdra9jpg>
cplxdra9.jpg
2016-03-15-22-21 record start 
2016-03-10-08-58
square mesh draw cexpf(z) and set equal z points 
0 1 2 to 8 9 10 11 198 199 200 to 203 204 205 
get constant y 0 1 2 3 4 5 6 7 8 9 10 11 and
get constant x 198 199 200 201 202 203 204 205 
2016-03-10-09-08 http://freeman2.com/cplxdra9.jpg

Delete check points 
[ numbers ] delete 
blue straight lines

square mesh input z
red curves cexpf(z)
blue lines equal z.
Please visit webpage
http://freeman2.com/
cplxdraw.htm
2016-03-10-09-20
Liu,Hsinhan 劉鑫漢 
2016-03-15-22-22 record stop 

<a name=a503300933>
2016-03-30-09-33 start 
When LiuHH write complex calculator 
http://freeman2.com/complex4.htm
get numerical answer of i^i, where 
i=sqrt(-1). That time LiuHH did not 
know i^i=exp(-PI/2). Now 2016-03-30 
read cvch1.pdf which Liu,Hsinhan 
2009-03-17-11-15 download from 
http://personal.maths.surrey.ac.uk/st/C.Wulff/Modules/MS224/cvch1.pdf
2009-03-17-11-17 receive cvch2.pdf
2009-03-17-11-19 receive cvch3.pdf
2009-03-17-11-20 receive cvch4.pdf
2009-03-17-11-23 receive cvch5.pdf

<a name=a503300934>
Next record cvch1.pdf by Dr Claudia Wulff
Department of Mathematics
Faculty of Engineering and Physical Sciences
University of Surrey
cvch1.pdf page 18/18 section 1.7.3 
i^i=exp(-PI/2) ---eqn.BW200
The following is study notes, that is follow 
Dr Claudia Wulff lecture and write down with 
LiuHH own words (alert! may contain error!).
2016-03-30-09-52 here 
<a name=a503300935>
We know exp() and log() are two inverse 
functions. That is 
f(z)=exp(log(f(z))) ---eqn.BW201
and  
f(z)=log(exp(f(z))) ---eqn.BW202
Since log() involve multiple phase angles 
eqn.BW201 and eqn.BW202 are true for 
principal phase angle.
<a name=a503300936>
Let z and w both be complex numbers. 
z=a+b*i ---eqn.BW203
w=c*d*i ---eqn.BW204
where a,b,c,d are all real numbers. 
Let 
f(z,w)=z^w (=zw) ---eqn.BW205
Follow eqn.BW201, get 
z^w=exp(log(z^w)) ---eqn.BW206
Next is one log() function rule 
log(m^n)=n*log(m) ---eqn.BW207
Use eqn.BW207 change eqn.BW206 to next 
z^w=exp(w*log(z)) ---eqn.BW208
z and w are general complex numbers. 

<a name=a503300937>
Since eqn.BW200 study i^i, in eqn.BW203
and eqn.BW204, we set 
a=0,b=1,c=0,d=1 change z^w to i^i . 
Red i come from z, blue i come from w. 
eqn.BW208 become 
i^i=exp(i*Log(i)) ---eqn.BW209
here 
i=√-1=sqrt(-1)=√(-1) ---eqn.BW210
In eqn.BW209, Log(i) has many values 
Log(i)=log(|i|) + i*(arg(i)+ 2k*PI) ---eqn.BW211
<a name=a503300938>
Uppercase Log(i) represent multiple values 
of log function of 'i'. 'Multiple' come 
from eqn.BW211 k value be arbitrary 
positive integer value, infinite many. 
Principal value is defined to be k=0

<a name=a503300939>
Log(i) is a complex number, 
Log(i) real part is log(|i|) in eqn.BW211
|i|=1, log(|i|)=log(1)=0. Real part is 0.
Log(i) imaginary part is i*(arg(i)+ 2k*PI) 
in eqn.BW211. Now study the principal value 
set k=0, Log(i) imaginary part is i*(arg(i))
arg(i) is phase angle of z=i, which is PI/2
Rotate complex 1+0*i 90 degree (PI/2) get 
new complex number 0+1*i. 
1+0*i has phase angle = 0 rad = 0 deg.
0+1*i has phase angle = PI/2 rad = 90 deg.
Therefore Log(i) principal value log(i) is 
log(i)=0+i*(PI/2) ---eqn.BW212

<a name=a503300940>
After study Log(i) eqn.BW211, back to 
eqn.BW209 which is i^i. Substitude eqn.BW211 
to eqn.BW209, get next multiple value equation
ii=i^i=ei*Log(i) ---eqn.BW213
Substitude principal value eqn.BW212 to 
eqn.BW209, get next principal value equation
ii=i^i=ei*(i*(PI/2)) ---eqn.BW214
 Red i come from  red i (z=0+i), 
Blue i come from blue i (w=0+i). 
In eqn.BW214, i*i is -1, 
eqn.BW214 tell us i^i principal value is 
i^i=exp(-PI/2) ---eqn.BW200
2016-03-30-10-50 here 

<a name=a503300941>
In http://freeman2.com/complex4.htm#Box3JS
Box3, JS command input next three lines 
[[
'i'^'i'
cexpf((cdivf(cnegf(PI),2)))
E^(-PI/2)
]]
<a name=a503300942>
Click [eval Box3] button, get 
[[
cpowf(('i'),('i'))
0.20787957635076193,0
cexpf((cdivf(cnegf(PI),2)))
0.20787957635076193,0
cpowf(E,((cdivf(cnegf(PI),2))))
0.20787957635076193,0
]]
They are the same, confirmed eqn.BW200

<a name=a503300943>
complex4.htm#Box3JS ERROR input are next
OK cexpf((cdivf(cnegf(PI),2)))
OK exp(cgetr(cdivf(cnegf(PI),2)))
NO exp((cdivf(cnegf(PI),2))) because 
real function exp() not read complex number.
NO exp(cabsf(cdivf(cnegf(PI),2))) because 
cabsf() change E^(-PI/2) to E^(+PI/2)

<a name=a503300944>
OK E^(-PI/2) [eval Box3] button OK
NO E^(-PI/2) [test Box3] button not read '^'
NO e^(-PI/2) lowercase 'e' is undefined.
2016-03-30-11-12 stop

<a name=a503310726>
2016-03-31-07-26 start 
2009-03-17-11-17 receive cvch2.pdf
http://personal.maths.surrey.ac.uk/st/C.Wulff/Modules/MS224/cvch2.pdf

cvch2.pdf page 11/12
Theorem 2.19 
Suppose that f is analytic on a set S; then 
f cannot depend on conj(z)=z for z in set S. 

Proof calculate ∂f(z)/[∂(conj(z))] apply 
Cauchy-Riemann Equations get 
∂f(z)/[∂(conj(z))] = 0  ---eqn.BW215
then f cannot depend on conj(z). Detail next. 

<a name=a503310727>
Let complex number z be 
z=x+i*y ---eqn.BW216
where both x and y are real numbers. 
i=√-1=sqrt(-1)=√(-1) ---eqn.BW210
Complex conjugaste of z is 
z=x-i*y=conj(z)  ---eqn.BW217
<a name=a503310728>
 
∂f(z)

z
=
∂f(x+i*y)

z
=
∂[u(x,y)+i*v(x,y)]

z
=
∂[u(x,y)]

z
+ i*
∂[v(x,y)]

z
---eqn.BW218
 
∂f(z)

z
=
∂[u(x,y)]

∂x
∂x

z
+
∂[u(x,y)]

∂y
∂y

z
+ i*
∂[v(x,y)]

∂x
∂x

z
+
∂[v(x,y)]

∂y
∂y

z
---eqn.BW219

width of above equation
<a name=a503310729>
2016-03-31-08-19 here
In eqn.BW219, 
define ∂[u(x,y)]/∂x = ux ---eqn.BW220 
define ∂[u(x,y)]/∂y = uy ---eqn.BW221
define ∂[v(x,y)]/∂x = vx ---eqn.BW222
define ∂[v(x,y)]/∂y = vy ---eqn.BW223
<a name=a503310730>
Following are ERROR! 
Because 
z=x-i*y=conj(z)  ---eqn.BW217
Differentiation get 
∂[z]/∂x = ∂[x-i*y]/∂x = 1 ---eqn.BW224
and
∂[z]/∂y = ∂[x-i*y]/∂y =-i ---eqn.BW225
Inverse get 
∂x/∂[z] = 1/1 = 1 ---eqn.BW226
and
∂y/∂[z] =1/(-i)=i ---eqn.BW227
Substitute eqn.BW220, eqn.BW221, eqn.BW222, 
eqn.BW223, eqn.BW226, eqn.BW227 to eqn.BW219 
get 
<a name=a503310731>
 
∂f(z)

z
=
∂[u(x,y)]

∂x
*1
+
∂[u(x,y)]

∂y
*i
+
i*
∂[v(x,y)]

∂x
*1
+
∂[v(x,y)]

∂y
*i
---eqn.BW228
 
∂f(z)

z
=
∂[u(x,y)]

∂x
+
∂[v(x,y)]

∂y
*i*i
+
i*
∂[v(x,y)]

∂x
+
∂[u(x,y)]

∂y
---eqn.BW229

width of above equation
<a name=a503310732>
2016-03-31-09-02 here
Cauchy-Riemann Equations are 
∂u(x+iy)/∂x = ∂v(x+iy)/∂y ---eqn.BW138 
and 
∂u(x+iy)/∂y =▬∂v(x+iy)/∂x ---eqn.BW139 

eqn.BW138 let eqn.BW229 left square bracket be zero. 
Remind, i*i=-1 
eqn.BW139 let eqn.BW229 right square bracket be zero. 
Cauchy-Riemann Equations let ∂f(x+iy)/∂z become zero 
identically. 
Above proved Theorem 2.19.
<a name=a503310733>
Above is Liu,Hsinhan's work after reading cvch2.pdf.
Above work HAS ERROR.  why error
From eqn.BW224 inverse to eqn.BW226 
From eqn.BW225 inverse to eqn.BW227 are WRONG. 
cvch2.pdf use 
x=(z+z)/2 ---eqn.BW230
y=(z-z)/(2i) ---eqn.BW231
cvch2.pdf not use inverse operation. 
cvch2.pdf derivation = half Liu's derivation 
because final answer is zero. Whole zero equal 
half zero. Liu's derivation look ok, but in 
fact wrong. LiuHH do not know why inverse 
operation get double value. Remark here, let 
reader re-do the right steps. 
2016-03-31-09-22

<a name=a503310734>
2016-03-31-09-30
LiuHH find one reason, explain why eqn.BW224 
to eqn.BW227 are wrong. 
Theorem 2.19 say that if f(z) is analytic , 
then 
∂f(z)/[∂z] = 0 ---eqn.BW215
eqn.BW215 is target equation, in eqn.BW215 
independent variables are z and z.
<a name=a503310735>
∂x/∂z treat x as dependent variable and z be 
independent, which is consistent with given 
problem. On the other hand, 
∂z/∂x treat x as independent variable and 
z be dependent variable, which is different 
from given problem. 
<a name=a503310736>
cvch2.pdf use 
x=(z+z)/2 ---eqn.BW230
and
y=(z-z)/(2i) ---eqn.BW231
differentiate 
∂x/∂z = ∂[(z+z)/2]/∂z = +1/2 ---eqn.BW232
and
∂y/∂z = ∂[(z-z)/(2i)]/∂z = -1/(2*i) ---eqn.BW233
Above are correct. 
<a name=a503310737>
Liu,Hsinhan use 
∂[z]/∂x = ∂[x-i*y]/∂x = 1 ---eqn.BW224
and
∂[z]/∂y = ∂[x-i*y]/∂y =-i ---eqn.BW225
Inverse get 
∂x/∂[z] = 1/1 = 1 ---eqn.BW226
and
∂y/∂[z] =1/(-i)=i ---eqn.BW227
result double the value, that is wrong. 
2016-03-31-09-47 stop

<a name=a504021036>
2016-04-02-10-36 pay attention to 
Cauchy's integral formula. see 
http://personal.maths.surrey.ac.uk/st/C.Wulff/Modules/MS224/cvch3.pdf
page 9/11
Because of Cauchy's integral formula 
derived that if |f(z)| is bounded for 
any z, conclude f(z)=constant. 
But in real function, no such theorem 
because sin(infinity) vary between -1 and 1 
real sin(infinity) is not a constant, 
complex sin(z) ==> infinity when z==> infinity.
2016-04-02-10-41 record 

<a name=a504021256>
2016-04-02-12-56
http://personal.maths.surrey.ac.uk/st/C.Wulff/Modules/MS224/cvch3.pdf 
page 9/11 bottom (page 36) say 
Cauchy's integral formula is remarkable in 
that it express the value of f at points 
interior to gamma in terms of f(z) on the 
countour gamma. There is no analogue in 
real-variable calculus, knowing f(x) on 
some interval of the x-axis does not determine 
f for values of x outside that interval. 
The formula is alo very useful for both 
pratical and theoretical purposes. 
2016-04-02-13-01

<a name=a504021320>
2016-04-02-13-20 in
C:\$fm\math\ebook\complex1
search "Cauchy's integral formula"
C:\$fm\math\ebook\complex1\ash_CV2.pdf
[[
2.2.8 Definition
Let C(z0, r) denote the circle with center z0 
and radius r. then
∫[C(z0,r)] f(z) dz is defined as
∫[γ] f(z) dz where γ(t) = z0 + r*exp(i*t), 
0 ≤ t ≤ 2π.
The following result provides the essential
equipment needed for the theory of power 
series. In addition, it illustrates 
the striking difference between the concept 
of differentiability of complex functions 
and the analogous idea in the real case. 
We are going to show that if f is analytic 
on a closed disk, then the value of f at 
any interior point is completely determined 
by its values on the boundary, and furthermore 
there is an explicit formula describing the 
dependence.

2.2.9 Cauchy’s Integral Formula for a Circle
]]
pay attention to 
[[
the striking difference between the concept 
of differentiability of complex functions 
and the analogous idea in the real case. 
]]

<a name=a504021437>
2016-04-02-14-37
http://personal.maths.surrey.ac.uk/st/C.Wulff/Modules/MS224/cvch3.pdf 
page 6/11 bottom (page 33) 

3.3 Cauchy's Theorem
We now come to what is arguably one of the 
most important results in complex analysis
and which opens the way to many interesting 
and useful developments, namely Cauchy's Theorem. 
There are many versions of this Theorem; we 
give a simple form which is adequate for our 
purposes.
Theorem 3.10 Suppose that f is analytic in a 
simply-connected domain D. Then for any closed 
contour γ in D,
∫γf(z)dz = 0
2016-04-02-14-40

<a name=cplxdr10jpg>
cplxdr10.jpg
2016-05-06-15-00 start 
cplxdr10.jpg has text as following 
above mesh around singular point
Graph below confirm graph above.

above http://freeman2.com/cplxdr10.jpg
above http://freeman2.com/cplxdraw.htm
both draw cplxFunction(z)=1/z=cdivf(1,z)
Confirm cplxdraw.htm graph from other source
below http://i.stack.imgur.com/kPZVZ.png 
2016-04-10-16-51
2016
 -04
 -10
 -16
 -51

2016-05-06-15-07 record here
<a name=cplxdr11jpg>
cplxdr11.jpg
2016-04-28-15-54 
This graph is complex zeta function czeta(z) 
Input is x=0.5 y=t purple vertical critical 
line. t bgn=-20 ; t end=60 ; t step=0.02
Output is red curve czeta(z) z=x+y*i;i=√(-1)
On critical line 1st zero is z0=0.5+14.1347i
First twelve zero imaginary numbers are next
14.13472514, 21.02203964, 25.01085758, 
30.42487613, 32.93506159, 37.58617816, 
40.91871901, 43.32707328, 48.00515088, 
49.77383248, 52.97032148, 56.4462477
Blue lines are equal z lines. Link input z0
and output czeta(z0)=0+0*i Please goto 
http://freeman2.com/cplxdraw.htm#graph_Div 
click [zetAll] button, you get this graph. 
Search online "Zeta function zero image" you 
will find many graph similar to this one. 
Other graph, you are visitor, only watch. 
With cplxdraw.htm in hand you can draw any 
range you want. You can draw most function 
Many thanks to Zeta function original author
http://www.robertelder.ca/calculatevalue/
Many thanks to xygraph author J. Gebelein
www.structura.info/XYGraph/XYGraphDemo.htm
This graph http://freeman2.com/cplxdr11.jpg
Drawing board freeman2.com/cplxdraw.htm#zt0
2016-04-28-16-20 Liu,Hsinhan 劉鑫漢 

2016-05-06-15-11 record here 
<a name=cplxdr12jpg>
cplxdr12.jpg
z0=0.5+14.1347i, z1=0.5+21.0220i be root of zeta(z) 
Input mesh is 田 shape. First ten output mesh are in
this graph. All 10 十 center sit on (0,0) Ten output 
mesh marked 0,1,...,9. Can you see any relation 
in this graph? Liu,Hsinhan 2016-05-05-20-21      劉
drawing board  http://freeman2.com/cplxdraw.htm  鑫
this graph     http://freeman2.com/cplxdr12.jpg  漢

2016-05-06-15-12 record stop

<a name=cplxdr13jpg>
2016-05-15-18-23 record start 
cplxdr13.jpg
Graph file 
http://freeman2.com/cplxdr13.jpg
has the following text
[[
//2016-05-13-18-05 czeta iteration 1000000
step=0.01;bgn0=-15;end0=15;oupStr=''; 
for(t=bgn0;t<end0;t+=step)  { 
i1=0.5+'+'+t+'i';z=czeta(i1); x=z[0]; 
y=z[1];oupStr+=t+','+x+';'+t+','+y+';\n';} 
oupStr; //2016-05-13-18-42 done 
]]

above done with http://freeman2.com/cplxdraw.htm
below 2016-05-13-19-01 done with 
https://www.wolframalpha.com/input/?i=plot+zeta
(1%2F2+%2B+i*t)&lk=3

<a name=a505131928>
2016-05-13-19-28
Zeta function first root is 0.5+i*14.134725
At root real (purple) = imag (red) = 0
Both graph show 0.5±i*14.134725 are roots.
Top draw czeta(z) from z=0.5-15i to 0.5+15i
bottom draw zeta(z) from z=0.5-20i to 0.5+20i
This graph http://freeman2.com/cplxdr13.jpg 
main point is to confirm cplxdraw.htm graph 
output with wolframalpha.com graph output. 
2016-05-13-19-39 Liu,Hsinhan 劉鑫漢

<a name=a505151827>
2016-05-15-18-27
If change 
[[
step=0.01;bgn0=-15;end0=15;oupStr=''; 
for(t=bgn0;t<end0;t+=step)  { 
i1=0.5+'+'+t+'i';z=czeta(i1); x=z[0]; 
y=z[1];oupStr+=t+','+x+';'+t+','+y+';\n';} 
oupStr; 
]]
to
[[
step=0.01;bgn0=-15;end0=15;oupStr=''; 
for(t=bgn0;t<end0;t+=step)  { 
i1=0.5+'+'+t+'i';z=czeta(i1); x=z[0]; 
y=z[1];oupStr+=x+','+y+';\n';} 
oupStr; 
]]
then output data draw circle path
2016-05-15-18-45 stop

<a name=a505170648>
2016-05-17-06-48 start 
Complex number drawing board
http://freeman2.com/cplxdraw.htm
update 2016-05-16 created    and  examples.
They are the same problem, but set ╋ differently.
<a name=a505170649>
If click [i7], Box71 has data below 
[[
xy=0.,0.;0.8,0.8;30
cf=exp(-1/z/z/z/z)=cexpf(cnegf(cpowf(z,-4)))
cf:x=0,y=0,fv='0+0i'
uf=exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
vf=exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
ux= (exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))))*((-4*x*x*x+12*x*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)+(-x*x*x*x+6*x*x*y*y-y*y*y*y)*(-1)*(8*x*x*x*x*x*x*x+24*x*x*x*x*x*y*y+24*x*x*x*y*y*y*y+8*x*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+ (exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))*(-sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))*((4*(3*x*x*y-y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+(4*(x*x*x*y-x*y*y*y)*(-1)*(8*x*x*x*x*x*x*x+24*x*x*x*x*x*y*y+24*x*x*x*y*y*y*y+8*x*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
uy= (exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))))*(((-0+12*x*x*y-4*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+(-x*x*x*x+6*x*x*y*y-y*y*y*y)*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+ (exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)) )* (-sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))*((4*(x*x*x-x*3*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+(4*(x*x*x*y-x*y*y*y)*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y))/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
vx=exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) * ((-4*x*x*x+12*x*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)+(-x*x*x*x+6*x*x*y*y-y*y*y*y)*(-1)*(8*x*x*x*x*x*x*x+24*x*x*x*x*x*y*y+24*x*x*x*y*y*y*y+8*x*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))*((4*(3*x*x*y-y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+(4*(x*x*x*y-x*y*y*y)*(-1)*(8*x*x*x*x*x*x*x+24*x*x*x*x*x*y*y+24*x*x*x*y*y*y*y+8*x*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
vy= exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))*(((-0+12*x*x*y-4*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+(-x*x*x*x+6*x*x*y*y-y*y*y*y)*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+ exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*((cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))*((4*(x*x*x-x*3*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+(4*(x*x*x*y-x*y*y*y)*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y))/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
tt=i7: Petra Bonfert-Taylor Lecture 3.5 exp(-1/z/z/z/z) CR NO a505160642
]]

<a name=a505170650>
The following is working record, show how 
LiuHH get the long formula ux,uy,vx,vy. 

2016-05-14-15-37 to 2016-05-14-16-01 
Liu,Hsinhan watched Professor Petra 
Bonfert-Taylor lecture
3.5 First Properties of Analytic Functions
https://www.youtube.com/watch?v=iBBsUanNYvg
At video file time 16:35 Professor Petra 
Bonfert-Taylor introduced one example 
f(z)=exp(-1/z/z/z/z) when z!=0
f(z)=0 when z=0
This f(z) satisfy Cauchy-Riemann Equations
at z=0 but not differentiable at z=0. 

<a name=a505141724>
2016-05-14-17-24
Original problem f(z)=exp(-z^(-4))
Find (a+b)^4 formula
(a+b)^2=aa+2ab+bb
(a+b)^4=(aa+2ab+bb)*(aa+2ab+bb)
=aaaa+2abaa+bbaa 
+2abaa+2ab2ab+2abbb
+aabb+2abbb+bbbb
(a+b)^4=aaaa+4abaa+6aabb+4abbb+bbbb

<a name=a505141725>
(x+iy)^4=x^4+4*x^3*(iy)+6xxiyiy+4xiyiyiy+iyiyiyiy
(x+iy)^4=xxxx+4xxxiy+6xxiyiy+4xiyiyiy+iyiyiyiy
(x+iy)^4=xxxx+4xxxiy-6xxyy-4xyyyi+yyyy
(x+iy)^4=xxxx-6xxyy+yyyy+4xxxiy-4xyyyi
(x+iy)^4=xxxx-6xxyy+yyyy+4i(xxxy-xyyy)
Above is (a+b)^4 
below is (a+b)^(-4)
(x+iy)^(-4)=1/[xxxx-6xxyy+yyyy+4i(xxxy-xyyy)]
(x+iy)^(-4)=[xxxx-6xxyy+yyyy-4i(xxxy-xyyy)]/[(xxxx-6xxyy+yyyy)^2+16(xxxy-xyyy)^2]

-(x+iy)^(-4)=[-xxxx+6xxyy-yyyy+4i(xxxy-xyyy)]/[(xxxx-6xxyy+yyyy)^2+16(xxxy-xyyy)^2]

<a name=a505141726>
Next is exp(z), see tute0006.htm#cexp01
exp(s+it)=exp(s)*[cos(t)+i*sin(t)]
exp(-z^(-4))=
exp([-xxxx+6xxyy-yyyy+4i(xxxy-xyyy)]/[(xxxx-6xxyy+yyyy)^2+16(xxxy-xyyy)^2])

exp(-z^(-4))= // exp(s)  *[cos(t)  +i*sin(t)]
exp([-xxxx+6xxyy-yyyy]/[(xxxx-6xxyy+yyyy)^2+16(xxxy-xyyy)^2])
*[ cos(4(xxxy-xyyy)/[(xxxx-6xxyy+yyyy)^2+16(xxxy-xyyy)^2]) 
+i*sin(4(xxxy-xyyy)/[(xxxx-6xxyy+yyyy)^2+16(xxxy-xyyy)^2])]

<a name=a505141727>
exp(-z^(-4))=u(x,y)+i*v(x,y)

u(x,y)=
exp([-xxxx+6xxyy-yyyy]/[(xxxx-6xxyy+yyyy)^2+16(xxxy-xyyy)^2])
*[ cos(4(xxxy-xyyy)/[(xxxx-6xxyy+yyyy)^2+16(xxxy-xyyy)^2]) 

v(x,y)=
exp([-xxxx+6xxyy-yyyy]/[(xxxx-6xxyy+yyyy)^2+16(xxxy-xyyy)^2])
+sin(4(xxxy-xyyy)/[(xxxx-6xxyy+yyyy)^2+16(xxxy-xyyy)^2])]

<a name=a505141728>
[[
to solve (xxxx-6xxyy+yyyy)^2, try
(a+b+c)*(a+b+c)=aa+2ab+bb  +2(a+b)*c +cc
=aa+bb+cc+2ab+2bc+2ca
]]

(xxxx-6xxyy+yyyy)^2+16(xxxy-xyyy)^2
=xxxxxxxx
+36xxyyxxyy
+yyyyyyyy
+2xxxx(-6xxyy)
+2yyyy(-6xxyy)
+2xxxxyyyy
+16(xxxyxxxy-2xxxyxyyy+xyyyxyyy)
=xxxxxxxx
+2xxxxyyyy
-12xxxxxxyy
-12yyyyxxyy
+yyyyyyyy
+36xxyyxxyy
+16xxxyxxxy
-32xxxyxyyy
+16xyyyxyyy
=xxxxxxxx
-12xxxxxxyy
+16xxxxxxyy

+2xxxxyyyy
+36xxxxyyyy
-32xxxxyyyy

+16xxyyyyyy
-12xxyyyyyy
+yyyyyyyy

=xxxxxxxx
+4xxxxxxyy
+6xxxxyyyy
+4xxyyyyyy
+yyyyyyyy

therefore
(xxxx-6xxyy+yyyy)^2+16(xxxy-xyyy)^2
=
xxxxxxxx+4xxxxxxyy+6xxxxyyyy+4xxyyyyyy+yyyyyyyy

<a name=a505141729>
substitute above result back to u(x,y) and v(x,y)
u(x,y)=
exp([-xxxx+6xxyy-yyyy]/[xxxxxxxx+4xxxxxxyy+6xxxxyyyy+4xxyyyyyy+yyyyyyyy])
*[ cos(4(xxxy-xyyy)/[xxxxxxxx+4xxxxxxyy+6xxxxyyyy+4xxyyyyyy+yyyyyyyy]) 

v(x,y)=
exp([-xxxx+6xxyy-yyyy]/[xxxxxxxx+4xxxxxxyy+6xxxxyyyy+4xxyyyyyy+yyyyyyyy])
+sin(4(xxxy-xyyy)/[xxxxxxxx+4xxxxxxyy+6xxxxyyyy+4xxyyyyyy+yyyyyyyy])]
2016-05-14-18-14

<a name=a505141814>
need to verify u(x,y) and v(x,y)

exp(-z^(-4))=u(x,y)+i*v(x,y)

Change square bracket to circle bracket 
u(x,y)=
exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)) 

v(x,y)=
exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
+sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
2016-05-14-18-22

<a name=a505141815>
string to one line 
exp(-z^(-4))=u(x,y)+i*v(x,y)

u(x,y)=
exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)) 
v(x,y)=
exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
2016-05-14-18-24

<a name=a505141816>
Assign x,y value and calculate 
x=1.1; y=0.6; 
z="'"+x+"+"+y+"i'"
z4=cpowf(z,-4)
z4=cnegf(z4)
cexpf(z4)
2016-05-14-18-29
cexpf(z4)
1.1030175828698711,0.42698032505304306

<a name=a505141817>
above is calculation of exp(-1/z/z/z/z)
below is calculation of u(x,y)+i*v(x,y)
Because exp(-z^(-4))=u(x,y)+i*v(x,y) 
above answer should be same as below answer

x=1.1; y=0.6; 
//u(x,y)=
u=exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) 
//v(x,y)=
v=exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
u
v
<a name=a505141818>
2016-05-14-18-36
u
1.1030175828698711
v
0.42698032505304323
compare with
2016-05-14-18-29
cexpf(z4)
1.1030175828698711,0.42698032505304306

<a name=a505141819>
Numerical verify at one point z=1.1+0.6i
success exp(-z^(-4))=u(x,y)+i*v(x,y) is OK
Especially, u(x,y) formula and v(x,y) formula 
are OK. Next draw 
exp(-z^(-4))=cexpf(cnegf(cpowf(z,-4)))

<a name=a505141820>
x=1.1; y=0.6; 
z="'"+x+"+"+y+"i'"
z4=cpowf(z,-4)
z4=cnegf(z4)
cexpf(z4)
2016-05-14-18-43
f(z)=exp(-1/z/z/z/z) is next 
cexpf(cnegf(cpowf(z,-4)))

2016-05-14-18-54 done draw 
cexpf(cnegf(cpowf(z,-4)))

<a name=a505141917>
Next is hard work, calculate ux=d[u(x,y)]/dx

//u(x,y)=
u=exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) 
//v(x,y)=
v=exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
2016-05-14-19-17

because exp(s+it)=exp(s)*[cos(t)+i*sin(t)]
therefore 
u(x,y)=exp(s)*cos(t)
v(x,y)=exp(s)*sin(t)

<a name=a505141918>
ux=d[u(x,y)]/dx // =d[exp(s)*cos(t)]/dx
=
d[exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) 
]/dx
=  // above d[exp(s)*cos(t)]/dx
 [exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) 
 ]*  
d[(-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)]/dx
  // above cos(t)*d[exp(s)]/dx=exp(s)*cos(t)*d[s]/dx
  // below exp(s)*d[cos(t)]/dx
+ [exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
 ]*
d[cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))]/dx

<a name=a505141919>
ux=
 [exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) 
 ]*
 [(-4*x*x*x+12*x*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
+
(-x*x*x*x+6*x*x*y*y-y*y*y*y)
*(-1)*(8*x*x*x*x*x*x*x+24*x*x*x*x*x*y*y+24*x*x*x*y*y*y*y+8*x*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
]
<a name=a505141920>
  // above exp(s)*cos(t)*d[s]/dx
  // below exp(s)*d[cos(t)]/dx=exp(s)*[-sin(t)]*d[t]/dx
  // both d[s]/dx and d[t]/dx have d[numerator]/dx and
  // d[denominator]/dx ➝ cause (-1)* and /(...)/(...)
+ [exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
 ]*
 [-sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))]
*
[(4*(3*x*x*y-y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))

+(4*(x*x*x*y-x*y*y*y)
*(-1)*(8*x*x*x*x*x*x*x+24*x*x*x*x*x*y*y+24*x*x*x*y*y*y*y+8*x*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
]
2016-05-14-19-36

<a name=a505141921>
Next is verify calculation, assign x,y value
calculate ux, compare with d[u(x,0.6)]/dx 
find d[u(x,0.6)]/dx at cplxdraw.htm#differentiate
[[
x=1.1; y=0.6; 
ux= first time ux formula has error. omit error
ux
-0.8316884365137684
]]
2016-05-14-20-19
Later find 
-0.8316884365137684  error
-0.8092738265248729  correct 

above ux=d[u(x,y)]/dx
<a name=a505141922>
below vy=d[v(x,y)]/dy

exp(s+it)=exp(s)*[cos(t)+i*sin(t)]
u(x,y)=exp(s)*cos(t)
v(x,y)=exp(s)*sin(t)
2016-05-17-08-03 stop 
2016-05-17-10-40 start

<a name=a505141937>
2016-05-14-19-37
v=exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
vy=d[v(x,y)]/dy
=d[exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))]/dy

next
vy=d[exp(s)*sin(t)]/dy
vy=sin(t)*d[exp(s)]/dy
  +exp(s)*d[sin(t)]/dy

<a name=a505141938>
vy=
d[exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))]/dy

vy=
d[exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))]/dy
*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
+ exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*d[(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))]/dy

<a name=a505141939>
next
vy=sin(t)*exp(s)*d[s]/dy
  +exp(s)*d[sin(t)]/dy

vy=
 exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
*d[((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))]/dy

+ exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*d[(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))]/dy

<a name=a505141940>
next
vy=sin(t)*exp(s)*d[s]/dy
  +exp(s)*cos(t)*d[t]/dy

vy=
 exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
*(((-0+12*x*x*y-4*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
+(-x*x*x*x+6*x*x*y*y-y*y*y*y)
*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
)
// above  sin(t)*exp(s)*d[s]/dy ; d[s]/dy s=num1/den1
<a name=a505141941>
// below +exp(s)*cos(t)*d[t]/dy ; d[t]/dy t=num2/den2

+ exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*[(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
*((4*(x*x*x-x*3*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
+(4*(x*x*x*y-x*y*y*y)
*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y))
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
)
]

<a name=a505142020>
2016-05-14-20-20
[[
vy=
 exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
*(((-0+12*x*x*y-4*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
+(-x*x*x*x+6*x*x*y*y-y*y*y*y)
*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
)
<a name=a505142021>
+ exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(
(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
*(
(4*(x*x*x-x*3*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
+(4*(x*x*x*y-x*y*y*y)
*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y))
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
)
)
]]

<a name=a505142022>
2016-05-15-19-48 drop next from ux
(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))  )* 
because u=exp(a(x,y))*cos(b(x,y))
ux=exp(a(x,y))*{d[a(x,y)]/dx}*cos(b(x,y))
  +exp(a(x,y))*[-sin(b(x,y))]*d[b(x,y)]/dx
second line cannot have both cos and sin 
cos(b(x,y))*[-sin(b(x,y))]
2016-05-15-19-50


<a name=a505142050>
2016-05-14-20-50
[[
x=1.1; y=0.6; 
vy= exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))*(((-0+12*x*x*y-4*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+(-x*x*x*x+6*x*x*y*y-y*y*y*y)*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+ exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*((cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))*((4*(x*x*x-x*3*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+(4*(x*x*x*y-x*y*y*y)*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y))/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
vy
-0.8092738265248729  correct 

]]
<a name=a505142051>
compare with 2016-05-15-19-48
[[
x=1.1; y=0.6; 
ux= (exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))  )*((-4*x*x*x+12*x*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)+(-x*x*x*x+6*x*x*y*y-y*y*y*y)*(-1)*(8*x*x*x*x*x*x*x+24*x*x*x*x*x*y*y+24*x*x*x*y*y*y*y+8*x*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+ (exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))*(-sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))*((4*(3*x*x*y-y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+(4*(x*x*x*y-x*y*y*y)*(-1)*(8*x*x*x*x*x*x*x+24*x*x*x*x*x*y*y+24*x*x*x*y*y*y*y+8*x*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
ux
ux
-0.8092738265248728  correct 
]]
-0.8316884365137684  error
-0.8092738265248729  correct 
2016-05-15-20-00

above Cauchy-Riemann Equation ∂u/∂x=∂v/∂y 
<a name=a505152000>
below Cauchy-Riemann Equation ∂u/∂y=-∂v/∂x

//u(x,y)=
u=exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) 
//v(x,y)=
v=exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))

exp(s+it)=exp(s)*[cos(t)+i*sin(t)]
u(x,y)=exp(s)*cos(t)
v(x,y)=exp(s)*sin(t)

<a name=a505152144>
2016-05-15-21-44
uy=d[u(x,y)]/dy
=
d[exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) 
]/dy
<a name=a505152145>
=
 [exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) 
 ]*
d[(-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)]/dy
+ 
[exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
 ]*
d[cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))]/dy

<a name=a505152146>
uy=
 [exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) 
 ]
*(((-0+12*x*x*y-4*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
+(-x*x*x*x+6*x*x*y*y-y*y*y*y)
*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
)

<a name=a505152147>
+ [exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
 ]*
 [-sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))]
*
*((4*(x*x*x-x*3*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
+(4*(x*x*x*y-x*y*y*y)
*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y))
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
)
]

<a name=a505152154>
2016-05-15-21-54

uy=
 (exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
)

( 8 //above equation has 8 '('
) 8 //above equation has 8 ')'

<a name=a505152155>
*(((-0+12*x*x*y-4*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
+(-x*x*x*x+6*x*x*y*y-y*y*y*y)
*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
)

( 9
) 9

<a name=a505152156>
+ (exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
 )*
 (-sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
*((4*(x*x*x-x*3*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
+(4*(x*x*x*y-x*y*y*y)
*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y))
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
)

( 18
) 19 //debug unbalance '(' and ')' and corrected

<a name=a505152206>
2016-05-15-22-06

uy=
 (exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
)
*(((-0+12*x*x*y-4*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
+(-x*x*x*x+6*x*x*y*y-y*y*y*y)
*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
)
<a name=a505152206a>
+ (exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
 )*
 (-sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
*((4*(x*x*x-x*3*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
+(4*(x*x*x*y-x*y*y*y)
*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y))
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
)
<a name=a505152207>
2016-05-15-22-07
x=1.1; y=0.6; 
uy= (exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))))*(((-0+12*x*x*y-4*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+(-x*x*x*x+6*x*x*y*y-y*y*y*y)*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+ (exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)) )* (-sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))*((4*(x*x*x-x*3*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+(4*(x*x*x*y-x*y*y*y)*(-1)*(0+8*x*x*x*x*x*x*y+24*x*x*x*x*y*y*y+24*x*x*y*y*y*y*y+8*y*y*y*y*y*y*y))/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
uy

2016-05-15-22-07
uy
1.3006193713876972

<a name=a505152208>
above x=1.1; y=0.6; uy(x,y)= 1.3006193713876972
below verify calculation, assign x,y value
calculate ux, compare with d[u(1.1,y)]/dy
find d[u(1.1,y)]/dy at cplxdraw.htm#differentiate

<a name=a505152209>
u=exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) 

x=1.1; y=0.6; //in u(x,y) change x to 1.1 
  1.1; x=0.6; //in u(1.1,y) change to u(1.1,x)
              //because cplxdraw.htm#differentiate
              //accept variable x, not y

u=exp((-1.1*1.1*1.1*1.1+6*1.1*1.1*x*x-x*x*x*x)/(1.1*1.1*1.1*1.1*1.1*1.1*1.1*1.1+4*1.1*1.1*1.1*1.1*1.1*1.1*x*x+6*1.1*1.1*1.1*1.1*x*x*x*x+4*1.1*1.1*x*x*x*x*x*x+x*x*x*x*x*x*x*x))*(cos(4*(1.1*1.1*1.1*x-1.1*x*x*x)/(1.1*1.1*1.1*1.1*1.1*1.1*1.1*1.1+4*1.1*1.1*1.1*1.1*1.1*1.1*x*x+6*1.1*1.1*1.1*1.1*x*x*x*x+4*1.1*1.1*x*x*x*x*x*x+x*x*x*x*x*x*x*x))) 
<a name=a505152214>
2016-05-15-22-14
Numerical Answer : 1.300619114630308
x=1.1; y=0.6; u(x,y)=exp(s)*cos(t)
d[u(x,y)]/dy = 1.3006193713876972
uy           = 1.3006193713876972
2016-05-15-22-15 uy formula OK 

above find uy=∂u(x,y)/∂y
<a name=a505152215>
below find vx=∂v(x,y)/∂x

exp(s+it)=exp(s)*[cos(t)+i*sin(t)]
u(x,y)=exp(s)*cos(t)
v(x,y)=exp(s)*sin(t)


<a name=a505152219>
//v(x,y)=
v=exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
2016-05-15-22-19
vx=
d[exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) ]/dx
<a name=a505152220>
=
d[exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))]/dx
*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) 
+
exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*d[(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))]/dx
<a name=a505152221>
=
exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) 
*d[((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))]/dx

+
exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
*d[(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))]/dx
2016-05-15-22-23 stop
2016-05-16-06-12 continue 

<a name=a505160612>
vx=
exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) 
*
 [(-4*x*x*x+12*x*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
+
(-x*x*x*x+6*x*x*y*y-y*y*y*y)
*(-1)*(8*x*x*x*x*x*x*x+24*x*x*x*x*x*y*y+24*x*x*x*y*y*y*y+8*x*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
]

<a name=a505160613>
+
exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
*
[(4*(3*x*x*y-y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))

+(4*(x*x*x*y-x*y*y*y)
*(-1)*(8*x*x*x*x*x*x*x+24*x*x*x*x*x*y*y+24*x*x*x*y*y*y*y+8*x*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
]

<a name=a505160621>
2016-05-16-06-21
vx=
exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) 
*
 ((-4*x*x*x+12*x*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
+
(-x*x*x*x+6*x*x*y*y-y*y*y*y)
*(-1)*(8*x*x*x*x*x*x*x+24*x*x*x*x*x*y*y+24*x*x*x*y*y*y*y+8*x*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
)
<a name=a505160621a>
+
exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
*
((4*(3*x*x*y-y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))
+(4*(x*x*x*y-x*y*y*y)
*(-1)*(8*x*x*x*x*x*x*x+24*x*x*x*x*x*y*y+24*x*x*x*y*y*y*y+8*x*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)
)
<a name=a505160622>
2016-05-16-06-22
x=1.1; y=0.6; 
vx=exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(sin(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))) * ((-4*x*x*x+12*x*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)+(-x*x*x*x+6*x*x*y*y-y*y*y*y)*(-1)*(8*x*x*x*x*x*x*x+24*x*x*x*x*x*y*y+24*x*x*x*y*y*y*y+8*x*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+exp((-x*x*x*x+6*x*x*y*y-y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))*(cos(4*(x*x*x*y-x*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))*((4*(3*x*x*y-y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y))+(4*(x*x*x*y-x*y*y*y)*(-1)*(8*x*x*x*x*x*x*x+24*x*x*x*x*x*y*y+24*x*x*x*y*y*y*y+8*x*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)/(x*x*x*x*x*x*x*x+4*x*x*x*x*x*x*y*y+6*x*x*x*x*y*y*y*y+4*x*x*y*y*y*y*y*y+y*y*y*y*y*y*y*y)))
vx
-1.3006193713876967 OK
2016-05-16-06-24

<a name=a505160624>
f(z)=exp(-1/z/z/z/z)
z=x+iy 
exp(s+it)=exp(s)*[cos(t)+i*sin(t)]
upto here, next six terms all solved 
u(x,y)=exp(s)*cos(t)
v(x,y)=exp(s)*sin(t)
ux=∂u(x,y)/∂x
uy=∂u(x,y)/∂y
vx=∂v(x,y)/∂x
vy=∂v(x,y)/∂y

Next build click buttons 

<a name=a505160648>
2016-05-16-06-48 build  button
2016-05-16-06-58 build  button
2016-05-16-09-32 build  button
All three buttons draw f(z)=exp(-1/z/z/z/z) 
 set ╋ center at (0,0) path parallel to x,y axis
 set ╋ center at (0,0) rotate 45 deg from x,y axis
 set ╋ center at (-.8,1.0) path start at (0.9,0.8)

<a name=a505161347>
2016-05-16-13-47 build  button
Click [z4] fill "cexpf(cnegf(cpowf(z,-4)))" to 
[used function name﹕] box 
Here 
f(z)=exp(-1/z/z/z/z)=cexpf(cnegf(cpowf(z,-4)))

<a name=a505161348>
Please visit 
Professor Petra Bonfert-Taylor lectures 
"Analysis of a Complex Kind" 
3.5 First Properties of Analytic Functions
https://www.youtube.com/watch?v=iBBsUanNYvg
for more about f(z)=exp(-1/z/z/z/z)
2016-05-17-11-28 stop

<a name=cplxdr14jpg>
2016-05-24-21-40 record start 
cplxdr14.jpg
f(z)=exp(-1/z/z/z/z) approach to zero and approach to infinity at same time

2016-05-24-19-10
http://freeman2.com/cplxdraw.htm
Parametric function Box15 input
x(t)=(0.85+0.05*sin(48*t))*cos(t);
y(t)=(0.85+0.05*sin(48*t))*sin(t); 
t bgn=0, t end=1.55, t step=0.005
Click [z4] used function name﹕
cexpf(cnegf(cpowf(z,-4)))
check points=[ 151 176 ]
151 is upper blue line. gear teeth 
near x-axis, near y-axis f(z)⇒0+0i 
teeth near y=x line f(z)⇒infinity 
http://freeman2.com/cplxdr14.jpg
Liu,Hsinhan 2016-05-24-20-23
劉鑫漢 

<a name=a505241946>
2016-05-24-19-46
f(z)=exp(-1/z/z/z/z)=cexpf(cnegf(cpowf(z,-4)))
tt=i7: Petra Bonfert-Taylor Lecture 3.5 
z=r1+i*0=[r1,0];f(z)=[exp(-1/(r1^4)),-4*0]
r1⇒ε⇒0,f(z)⇒[exp(-1/(ε^4)),-4*0]⇒exp(-∞)⇒0+0i
phase -4*0⇒1, not alter '-' in 'exp(-1/(r1^4))'
z=0+i*r2=[r2,PI/2];f(z)=[exp(-1/(r2^4)),-4*PI/2]
r2⇒ε⇒0,f(z)⇒[exp(-1/(ε^4)),-2*PI]⇒exp(-∞)⇒0+0i
phase -2*PI⇒1, not alter '-' in 'exp(-1/(r2^4))'
z=r3+i*r3=[r,PI/4];f(z)=[exp(-1/(r^4)),-4*PI/4]
r⇒ε⇒0,f(z)⇒[exp(-1/(ε^4)),-PI]⇒exp(+∞)⇒∞
in f(z) phase -PI⇒-1 alter '-' to '+', change 
exp(-1/(ε^4)) to exp(+1/(ε^4))⇒exp(+1/0)⇒∞
2016-05-25-07-31 record stop



x0+i*y0 x0+∆x <a name="docA999"> <a name="NumberSetsChar"> ℂ Complex numbers ; 複數 ℍ Hello ; ℕ Natural numbers ; 自然數(正整數及零) ℙ Prime numbers ; 素數 ℚ Quotient, Rational numbers ; 有理數 ℝ Real numbers ; 實數 ℤ Zahl, Integers ; (from Zahl, German for integer) ; ℤ 整數(正整數及零及負整數) 2015-03-13-18-52






Javascript index
http://freeman2.com/jsindex2.htm   local
Save graph code to same folder as htm files.
http://freeman2.com/jsgraph2.js   local


file name tute0066.htm mean
TUTor, English, 66 th .htm
Chinese version is tutc0066.htm

2016-01-29-16-23 save as tute0066.htm


The address of this file is
http://freeman2.com/tute0066.htm
First upload 2016-02-14

Thank you for visiting Freeman's page.
Freeman Liu,Hsinhan 劉鑫漢
2016-02-14-04-48