﻿ Complex number study notes tute0066 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
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.

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

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
be not differentiable.
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.

Assume g(t) represent distance traveled.
Then g'(t) represent distance traveled in
one second. That is speed.
We cannot add length with speed.

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

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">t066link()
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">t066link()
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
Analytic function section
2016-02-12-13-15
What is the difference between
analytic complex function f(z) and
differentiable complex function f(z) ?

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">
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(reiθ0) - [f(r0eiθ0)] reiθ0 - r0eiθ0 ＝ 1 eiθ0 limit r→r0 u(r,θ0) +i*v(r,θ0) - [u(r0,θ0) +i*v(r0,θ0)] 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 eiθ0

<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(r,θ0) - u(r0,θ0) +i*[v(r,θ0) - v(r0,θ0)] 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(r,θ0) - u(r0,θ0) r - r0 ＋ e-iθ0 limit r→r0 i*[v(r,θ0) - v(r0,θ0)] 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 eiθ) - [f(r0 eiθ0 )] r0 eiθ - r0 eiθ0 ＝ 1 r0 limit θ→θ0 u(r0,θ) +i*v(r0,θ) - [u(r0,θ0) +i*v(r0,θ0) ] eiθ - eiθ0
---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) eiθ - eiθ0 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 (eiθ - eiθ0) 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 eiθ - eiθ0 }
---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(r0,θ0)}/{θ-θ0} is very
close to ∂u(r,θ)/∂θ and
limit[θ→θ0]{v(r0,θ)-v(r0,θ0)}/{θ-θ0} is very
close to ∂v(r,θ)/∂θ , IF (θ-θ0)/(eiθ - eiθ0)
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

 eiθ - eiθ0 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*eiθ0 ---eqn.BW153
Because (θ-θ0)/(eiθ - eiθ0) has a reciprocal limit
value i*eiθ0, substitute this value back to eqn.BW149

<a name="docA059">
f ＇(z0)＝

 1 i*eiθ0 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/eiθ0 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(r0,θ0) θ-θ0 ＋ i*i* v(r0,θ) -v(r0,θ0) θ-θ0 }
---eqn.BW155
width of above equation
<a name="docA062">
2016-02-13-16-22
In eqn.BW155
limit{u(r0,θ)-u(r0,θ0)]/[θ-θ0]} is uθ(r,θ)
limit{v(r0,θ)-v(r0,θ0)]/[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

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.
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
[[
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.

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

<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
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

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))

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

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))

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

∂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

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

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.
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
http://personal.maths.surrey.ac.uk/st/C.Wulff/Modules/MS224/cvch1.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
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
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

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

2016-03-31-07-26 start
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.
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.
]]

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

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
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.

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

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
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

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

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)
)
+ 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)

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 ')'

*(((-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

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
=
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)
)
+
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)))

Professor Petra Bonfert-Taylor lectures
"Analysis of a Complex Kind"
3.5 First Properties of Analytic Functions
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

<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


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