Complex function derivation
Study notes, file 6　tute0006.htm　 update 2013-04-03

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```
<a name=index01> Euler's Formula proof
Complex/real coefficient polynomial root, other page
problem 1 equation for complex1^complex2
complex1^complex2 general equation
from complex1^complex2 to x^i=i
problem 2, in i^x=x find x
problem 2 correct up to here
better numerical solution
i^x=x numerical solution look OK
complex1^complex2 program document
graph show: real^i
x^i=i has infinite many solution
i^x=x has infinite many solution
i^x=x stop at equation (B31) and (B32)
complex1^complex2 data box
i^x=x first zero point
i^x=x all zero points
<a name=index04> key for all; for arc
All func() change to exp(), sin() to exp()
complex power　　complex sine
complex cosine　　complex tangent
hyperbolic sine　　hyperbolic cosine
hyperbolic tangent　　complex log
Log z = u + iv = log(r) + i*(θ+2kπ) 9805151627 add
Log z1 + Log z2 = Log(z1*z2)
arc-sine　　arc-cosine　　arc-tangent
arc-hyperbolic sine　arc-hyperbolic cosine
arc-hyperbolic tangent

problem 1 in x^i = i -----(A00) find x

problem 2 in i^x = x -----(B02) find x

problem 3 in x^x = i -----(C00) find x
find problem 3 at "Ask Dr. Math" page
2009-03-13-09-55-27 LiuHH visited
http://mathforum.org/library/drmath/view/52253.html
All three problems have multiple answers.
Above mentioned answer is n=0 in 2*n*PI

<a name="prob0a01">　index
Freeman's work no one proofread. the
derivation may contain error. If reader
suspect any view point wrong, please
many critical question when you read.
Freeman Liu,Hsinhan 2009-03-04-16-04.
2009-03-04-11-03 start //a203191432 When write polynomial root program
http://freeman2.com/polyroot.htm
need find complex1 rise to power of
complex2, where complex1, complex2
are arbitrary complex numbers.
Assume
complex1=(a+b*i)
complex2=(p+q*i)
then (a+b*i)^(p+q*i) has what value?
After get general equation, assign
a,b,p,q value to solve x in
x^i = i -----(A00)
The following find general equation
of (a+b*i)^(p+q*i) as a start point.
（2009-03-04-11-09 here） //a203191602

2009-03-02-08-29 begin the following
calculation. All "a20319" lines are
//Cartesian form and polar form for same
//complex number. a,b given; c,d defined
//(a+b*i)≡c*exp(i*d) ----(AA1) //a203191610
(a+b*i)^(p+q*i) //a,b,c,d,p,q are all given
=[c*exp(i*d)]^(p+q*i) ----(A01)
// m^(s+t)=[m^s]*[m^t] ----(AA2) //a203191620
={[c*exp(i*d)]^p}*[c*exp(i*d)]^(q*i) ----(A02)
// [m*n]^p=(m^p)*[n^p] ----(AA3) //a203191624
// [exp(i*d)]^p=exp(i*d*p) ----(AA4) //a203191627
={[(c^p)*exp(i*p*d)]} ----(A03)
*{[c^(q*i)]*exp(i*d*q*i)}
//next line is Euler's Formula Key: m must be real
// exp(i*m)=cos(m)+i*sin(m) ----(AA5) //a203191629
//blue i can not be an answer, red i can.
={(c^p)*[cos(p*d)+i*sin(p*d)]} ----(A04)
*{[c^(q*i)]*exp(i*i*d*q)}

<a name="prob0a03">
let ww={(c^p)*[cos(p*d)+i*sin(p*d)]} ----(A05)
(A04)=ww  // i*i=-1 ----(AA6) //a203191639
*{[c^(q*i)]}*exp(-d*q) ----(A06)
=ww //2^(3*4)=2^12=(2^3)^4=8^4=4096 //a203191649
*{[(c^q)^i]}*exp(-d*q) ----(A07)
=ww //anyNumber=exp(log(anyNumber)) //a203191652
*{exp[log((c^q)^i)]}*exp(-d*q) ----(A08)
=ww //log(m^n)=n*log(m) ----(AA7) //a203191657
*{exp[i*log(c^q)]}*exp(-d*q) ----(A09)
<a name="prob0a04">
=ww  ----(A10) //apply (A91); m=log(c^q)
*{cos[log(c^q)]+i*sin[log(c^q)]}*exp(-d*q)
={(c^p)*[cos(p*d)+i*sin(p*d)]} //recover ww
*{cos[log(c^q)]+i*sin[log(c^q)]}
*exp(-d*q) ----(A11)
={(c^p)*exp(-d*q)}
*[cos(p*d)+i*sin(p*d)]
*{cos[log(c^q)]+i*sin[log(c^q)]} ----(A12)
//[cos(m)+i*sin(m)]*[cos(n)+i*sin(n)]=
// cos(m+n)+i*sin(m+n)----(AA8) //a203191712
[c*exp(i*d)]^(p+q*i) //re-write start eq.
={(c^p)*exp(-d*q)} ----(A13)
*{cos[p*d+log(c^q)]+i*sin[p*d+log(c^q)]}
2009-03-02-08-51 is this right result?
//a203191715 in (A13), i is not in exp()
//not in cos(), not in sin(), not in log()
//i is not in power, i is not in any

<a name="prob0a05">
Explain above derivation as following.

Euler's formula is next
exp(i*t)=cos(t)+i*sin(t) ----(A91)
If t is real, A91 has absolute value one.
If t is complex, A91 has modulus other
than one. modulus=absolute value a203191815

The proof of Euler's formula, please
refer to other source. One example is
Calculus and Analytic Geometry 6th ed.
George B. Thomas, Jr. and Ross L. Finney
ISBN 0-201-16290-3 Page 668.
Use Taylor expansion to prove.
Here has proof of Euler's formula.

<a name="prob0a06">
Euler's formula is a very important equation.
This equation change Cartesian coordinates
cos(t)+i*sin(t) to polar form exp(i*t) .
Euler's formula require complex number modulus
be one. If not, modify as next. Assume complex
number be (a+b*i). Assume its modulus (absolute
value) be L not=1
Let L=sqrt(a*a+b*b)
(a+b*i) = (a+b*i)*1 = (a+b*i)*L/L
(a+b*i) = L*(a+b*i)/L = L*((a/L) +(b/L)*i)
Here (a/L) +(b/L)*i is a modulus=1 complex
number, apply Euler's formula is OK.

<a name="prob0a06a">
First step, in (a+b*i)^(p+q*i) change
Cartesian (a+b*i) to polar c*exp(i*d)
(a+b*i)^(p+q*i)
=(c*exp(i*d))^(p+q*i) ----(A01)
a+b*i is identical with c*exp(i*d)
Because //cvch1.pdf page 6, page 9/18
a = c*cos(d)      //x=r*cos(θ) eqn.1.2
b = c*sin(d)      //y=r*sin(θ) eqn.1.2
c = sqrt(a*a+b*b) //r=sqrt(x*x+y*y) eqn.1.3
d = arctan(b/a)   //tan(θ)=y/x eqn.1.3
Power index complex (p+q*i) still use
Cartesian , (p+q*i) not change to ploar.
because power calculation need (p+q*i)
in Cartesian form. Above c = length L .
a203191850 here
2013-03-19-18-53 start
After First step, we move toward what
we move toward equation A13
Sure, equation A13 is the end of proof.
But, what make equation A13 unique?
What is the difference between eq.A13
and its earlier cousin eq.A01 to eq.A12?
eq.A13 let imaginary i=sqrt(-1) stand out
not involve with other calculation.
eq.A01 to eq.A12 all have i=sqrt(-1) mixed
with other calculation. We want answer be
real1 + i*real2 , only eq.A13 satisfy.
LiuHH raise quastion and answer it.
2013-03-19-19-03 stop
//a203191903 stop
//a203211517 start
<a name="prob0a07">
Next is step 2  //(a+b*i)^(p+q*i)
=(c*exp(i*d))^(p+q*i) ----(A01)
={[c*exp(i*d)]^p}*[c*exp(i*d))]^(q*i) ----(A02)
From (A01) to (A02) change power (p+q*i)
to real p and imaginary q*i .
Base (c*exp(i*d)) not change.
From (A01) to (A02) use next equation
m^(s+t)=[m^s]*[m^t] ----(AA2)

<a name="prob0a07a">
Step 3 //(a+b*i)^(p+q*i)
={[(c^p)*exp(i*p*d)]} ----(A03)
*{[c^(q*i)]*exp(i*d*q*i)}
merge real power p and imaginary
power q*i to base (c*exp(i*d))
Applied relation is next
(m^n)^o = m^(n*o)  ----(A92)
that is
[exp(i*d)]^p≡[2.718281828459045^(i*d)]^p
[exp(i*d)]^p≡ 2.718281828459045^(i*d*p)
same thing for another term
[exp(i*d)]^(q*i)≡[2.718281828459045^(i*d)]^(q*i)
[exp(i*d)]^(q*i)≡ 2.718281828459045^[(i*d)*(q*i)]

<a name="prob0a08">
explain (A92) as next (2009-03-04-11-33 here)
〔〔
2009-03-01-22-42
2^3=8
(2^3)^2=8*8=64
on the other hand
(2^3)^2=2^(3*2)=2^6=64
〕〕

<a name="prob0a09">
step 4  //(a+b*i)^(p+q*i)
={[(c^p)*exp(i*p*d)]} ----(A03)
*{[c^(q*i)]*exp(i*d*q*i)}
={(c^p)*[cos(p*d)+i*sin(p*d)]} ----(A04)
*{[c^(q*i)]*exp(i*i*d*q)}
Apply Euler's formula expand blue term
[(c^p)*exp(i*p*d)] to next line
{(c^p)*[cos(p*d)+i*sin(p*d)]}
Above line move i from power exp(i*p*d)
to coefficient cos()+i*sin() //why?
Second term [c^(q*i)]*exp(i*d*q*i) put
two i together, change i in power to
-1 in power (i disappear).

Step 5, define ww be next line,
ww={(c^p)*[cos(p*d)+i*sin(p*d)]} ----(A05)
simplify future calculation.
<a name="prob0a10a">
Step 6, re-write equation with ww and
-1, remember -1 come from i*i.
(a+b*i)^(p+q*i) =ww
*{[c^(q*i)]}*exp(-d*q) ----(A06)

<a name="prob0a11">
Step 7, apply (A92) to isolate i allow
step 10 apply Euler's formula.
(a+b*i)^(p+q*i) =ww
*{[(c^q)^i]}*exp(-d*q) ----(A07)

Euler's formula is
exp(i*t)=cos(t)+i*sin(t) ----(A91)
but, [(c^q)^i] in (A07) do not have exp()
We need put [(c^q)^i] into exp() We
know that exp() and log() are two operators
opposite to each other, that is
exp(log(Y))=Y ----(A93)
[(c^q)^i] in (A07) do not in exp() ?
easy !

<a name="prob0a12">
step 8
(a+b*i)^(p+q*i) =ww
*{exp[log((c^q)^i)]}*exp(-d*q) ----(A08)
Apply (A93) put [(c^q)^i] into exp()

but Euler's formula is
exp(i*t)=cos(t)+i*sin(t) ----(A91)
require i multiply with another number t
Equation (A08) is
exp[log((c^q)^i)]
in (c^q)^i imaginary i sit in power seat
and the function is log() , we know that
log((c^q)^i) can be written as
i*log(c^q)

<a name="prob0a13">
This is step 9
(a+b*i)^(p+q*i) =ww
*{exp[i*log(c^q)]}*exp(-d*q) ----(A09)

up to here, step 6 [c^(q*i)] change to
exp[i*log(c^q)]. Next is
step 10, apply Euler's formula get
(a+b*i)^(p+q*i) =ww  ----(A10)
*{cos[log(c^q)]+i*sin[log(c^q)]}*exp(-d*q)
Key observation: move i=sqrt(-1)
from power (A06) to coefficient (A10)

<a name="prob0a14">
Step 11, recover ww in step 5.
(a+b*i)^(p+q*i)
={(c^p)*[cos(p*d)+i*sin(p*d)]}
*{cos[log(c^q)]+i*sin[log(c^q)]}
*exp(-d*q) ----(A11)

Step 12, simplify equation.
(a+b*i)^(p+q*i) ={(c^p)*exp(-d*q)}
*[cos(p*d)+i*sin(p*d)]
*{cos[log(c^q)]+i*sin[log(c^q)]} ----(A12)
〔hint﹕cos(p*d) multiply cos(log(c^q))
add two phase angles get cos[p*d+log(c^q)]
Same method apply to sin(). In polar
expression, multiplication of two
complex numbers, add two phase angles.
9803111111 〕

Step 13 //include left hand side
(a+b*i)^(p+q*i)=(c*exp(i*d))^(p+q*i)
final equation //right hand side [9803131121 add]
={(c^p)*exp(-d*q)} ----(A13)
*{  cos[p*d+log(c^q)]
+i*sin[p*d+log(c^q)]
}
As a solution, i is not in any function.
2009-03-02-08-51 is this a right answer? OK
review early notes
2009-03-04-11-56 record here
2013-03-21-16-46 stop, a203211747 start

2009-03-14-09-58 start
are periodic function, add or subtract
any integer of 2*PI not change the value
of answer. Then if n is an integer, next
equations are true
cos(x)=cos(x+2*n*PI) ----(nPI1)
sin(x)=sin(x+2*n*PI) ----(nPI2)
after include 2*n*PI final equation is
(a+b*i)^(p+q*i)=(c*exp(i*d))^(p+q*i)
*{  cos[p*d+log(c^q)+2*n*PI]
+i*sin[p*d+log(c^q)+2*n*PI]
}　　　where n is an integer.
2009-03-14-10-04 stop

2009-03-15-23-51 start
Above +2*n*PI do not help.
Add 2*n*PI to phase angle p*d+log(c^q) ,
if answer involve phase angle in
subtract 2*n*PI not change trigonometric
function value, for example
exp(ix)=cos(x)+i*sin(x)
compare with
exp(i(x+2*n*PI)) = cos(x+2*n*PI)
+i*sin(x+2*n*PI)
Equation left side has many different
input value (different n value), the
〔 cos(x+2*n*PI) and cos(x) have same
numerical value. 〕
That is, in answer no need to consider
2*n*PI .

On the other hand, if answer has phase
angle and not involve trigonometric
function, for example, log function
log(z) = log(abs(z)) + i*(arg(z)+2*n*PI)
in this case different n in 2*n*PI get
different answer, therefore we can not
ignore 2*n*PI in log function.

It is long time not study complex variable
recent few days paid attention to the name
"phase angle" in Chinese. and phase angle
is written as arg() . Today 2009-03-15
write complex computer program, paid
attention to the difference of
one to many complex number function and
many to one complex number function use
2*n*PI differently.

This file is Liu,Hsinhan's study notes,
output may not be correct. Please verify
output first. Study notes view point may
not be correct, please suspect each step.
2009-03-16-00-18 stop

<a name="prob0a16">
Equation (A13) is a general equation. If
assign coefficient value, it is possible
to get the answer of x^i = i where x is
unknown. [i=sqrt(-1)]

Next is another working record.
2009-03-04-11-59 record here
2013-03-21-18-17 translate here

2009-03-03-22-22
x^i=i
find x
[[
2009-03-02-09-06 final equation is
(a+b*i)^(p+q*i)
=(c*exp(i*d))^(p+q*i)
={(c^p)*exp(-d*q)} ----(A13)
*{cos[p*d+log(c^q)]+i*sin[p*d+log(c^q)]}
]]

<a name="prob0a18">　index
Problem is x^i=i -----(A00)
Left hand side general expression is
(c*exp(i*d))^(p+q*i)
assign c, d, p, q to match x^i
Let
p=0, q=1 -----(A21)
left hand side become
(c*exp(i*d))^(0+1*i)
that is
(c*exp(i*d))^(i) -----(A22)
match power i in (A00)
c and d in (A22) are unknown and they
are our target.

<a name="prob0a19">
General expression right hand side is
={(c^p)*exp(-d*q)} ----(A13)
*{cos[p*d+log(c^q)]+i*sin[p*d+log(c^q)]}
Problem (A00) right hand side is
i -----(A23)
Use (A21) re-write (A23) as following
right hand side = i = 1*(0+1*i) -----(A24)
//(A24) is coefficient*[cos()+i*sin()]

<a name="prob0a20">
Compare (A13) with (A24) they must have
the following relation
(c^p)*exp(-d*q)=1 -----(A25)
p*d+log(c^q)=PI/2 -----(A26)
[ hint: cos(PI/2)=0, sin(PI/2)=1 ]

Substitute p=0, q=1 -----(A21)
to (A25) get
(c^0)*exp(-d*1)=1 -----(A27)

Substitute p=0, q=1 to (A26) get
0*d+log(c^1)=PI/2 -----(A28)

from (A27) get
1*exp(-d*1)=1  ==> -d=0  (e^0=1) -----(A29)

from (A28) and (A29) get
0*d+log(c^1)=PI/2 ==> c=exp(PI/2) -----(A30)

Final result is
d=0   -----(A31)
Substitute (a+b*i)=c*exp(i*d) get
(a+b*i)=c*exp(i*d)=exp(PI/2)*exp(i*0)
(a+b*i)=exp(PI/2)*1
(a+b*i)=exp(PI/2) -----(A32)

<a name="prob0a22">　index
problem
x^i = i
find x
correct solution is
x=exp(PI/2)=4.810477380965351 -----(A33)
2009-03-03-22-41 correct

<a name="prob0a23">
http://freeman2.com/complex4.htm#testSet3
Find click button
[c1 power to c2] [c1] [c2] cpowf(c1,c2,nB,nE)
In [c1] box fill in 4.810477380965351
In [c2] box fill in  i
click [c1 power to c2] button, output to
6.123233995736766e-17 , 1
This number is 0 , 1
This number is 0 + i
program verify correct.
2009-03-04-12-24 record stop.

<a name="a20321a01">
2013-03-21-18-47
Above said
Final result is
d=0
What we get if n is NOT zero? Now set
n=1             //PI=3.141592653589793
nn=2*n*PI+PI/2  //nn=7.853981633974483
exp(nn)
2575.9704965975697

<a name="a20321a02">
In [c1] box fill in 2575.9704965975697
In [c2] box fill in  i
click [c1 power to c2] button, output to
2575.9704965975697^i=
3.061616997868383e-16 , 1
confirm correct when n=1
2013-03-21-18-53

Freeman's work no one proofread. the
derivation may contain error. If reader
suspect any view point wrong, please
many critical question when you read.
Freeman Liu,Hsinhan 2009-03-04-16-04.

<a name="a20321b01">
2013-03-21-22-21 start
Above is problem 1
x^i = i -----(A00)
This is a fabricated problem. Next,
fabricate another problem
x^i = 2*i -----(X00)
//x^i = 2i allow LiuHH search "= 2i"
2013-03-21-22-31 draw wrong conclusion
a20321err_a20326ok

2013-03-26-15-48 start
How to solve
x^i = 2*i -----(X00)
for unknown x ?
compare (X00) with
(a+b*i)^(p+q*i)=(c*exp(i*d))^(p+q*i)
={(c^p)*exp(-d*q)} ----(A13)
*{  cos[p*d+log(c^q)]
+i*sin[p*d+log(c^q)]
}
<a name="a20321b02a">
x be (c*exp(i*d)) may be a complex.
require (p+q*i) be i, then
p=0, q=1  -----(X01)
(A13) right side is
{(c^0)*exp(-d*1)} ----(X02)
*{  cos[0*d+log(c^1)]
+i*sin[0*d+log(c^1)]
}
<a name="a20321b02b">
simplify to
{exp(-d)} ----(X03)
*{  cos[log(c)]
+i*sin[log(c)]
}
(X00) require (X03)=2i
then exp(-d)=2   ----(X04)
and  log(c)=PI/2 ----(X05)
d=-log(2)   ----(X06)
c=exp(PI/2) ----(X07)
then
<a name="a20321b02c">
x=c*exp(i*d)
x=exp(PI/2)*exp{i*[-log(2)]}
x=exp(PI/2)*{cos[-log(2)]+i*sin[-log(2)]}
x= exp(PI/2)*(cos(-log(2)))
+i*exp(PI/2)*(sin(-log(2)))

<a name="a20321b02d">
The answer for x^i = 2*i -----(X00)
is x=
3.7004063355700247-3.073708767019492i
x^i=
1.2246467991473532e-16 , 2
2013-03-26-16-10 solve x^i = 2*i

<a name="a20321b03">
2013-03-21-22-31 here
Above solve
x^i = i -----(A00)
and solve
x^i = 2*i -----(X00)
the regular way, it is longer and it
is a must understand method.
For the special case, right hand side
be i. There is a short cut to solve
(A00).

<a name="a20321b04">
Euler's formula is next
exp(i*t)=cos(t)+i*sin(t) ----(A91)
Compare (A00) right hand side i with
(A91) right hand side cos(t)+i*sin(t)
If cos(t)=0 and sin(t)=1, then two
right hand side match.
For this purpose choose
t=PI/2 --------(AA9)
Because we know
cos(PI/2)=0 and sin(PI/2)=1 ----(AB0)

<a name="a20321b05">
Next turn attention to (A00) left hand
side x^i . Apply
exp(log(Y))=Y ----(A93)
to x^i . Find
x^i=exp(log(x^i)) ----(AB1)
x^i=exp(i*log(x)) ----(AB2)
x^i=cos(log(x))+i*sin(log(x)) ----(AB3)
on the other hand, problem given
x^i= i  =   0 + i* 1  -----(A00)
compare (AB3) with (A00)
log(x) must be PI/2
<a name="a20321b06">
log(x)=PI/2 ----(AB4)
Take exp() for (AB4), solve x as
x=exp(PI/2) ----(AB5)
Answer (AB5) is same as (A33)
This short solution is for special case
x^i=i only.

<a name="a20321b07">
Several special case answer are next
x^i = +1 ans. x=exp( 0)= 1.0
x^i = +i ans. x=exp(PI/2)=4.810477380965351
x^i = -1 ans. x=exp(PI)=23.140692632779267
Next two different x give same result -i
x^i = -i ans. x=exp(3*PI/2)=111.31777848985621
x^i = -i ans. x=exp(-PI/2)=0.20787957635076193
2013-03-21-23-08 stop

<a name="a20321b08">
//put next two lines to complex4.htm box 3
//click [test box3 command] get answer
exp(-PI/2)            //=0.20787957635076193
cpowf(exp(-PI/2),'i') //6.123233995736766e-17,-1
//6.123233995736766e-17 is zero, -1 is -i

exp(3*PI/2) //=111.31777848985621
cpowf(111.31777848985621,'i')
-1.8369701987210297e-16,-1
=0,-1  , that is
=-i
2013-03-21-23-11 stop

<a name="prob0b01">
2009-03-05-19-05 start //a203221014 start
Above is problem 1.
given i = sqrt(-1)
in equation
x^i = i -----(A00)
find x
x=exp(PI/2)=4.810477380965351

Next is problem 2, in problem 1
exchange x, i to form problem 2
given i = sqrt(-1)  -----(B01)
in equation
i^x = x  -----(B02)
find x
x=(p+q*i)=  -----(B03)
(0.4382829367270321+0.36059247187138543*i)
This number has the following relation
i^(p+q*i)=(p+q*i)  -----(B04)

<a name="prob0b03">
Explain as next.

[[
2009-03-02-09-06
complex power general equation is
(a+b*i)^(p+q*i)
=(c*exp(i*d))^(p+q*i)
={(c^p)*exp(-d*q)} ----(A13)
*{cos[p*d+log(c^q)]+i*sin[p*d+log(c^q)]}
]]
2009-03-05-15-16
How to solve i^x=x ?
<a name="prob0b04">
Base number is i therefore
(a+b*i)=0+i  -----(B05)
Above is base number Cartesian expression.
Next is base number polar expression.
c*exp(i*d)=  -----(B06)
1*exp(i*1.5707963267948965)
=1*exp(i*PI/2)
=cos(PI/2)+i*sin(PI/2)=0+i
<a name="prob0b05">
Base complex general expression c*exp(i*d)
Base complex special expression
i = 1*exp(i*PI/2)
general expression = special expression
i = c*exp(i*d)= 1*exp(i*PI/2)  -----(B07)
compare coefficient get
c=1      -----(B08)
d=PI/2   -----(B09)

<a name="prob0b06">
Above is i^x=x base complex number i
Below is i^x=x power x and right hand side x

Complex power x is (p+q*i)  -----(B10)
General equation (A13) tell us right hand
side is
{(c^p)*exp(-d*q)} -----(B11)
*{cos[p*d+log(c^q)]+i*sin[p*d+log(c^q)]}

Under problem 2 equation i^x=x
must have
complex power x = (p+q*i) and
right hand side x =
{(c^p)*exp(-d*q)}
*{cos[p*d+log(c^q)]+i*sin[p*d+log(c^q)]}
be identical, therefore

<a name="prob0b08">
following relation is true
(again problem 2 i^x=x special case)
(p+q*i)=           -----(B12)
{(c^p)*exp(-d*q)}
*{cos[p*d+log(c^q)]+i*sin[p*d+log(c^q)]}

<a name="prob0b09">
For complex1 = complex2
real_one=real_two and
imagine_one=imagine_two
demand next equality
p=(c^p)*exp(-d*q)*cos[p*d+log(c^q)] -----(B13)
q=(c^p)*exp(-d*q)*sin[p*d+log(c^q)] -----(B14)
p and q twist in two equations.
Here, c,d are known, p,q are unknown.
2009-03-05-15-20 solve here
2009-03-05-19-26 document here
2013-03-22-10-37 translate here

<a name="prob0b10">
Substitute
c=1      -----(B08)
d=PI/2   -----(B09)
to (B13) and (B14) get
p=(1^p)*exp(-q*PI/2)*cos[p*PI/2+log(1^q)]
q=(1^p)*exp(-q*PI/2)*sin[p*PI/2+log(1^q)]
simplify
p=1*exp(-q*PI/2)*cos[p*PI/2+0]
q=1*exp(-q*PI/2)*sin[p*PI/2+0]
<a name="prob0b11">
simplify again
p=exp(-q*PI/2)*cos[p*PI/2] -----(B15)
q=exp(-q*PI/2)*sin[p*PI/2] -----(B16)
how to find p ? how to find q ?
Square sum eliminate sin() and cos()
p*p=exp(-q*PI/2)*exp(-q*PI/2)*cos[p*PI/2]*cos[p*PI/2]
q*q=exp(-q*PI/2)*exp(-q*PI/2)*sin[p*PI/2]*sin[p*PI/2]

p*p+q*q=exp(-q*PI/2)*exp(-q*PI/2) -----(B17)
2009-03-05-15-25 achieve any progress?
look like exp(-q*PI/2) relate to the
length of complex (p+q*i)
exp(-q*PI/2)=abs(p+q*i) -----(B18)
Put (B18) into (B15),(B16) get
p=abs(p+q*i)*cos[p*PI/2] -----(B19)
q=abs(p+q*i)*sin[p*PI/2] -----(B20)
2009-03-05-15-30
a203221043 stop
a203221300 start

<a name="prob0b13">　index
2009-03-05-15-50 next equation are correct
apply (B17) to (B19) and (B20) get
p=exp(-q*PI/2)*cos[p*PI/2] -----(B15)
q=exp(-q*PI/2)*sin[p*PI/2] -----(B16)

In exp() and in sin() both present PI/2
now multiply PI/2 to (B15),(B16) get
p*PI/2=exp(-q*PI/2)*cos[p*PI/2]*PI/2 -----(B21)
q*PI/2=exp(-q*PI/2)*sin[p*PI/2]*PI/2 -----(B22)
<a name="prob0b14">
2009-03-05-15-54 here
let p*PI/2=m -----(B23)
let q*PI/2=n -----(B24)
then (B21) and (B22) become
m=exp(-n)*cos(m)*PI/2 -----(B25)
n=exp(-n)*sin(m)*PI/2 -----(B26)

<a name="prob0b15">
(B25)/(B26) get
m/n=cos(m)/sin(m) -----(B27)
(B25) and (B26) square sum is
m*m+n*n=exp(-n)*exp(-n)*(PI*PI/2/2) -----(B28)
from (B27) find n
n=m*sin(m)/cos(m) -----(B29)
<a name="prob0b16">
substitute n into (B28)
m*m+[m*sin(m)/cos(m)]*[m*sin(m)/cos(m)]
=exp(-m*sin(m)/cos(m))*exp(-m*sin(m)/cos(m))
*(PI*PI/2/2) -----(B30)
up to here (B30) has one unknown m
let (B30) be
f(m)=m*m+(m*sin(m)/cos(m))*(m*sin(m)/cos(m))
-exp(-m*sin(m)/cos(m))*exp(-m*sin(m)/cos(m))*(PI*PI/2/2)

the following is numerical solution.
2009-03-05-16-02 change m to t , because
program set t as variable.
<a name="prob0b18">　index
f(t)=
t*t+(t*sin(t)/cos(t))*(t*sin(t)/cos(t))-exp(-t*sin(t)/cos(t))*exp(-t*sin(t)/cos(t))*(PI*PI/2/2)
from graph, estimate
m=t=0.7 is one solution.
let p*PI/2=m
p=m*2/PI=0.7*2/PI=0.44563384065730693

<a name="prob0b19">
http://freeman2.com/graph09e.htm

[[
later part of this file has drawing tool
no need go to graph09e.htm

this file following section has click
button modify 602
Click it get desired curve
m=0.7 approximate solution
]]

Click "modify 602" button
in "Del function 1 2 3 4 5 ALL" click "ALL"
In function 1 x(t) fill 't' //fill t, not use [']
In function 1 y(t) fill next line
t*t+(t*sin(t)/cos(t))*(t*sin(t)/cos(t))-exp(-t*sin(t)/cos(t))*exp(-t*sin(t)/cos(t))*(PI*PI/2/2)
next click yellow stripe side "draw" button.
Do not click "draw 602"
"draw 602" button draw default equation
Curve show a zero point near m=t=0.7

<a name="prob0b20">
Let q*PI/2=n
m=0.7
n=m*sin(m)/cos(m)
q=n*2/PI
q=0.375352205926785

<a name="prob0b21">
use iteration loop, tool is next page
http://freeman2.com/eyefoot2.htm
2009-03-05-16-13 use following code
[[
var oStr='';
var aa=0;
var t=0;
for(t=0.688453;t<0.688456;t=t+0.0000001){aa=t*t+(t*sin(t)/cos(t))*(t*sin(t)/cos(t))-exp(-t*sin(t)/cos(t))*exp(-t*sin(t)/cos(t))*(PI*PI/2/2);oStr+=t+', '+aa+'\n';}
oStr
]]

t=0.6884532999999998
t*t+(t*sin(t)/cos(t))*(t*sin(t)/cos(t))-exp(-t*sin(t)/cos(t))*exp(-t*sin(t)/cos(t))*(PI*PI/2/2)
= 4.927305469193399e-7
(goal, push 4.9e-7 to zero)

<a name="prob0b23">
2009-03-05-16-25
m=t=0.68845322722
t*t+(t*sin(t)/cos(t))*(t*sin(t)/cos(t))-exp(-t*sin(t)/cos(t))*exp(-t*sin(t)/cos(t))*(PI*PI/2/2)
= 7.59100893255038e-10

<a name="prob0b24">
let p*PI/2=m
let q*PI/2=n
2009-03-05-16-27
m=0.68845322722
n=m*sin(m)/cos(m)
q=n*2/PI
q= 0.3605924720127381
p=m*2/PI
p= 0.4382829367985232

<a name="prob0b25">
p+qi=
0.4382829367985232+0.3605924720127381i

i^(p+qi)
i^[0.4382829367985232 +0.3605924720127381i]
equal         |||                |||  error
0.43828293658922373,0.3605924718405392
2009-03-05-16-31 //smaller error

<a name="prob0b26">
What is i^x=x analytic solution?
[[
p=abs(p+q*i)*cos[p*PI/2]
q=abs(p+q*i)*sin[p*PI/2]
p*PI/2=exp(-q*PI/2)*cos[p*PI/2]*PI/2
q*PI/2=exp(-q*PI/2)*sin[p*PI/2]*PI/2
2009-03-05-15-54
let p*PI/2=m
let q*PI/2=n
m=exp(-n)*cos(m)*PI/2
n=exp(-n)*sin(m)*PI/2
]]
look like it is possible find
analytic solution for i^x=x
2009-03-05-16-33
sqrt(p*p+q*q)
= 0.5675551634519729
Solution absolute value is not one.

2009-03-05-16-59
analytic solution is true solution,
numerical solution is side proof.

<a name="prob0b28">
2009-03-05-17-05
sqrt(p*p+q*q) = exp(-q*PI/2)
(p*p+q*q) = [exp(-q*PI/2)]^2
(p*p+q*q) =  exp[(-q*PI/2)*2]
(p*p+q*q) =  exp(-q*PI)

<a name="prob0b29">
2009-03-05-17-07
atan2(q,p)  =  p*PI/2

2009-03-05-17-12 try start from m, n
[[
let p*PI/2=m
let q*PI/2=n
m=exp(-n)*cos(m)*PI/2
n=exp(-n)*sin(m)*PI/2
2009-03-05-15-55
m/n=cos(m)/sin(m)
m*m+n*n=exp(-n)*exp(-n)*(PI*PI/2/2)
n=m*sin(m)/cos(m)
]]

<a name="prob0b30">　index
2009-03-05-18-05
m=t=0.6884532271077
t*t+(t*sin(t)/cos(t))*(t*sin(t)/cos(t))-exp(-t*sin(t)/cos(t))*exp(-t*sin(t)/cos(t))*(PI*PI/2/2)
= -1.4988010832439613e-14
t=0.688453227107703
= 5.218048215738236e-15
t=0.688453227107702105
= -3.3306690738754696e-16
(goal, push 3.33e-16 to zero)
2009-03-05-18-08 find minimum error
solution.
m=t=0.688453227107702105
function value = -3.3306690738754696e-16

<a name="prob0b31">
2009-03-05-18-10
m,n are auxiliary variables.
p,q are coefficients in i^(p+q*i)=(p+q*i)

m=0.688453227107702105
n=m*sin(m)/cos(m)
q=n*2/PI
p=m*2/PI
p
= 0.4382829367270321
q
= 0.36059247187138543

i^(p+q*i)=(p+q*i)

<a name="prob0b32">
2009-03-05-18-12
i^(0.4382829367270321+0.36059247187138543*i)
=(p+q*i)
open older page
http://freeman2.com/polyroot.htm#testFunc
Next is better page
http://freeman2.com/polyroot.htm#testSet3
<a name="prob0b33">　index
[c2 power to c2]　〔　　〕　〔　　〕　cpowf(c1,c2)
in left 〔　　〕 fill one byte i
in right 〔　　〕 fill next line complex number
0.4382829367270321+0.36059247187138543*i
Click [c2 power to c2] button
page expand three answer boxes //
LiuHH: 0.4382829367270322,0.3605924718713855
|                  | error
input: 0.4382829367270321+0.36059247187138543*i
input data is power  i^(p+q*i)=(p+q*i)
LiuHH ans is i^(p+q*i)=(p+q*i)
the error is close to zero.
bigger error, smaller error
2009-03-05-20-04 document here

<a name="prob0b34">
Problem 2
i^x=x
analytic solution is true solution,
numerical solution is side proof.
Freeman ask math experts, whether it is
possible to find analytic solution for
Problem 2 ?

Freeman 2009-03-05-20-07

Freeman's work no one proofread. the
derivation may contain error. If reader
suspect any view point wrong, please
many critical question when you read.
Freeman Liu,Hsinhan 2009-03-04-16-04.
2013-03-22-14-26 translate stop
2013-03-22-16-02 translate start
2009-03-06-20-32 start
After solve problem 1
x^i = i -----(A00)
LiuHH hope to see the curve for
(c*exp(i*d))^(p+q*i) ----(A01)
Curve is one dimension space, allow only
one variable. Now (A01) has c,d,p,q
four parameters, one of which is free
to vary. It is possible to draw one of
next four
c=c(t)
d=d(t)
p=p(t)
q=q(t)
It is possible to change two variables
at related pace. But use what constraint
for those two? Without constraint, now
let c,d,p,q change individually.
2009-03-06-18-17 start write drawing command
2009-03-06-20-06 done drawing command
After draw, LiuHH can not find any secret.
Now upload to internet let everyone take
a look, maybe someone can find a key
observation. Graph is here
Thank you for visiting freeman2.com
Freeman 　Liu,HsinHan　劉鑫漢
2009-03-06-20-39

<a name="980307a">
2009-03-07-12-22
http://freeman2.com/graph09e.htm
Now this file tute0006.htm added drawing
code and click button. Please goto
a name=graph02
click "draw 602" get curve directly.
Estimate m=0.7 close to solution, from
m=0.7 to find better solution
m=0.688453227107702105
Freeman's method is not best method.
Better method example is Newton's
iteration automatic find better solution.
Freeman cannot use Newton's iteration
because LiuHH do not have the algorithm.

<a name="980307b">　index
Another point reader may try is to draw
real^i that is
f(x)=x^i
base complex: c  *exp(i*d  )
power index complex: p  +i*q 
fill the following to four boxes
2 ,c box
0 ,d box
0 ,p box
1 ,q box
<a name="980307c">
Equation [c*exp(i*d)]^(p+i*q)
become   [2*exp(i*0)]^(0+i*1)
that is 2^i
Click "c vary", 2^i become t^i
Program change 2 to t
Click "c vary" button, get one curve.
This file problem 1 answer is
4.810477380965351^i=i
therefore
<a name="980307d">
variable step size [  ] start [  ] end [  ]
start to end must cover 4.810477380965351
If assign end at 4.82 Click "c vary",
curve stop at (0,1) this is i . This i
is '=' right side i in next equation
4.810477380965351^i=i

'=' left side i is defined at
0 ,p box ('i' has real be zero)
1 ,q box ('i' has coefficient 1)
Freeman 2009-03-07-12-43
2013-03-22-16-36 stop
2013-03-22-17-33 start

<a name="980307e">　index
2009-03-07-19-18 start
Add draw graph function hope to find
some hint from curve. Result is positive.
Above said
[[
fill the following to four boxes
2 ,c box
0 ,d box
0 ,p box
1 ,q box
Equation [c*exp(i*d)]^(p+i*q)
become   [2*exp(i*0)]^(0+i*1)
that is 2^i
Click "c vary", 2^i become t^i
Program change 2 to t
]]
<a name="980307f">
LiuHH paid attention that increase t
value, curve t^i go along a circle,
therefore curve will pass i=(0,1)
many times, in other words
x^i=i
has infinite many answers. The following
is working record.

<a name="980307g">
2009-03-07-18-53

From (A28) and (A29) get
0*d+log(c^1)=PI/2 ==> c=exp(PI/2) -----(A30)

Adding 2*n*PI to equation not change
sin() value and not change cos() value
0*d+log(c^1)= 2*n*PI + PI/2
therefore, for any integer n
x=c=exp(2*n*PI+PI/2) -----(A30A)
is general solution.
2009-03-07-18-56

n=0 : 2*n*PI+PI/2=
1.5707963267948965
exp(1.5707963267948965)=
4.810477380965351 this is a solution
when n=0 , check its value
4.810477380965351^i=
6.123031769111886e-17,1 result OK
IS 0,1 and is 0+1*i and is i
whole equation is
exp(1.5707963267948965)^i=i
or 4.810477380965351^i=i
This is problem 1 x^i=i solution.
x=4.810477380965351 is one solution.
2009-03-11-14-58
〕

n=1 : 2*n*PI+PI/2=
7.853981633974483
exp(7.853981633974483)=
2575.9704965975697
this is a solution when n=1 ,
check its value
2575.9704965975697^i=
3.061515884555943e-16,1 confirmed.

<a name="980307i">
n=2 : 2*n*PI+PI/2=
14.137166941154069
exp(14.137166941154069)=
1379410.7058059827
this is a solution when n=2 ,
check its value
1379410.7058059827^i=
5.510728592200698e-16,1 confirmed.
2009-03-07-19-03

n=-1 :
2*n*PI+PI/2
= -4.71238898038469
exp(-4.71238898038469)=
0.008983291021129429
this is a solution when n=-1 ,
check its value
0.008983291021129429^i=
-1.836909530733566e-16,1 confirmed.
2009-03-07-19-06
2009-03-07-19-37 record stop
2013-03-22-18-13 start
To check program correctness, copy
next five lines
cpowf(4.810477380965351,'i')  //n=0
cpowf(2575.9704965975697,'i') //n=1
cpowf(1379410.7058059827,'i') //n=2
cpowf(0.008983291021129429,'i') //n=-1
cpowf(123.45678901234567,'i') //arbitrary
paste above file lines to complex4.htm
box3, then click [test box3 command]
Output to box4 the following result
[[
cpowf(4.810477380965351,'i')  //n=0
6.123233995736766e-17,1       //0,1 IS 0+1i IS i
cpowf(2575.9704965975697,'i') //n=1
3.061616997868383e-16,1       //ans=i
cpowf(1379410.7058059827,'i') //n=2
5.51091059616309e-16,1        //ans=i
cpowf(0.008983291021129429,'i') //n=-1
-1.8369701987210297e-16,1     //ans=i
cpowf(123.45678901234567,'i') //arbitrary
0.10331752863255208,-0.9946484244582413
]]
Last blue answer is NOT i
123.45678901234567^i is NOT i
2013-03-22-18-20 stop

<a name="980308a">　index
2009-03-08-10-07 start
When start solve problem, attention is
there exist answer or not. After find
solution, attention change to solution
is unique or multiple? or infinite many?

2009-03-07-18-53 paid attention
x^i=i has infinite many solution.

2009-03-07-21-32 paid attention
x^i=i has infinite many solution,
question is that
i^x=x has how many solution?

<a name="980308b">
This moment review "draw 602" and
write "draw 603"

"draw 602" draw one solution point,
"draw 603" draw many solution points.
"draw 602" and "draw 603" draw same equation
but different output range.

In i^x=x solve for x need next two equations
m=exp(-n)*cos(m)*PI/2 -----(B25)
n=exp(-n)*sin(m)*PI/2 -----(B26)
<a name="980308c">
In m,n eliminate n left m only equation.
This m only equation is "draw 602" and
"draw 603" default equation as next
(change m to t)
t*t+(t*sin(t)/cos(t))*(t*sin(t)/cos(t))
-exp(-t*sin(t)/cos(t))
*exp(-t*sin(t)/cos(t))*(PI*PI/2/2) -----(B30)
Above three lines is one equation and
should be string to one line.
<a name="980308d">
Because curve shape close to Chinese
charcacter "冊" which mean book. Then
the curve is named as "book curve" .
//Public telephone hang one phone book.
//this phone book end view is "冊" .
//"冊" is a picture character in Chinese.
From book curve, LiuHH guess that i^x=x
has infinite many solutions. Guess is
not a proof. Until now, LiuHH still
think the analytic solution of i^x=x .
2009-03-08-10-20 stop

<a name="980309a">　index
2009-03-09-10-50 start
The following is partial working record
when solve i^x=x for x
x=p+i*q

[[
let R=p+i*q
let S=p-i*q
2009-03-08-22-32
]]

<a name="980309b">
2009-03-09-04-15
Next equation
R=exp(+i*R*PI/2) -----(B31)
is correct
how to find R ?

<a name="980309c">
2009-03-09-04-27
Next equation
S=exp(-i*S*PI/2) -----(B32)
is correct
how to find S ?

<a name="980309d">
2009-03-09-10-33
Spend one day time found
R=exp(+i*R*PI/2) -----(B31)
is correct
how to find R from (B31)?

[[
2009-03-05-15-50 next equations are correct
p=exp(-q*PI/2)*cos[p*PI/2] -----(B15)
q=exp(-q*PI/2)*sin[p*PI/2] -----(B16)
]]
2009-03-09-10-40
(B15) and (B16) combine to (B31)
Now LiuHH is able to conclude that (B15)
and (B16) combine to (B31). Because
(B15),(B16) and (B31) are one thing.
2009-03-09-11-03 stop

<a name="980309e">
exp(i*REAL) length is one.
exp(i*COMPLEX) length is NOT one !!
because result equation multiply by
extra term exp(-q*PI/2) which change
the length of cos[p*PI/2]+i*sin[p*PI/2]
2009-03-09-11-09 stop

2009-03-23-22-59 include some equation
derivation, they are from
http://freeman2.com/complex1.htm
2013-03-22-19-25 stop
2013-03-23-10-38 start

complex power
exp(z) = exp(x+iy) = exp(x)*exp(iy) =
exp(x)*(cos(y) + i*sin(y))
exp(iy) = cos(y) + i*sin(y)
is Euler's formula. The most important
equation. Only Euler's formula bring i
in function argument out to ground i
All complex function must change to exp()
before landing ! a203271136
<a name="cexp02">
2013-03-23-10-46 start
Above two lines calculation change from
exp(z) = exp(x+iy) -----(AC1)
to
exp(x)*(cos(y) + i*sin(y)) -----(AC2)
which is
exp(x)*cos(y) + i*exp(x)*sin(y)
main point is single out i from deep hidden
exp(x+iy).
<a name="cexp03">
An answer complex number must be
real1 + i*real2 -----(AC3)
Equation (AC1) is not in (AC3) form.
After simple calculation,
equation (AC2) is in (AC3) form, which

exp(iy) = (cos(y) + i*sin(y))
is Euler's formula
2013-03-23-10-52 stop

<a name="csin01">　index
2013-03-23-11-01 start  sin() to exp()
for real number t we know
exp(t)=2.718281828459045^t
exp(t)=1 + t + t*t/2! + t*t*t/3! + t*t*t*t/4!
+ ..... + t^n/n! + .....

for complex number z=x+y*i we DEFINE
exp(z)=1 + z + z*z/2! + z*z*z/3! + z*z*z*z/4!
+ ..... + z^n/n! + ..... ---eq.1.23
cvch1.pdf page 12 (page 15/18) eq.1.23
Similarly
for complex number z we DEFINE
sin(z)=z - z*z*z/3! + z^5/5! - z^7/7! + z^9/9!
+ ..... + (-1)^n*z^(2*n+1)/(2*n+1)! + ..... ---eq.1.25
for complex number z we DEFINE
cos(z)=1 - z*z/2! + z^4/4! - z^6/6! + z^8/8!
+ ..... + (-1)^n*z^(2*n)/(2*n)! + .....  ---eq.1.26
cvch1.pdf page 13 (page 16/18) eq.1.25 eq.1.26

<a name="csin03">
In eq.1.23 change complex z to complex iz get
exp(iz)=1 + iz + iz*iz/2! + iz*iz*iz/3! + iz*iz*iz*iz/4!
+ ..... + (iz)^n/n! + ..... ---(AC4)
simplify i*i, i*i*i, i*i*i*i get
exp(iz)=1 + iz - z*z/2! - i*z*z*z/3! + z*z*z*z/4!
+ ..... + (iz)^n/n! + ..... ---(AC5)
Similarly, change i to -i get
exp(-iz)=1 - iz + iz*iz/2! - iz*iz*iz/3! + iz*iz*iz*iz/4!
+ ..... - (iz)^n/n! + ..... ---(AC6)
<a name="csin04">
simplify i*i, i*i*i, i*i*i*i get
exp(-iz)=1 - iz - z*z/2! + i*z*z*z/3! + z*z*z*z/4!
+ ..... - (iz)^n/n! + ..... ---(AC7)
re-display (AC5)
exp(iz)=1 + iz - z*z/2! - i*z*z*z/3! + z*z*z*z/4!
+ ..... + (iz)^n/n! + ..... ---(AC5)
re-display eq.1.25
sin(z)=z - z*z*z/3! + z^5/5! - z^7/7! + z^9/9!
+ ..... + (-1)^n*z^(2*n+1)/(2*n+1)! + ..... ---eq.1.25
Find eq.1.25 link to (AC5) and (AC7) as next
<a name="csin05">
(AC5) - (AC7) get
exp(iz)-exp(-iz)=
0 + 2iz + 0 - 2i*z*z*z/3! - 0 + ..... ---(AC8)
on the other hand
2i*(eq.1.25) get
2i*sin(z)=
2i*z - 2i*z*z*z/3! + 2i*z^5/5! - .....
2i*(-1)^n*z^(2*n+1)/(2*n+1)! + .....  ---(AC9)
that is
<a name="csin06">
or we have
sin(z) = [exp(iz) - exp(-iz)]/(2*i) ---(AD1)
(AD1) is next 9803161750 start point.

Long time not study complex number, forget
everything.
2009-03-16 LiuHH see (AD1) as matter of course.
2013-03-23 LiuHH need derive (AD1) again.
2013-03-23-11-46 stop
2013-03-23-13-55 start
9803161750
Complex sine //(CS1) derivation
sin(z) = [exp(iz) - exp(-iz)]/(2*i) ---(CS1)
z=x+iy  ---(CS2)
The following calculation goal is change
from sin(x+iy) to real1 + i*real2 -----(AC3)
Done at (CS4) below
sin(x+iy)
= {exp[i(x+iy)] - exp[-i(x+iy)]}/(2*i)
= {exp(ix)*exp(iiy)
- exp(-ix)*exp(-iiy)}/(2*i)  ---(CS3)
= {exp(ix)*exp(-y)
- exp(-ix)*exp(+y)}/(2*i)
= {exp(-y)*[cos(x)+i*sin(x)]
- exp(+y)*[cos(-x)+i*sin(-x)]}/(2*i)
= {exp(-y)*[cos(x)+i*sin(x)]
- exp(+y)*[cos(x)-i*sin(x)]}*i/(i*i)/2
= {exp(-y)*cos(x)+i*exp(-y)*sin(x)
- exp(+y)*cos(x)+i*exp(+y)*sin(x)}*(-i)/2
= {(-i)*exp(-y)*cos(x)+(-i)*i*exp(-y)*sin(x)
- (-i)*exp(+y)*cos(x)+(-i)*i*exp(+y)*sin(x)}/2
= {(-i)*exp(-y)*cos(x)+exp(-y)*sin(x)
- (-i)*exp(+y)*cos(x)+exp(+y)*sin(x)}/2
= { -i*exp(-y)*cos(x)+exp(-y)*sin(x)
+i*exp(+y)*cos(x)+exp(+y)*sin(x)}/2
<a name="csin08">
sin(x+iy)
=  sin(x)*[exp(-y)+exp(+y)]/2
+i*cos(x)*[exp(+y)-exp(-y)]/2  ---(CS4)
9803161809

<a name="ccos01">　index
9803161931
complex cosine
(CC1) derivation please refer to (CS1)
(CS1) derive complex sine, verify (CC1) [that
is complex cosine] is reader's home work.
cos(z) = [exp(iz) + exp(-iz)]/(2) ---(CC1)
cos(x+iy)
= {exp[i(x+iy)] + exp[-i(x+iy)]}/(2)
= {exp(ix)*exp(iiy)
+ exp(-ix)*exp(-iiy)}/(2) ---(CC2)
= {exp(ix)*exp(-y)
+ exp(-ix)*exp(+y)}/(2)
= {exp(-y)*[cos(x)+i*sin(x)]
+ exp(+y)*[cos(-x)+i*sin(-x)]}/(2)
= {exp(-y)*[cos(x)+i*sin(x)]
+ exp(+y)*[cos(x)-i*sin(x)]}/2
= {exp(-y)*cos(x)+i*exp(-y)*sin(x)
+ exp(+y)*cos(x)-i*exp(+y)*sin(x)}/2
= {exp(-y)*cos(x)+exp(+y)*cos(x)
-i*[exp(+y)*sin(x)-exp(-y)*sin(x)]}/2
cos(x+iy)
=  cos(x)*[exp(-y)+exp(+y)]/2
-i*sin(x)*[exp(+y)-exp(-y)]/2 ---(CC3)
9803161941

<a name="ctan01">
complex tangent
tan(x+iy)=sin(x+iy)/cos(x+iy)  ---(CT1)

<a name="sinh01">　index
9803162117
complex hyperbolic sine is DEFINED by
sinh(z)=[(exp(z)-exp(-z)]/2 ---(HS1)
Definition see cvch1.pdf page 14 (page 17/18) line 1,2
sinh(x+iy)
=[ exp(x+iy)-exp(-x-iy)]/2
=[ exp(x)*exp(iy)-exp(-x)*exp(-iy)]/2
={ exp( x)*[cos(y) +i*sin(y)]
-exp(-x)*[cos(-y)+i*sin(-y)]}/2 ---(HS2)
={ exp( x)*[cos(y) +i*sin(y)]
-exp(-x)*[cos(y) -i*sin(y)]}/2
={ exp( x)*cos(y) +i*exp( x)*sin(y)
-exp(-x)*cos(y) -i*[-exp(-x)]*sin(y)}/2
={ exp( x)*cos(y) +i*exp( x)*sin(y)
-exp(-x)*cos(y) +i*exp(-x)*sin(y)}/2
<a name="sinh02">
sinh(x+iy)
= cos(y)*[exp( x)-exp(-x)]/2
+i*sin(y)*[exp( x)+exp(-x)]/2 ---(HS3)
9803162125

<a name="cosh01">　index
complex hyperbolic cosine is DEFINED by
cosh(z)=[(exp(z)+exp(-z)]/2 ---(HC1)
Definition see cvch1.pdf page 14 (page 17/18) line 1,2
cosh(x+iy)
=[ exp(x+iy)+exp(-x-iy)]/2
=[ exp(x)*exp(iy)+exp(-x)*exp(-iy)]/2
={ exp( x)*[cos(y) +i*sin(y)]
+exp(-x)*[cos(-y)+i*sin(-y)]}/2 ---(HC2)
={ exp( x)*[cos(y) +i*sin(y)]
+exp(-x)*[cos(y) -i*sin(y)]}/2
={ exp( x)*cos(y) +i*exp( x)*sin(y)
+exp(-x)*cos(y) -i*[+exp(-x)]*sin(y)}/2
={ exp( x)*cos(y) +i*exp( x)*sin(y)
+exp(-x)*cos(y) -i*exp(-x)*sin(y)}/2
<a name="cosh02">
cosh(x+iy)
= cos(y)*[exp( x)+exp(-x)]/2
+i*sin(y)*[exp( x)-exp(-x)]/2 ---(HC3)
9803162141

<a name="tanh01">
complex hyperbolic tangent
tanh(z)=sinh(z)/cosh(z)  ---(HT1)
9803242150
2013-03-23-14-27 stop
2013-03-24-06-38 start

<a name="980313">
2009-03-13-10-17-02 access
http://mathforum.org/library/drmath/view/63367.html

Log[z] = Log[Abs[z]] + I*(Arg[z]+2*Pi*k) ,

=====

<a name="980318a">　index
2009-03-17-11-15-56 access
http://personal.maths.surrey.ac.uk/st/C.Wulff/Modules/MS224/cvch1.pdf
If you think document below is tedious
which is concise and professional.
2009-03-18-14-15 start study notes
//the following most copy from cvch1.pdf
1.7.2 The logarithm of a complex number
If
z = e^w                  ---(LC1)
then w is defined to be a
natural logarithm of complex number z,
and we write
w = Log z                ---(LC2)
z is given, w is unknown.
i is in side of Log().
i inside a function can not be answer.
An answer complex number must be
real1 + i*real2        -----(AC3)
(AC3) i=sqrt(-1) is not in any function.
Our final goal must be
w = u + iv               ---(LC3)
w is unknown, then u,v are unknown.
z = r(cos θ + i sin θ)   ---(LC4)
z is given, r,θ are given.
//following find u,v in terms of r,θ
where -π < θ <= π        ---(LC5)
（π=3.141592653589793...）
Require -π < θ <= π that is same as
[ arg(z) has infinite many value.]
2013-03-24-07-48 stop
2013-03-24-11-48 start
<a name="980318b">
equation
z = e^w                  ---(LC1)
//red to red, blue to blue
//red z is given, blue w is unknown.
become
r(cos θ + i sin θ) = e^(u + iv)
= e^(u + iv)
= e^u*e^(iv)
= e^u*(cos v + i sin v) --(LC6)
whole equation is
r(cos θ + i sin θ)
= e^u*(cos v + i sin v) --(LC7)
Compare left side with right side
from non-trigonometric term equality
get //r is given, u is unknown
r = e^u                 ---(LC8)
from trigonometric term equality
get       //  next line  ---(LC9)
(cos θ + i sin θ) = (cos v + i sin v)
from 980515 doc get
v = θ+2kπ               ---(LCa)
k=0, +/-1, +/-2, +/-3 ....

<a name="980318c">
9803181436
Step (LCa) bring in infinite many answer,
because integer k has infinite many value.
in (LCa) set k=0 get
v = θ                   ---(LCb)
no problem, in (LC7) trigonometric term
equality is true.
But
v = θ+2kπ               ---(LCc)
then when k not =0, clearly
v not= θ                  ---(LCd)
how can we say
v = θ+2kπ               ---(LCc)
Because v and θ both used in sin() and
cos(). 2kπ is k complete circle. Adding
2kπ, sin(θ+2kπ) and cos(θ+2kπ) have
same value as sin(θ) and cos(θ).
After adding +2kπ we get complete
solution.
9803181449

r = e^u                 ---(LC8)
is same as
u = log(r)              ---(LCe)
//r is given, u is unknown
therefore, for given
z = r(cos θ + i sin θ)   ---(LC4)
Log z = w                ---(LC2)
w = u + iv               ---(LC3)
Log of a given complex z is equal to
Log z = u + iv =
log(r) + i*(θ+2kπ)  ---(LCf)
=real be given z log absolute
+imag be given z phase+2kπ
//r and θ are given.
in which log(r) = log(|z|)   ---(LCg)
r is absolute value of complex z .
θ=arg(z)=atan2(Im(z),Re(z)) ---(LCh)
k=0, +/-1, +/-2, +/-3 ....
π=3.141592653589793...
<a name="980318e">
Log z and
log z are different

Log z = log(r) + i*(θ+2kπ)  ---(LCf)
above k is arbitrary integer
below k=0
log z = log(r) + i*(θ+0)    ---(LCi)
Log z is general solution, infinite many.
log z is principal solution k=0 .
Solution can be written as
Log z = log z + i*2kπ       ---(LCj)
k is arbitrary integer

If given complex z=0, Log z is not
defined.
log(r) = log(|z|) = log(0) = -infinity

<a name="980318f">
Another point also worth your attention.
z = x + iy                 ---(LCk)
x < 0
in this case,
y approach 0 from positive imag axis, or
y approach 0 from negative imag axis,
lim[y->0+]{log(z)} = log(|x|) + i*PI ---(LCL)
lim[y->0-]{log(z)} = log(|x|) - i*PI ---(LCm)
therefore,
on negative real asix, y=0, x<0
log(z) is not continuous.
cvch1.pdf page 14 (17/18) line -8, line -7

9805151631 start
Above
(cos θ + i sin θ) = (cos v + i sin v)
result
v = θ+2kπ
k=0, +/-1, +/-2, +/-3 ....
Next, use numerical example to check
Assume
then // (2π radian = 360 degree)
v in (cos v + i sin v)
can be　60 deg  (k=0)
can be　60 deg + 1*360 deg = 420 deg (k=1)
can be　60 deg + 2*360 deg = 780 deg (k=2)
can be　60 deg - 3*360 deg =-1020 deg(k=-3)
etc.
60 deg +/- integer multiple of 360 degree
(cos v + i sin v) ≡ (cos θ + i sin θ)
trigonometric functions are periodic,
phase angle shift integer multiple of
period get same trigonometric value.
v = θ+2kπ where k is arbitrary integer,
there are infinity many k. This simple
observation tell us that
log(complex1) has one given phase θ
but it has infinite many output phase v .
Above numerical example can help entry
9805151645 stop

<a name="980318g">　index
Next is
complex1 * complex2 first,
then take log(complex1 * complex2)
compare with
two complex take log first, then sum
log(complex1) + log(complex2)
Two different approach bring us same
（LiuHH is very confuse, because log
function has multiple value 9803181510）

let
z1=r1*exp(i*θ1)
z2=r2*exp(i*θ2)
here both θ1 and θ2 are principal value.
(do not introduce multiple value at
start up point.)
then
Log z1 = log(r1) + i*(θ1+2nπ) // n is z1 variable
Log z2 = log(r2) + i*(θ2+2mπ) // m is z2 variable

<a name="980318h">
Take log first, then sum as next
Log z1 + Log z2 =
log(r1) + i*(θ1+2nπ)
+log(r2) + i*(θ2+2mπ) =
log(r1*r2) + i*(θ1+θ2+2nπ+2mπ)
=log(r1*r2) + i*(θ1+θ2+2(n+m)π)
=log(r1*r2) + i*(θ1+θ2+2pπ)  // p=(n+m)

on the other hand,
complex1 * complex2 first,
then take log(complex1 * complex2)
calculate as next
z1=r1*exp(i*θ1)
z2=r2*exp(i*θ2)
z1*z2=r1*r2*exp(i*(θ1+θ2))
then
Log(z1*z2)= log(r1*r2) + i*(θ1+θ2+2qπ) // q be arbitrary
compare with
Log z1 + Log z2 =
log(r1*r2) + i*(θ1+θ2+2pπ) // p be arbitrary
<a name="980318j">
Two calculation get ALMOST identical
result. We say
(Log z1 + Log z2) some p value answer
is same as

Log z1 + Log z2 = Log(z1*z2)

for arbitrary n
n*Log(z) is one of Log(z^n) .
cvch1.pdf page 14 (17/18) last six lines
cvch1.pdf page 15 (18/18) first five lines

<a name="980318k">
Complex log subtraction rule
Log z1 - Log z2 = Log(z1/z2)
has similar derivation as addition rule.

Note:
This file use
Log z = log(r) + i*(θ+2kπ) // k = any integer
log z = log(r) + i*(θ+0)   // k=0
Other author use Log() and log() the
other way around. As long as consistent
in one paper, different usage is not a
trouble.

Above is study notes for
http://personal.maths.surrey.ac.uk/st/C.Wulff/Modules/MS224/cvch1.pdf
Most are copy, few are LiuHH comments.
2009-03-18-15-36 stop
If you think document above is tedious
which is concise and professional.
2013-03-24-13-14 translation stop

2013-03-24-15-00 summary start //conclude
This file discuss
function(complex_input) = complex_output
for example
exp(x+iy) = exp(x)*(cos(y) + i*sin(y))
sin(x+iy) = sin(x)*[exp(-y)+exp(+y)]/2
+i*cos(x)*[exp(+y)-exp(-y)]/2 ---(CS4)
cosh(x+iy) = cos(y)*[exp( x)+exp(-x)]/2
+i*sin(y)*[exp( x)-exp(-x)]/2 ---(HC3)
Log[r(cos θ + i sin θ)]=         ---(LCf)
log(r) + i*(θ+2kπ)
(a+b*i)^(p+q*i)=[c*exp(i*d)]^(p+q*i)
={(c^p)*exp(-d*q)} ----(A13)
*{cos[p*d+log(c^q)]+i*sin[p*d+log(c^q)]}
<a name="a20324a2">
All of above equations (and following
future other functions) have one thing
in common.
Equality left side has i=sqrt(-1) in
function arguments, which is input, or
known value.
Equality right side is real1 + i*real2
<a name="a20324a3">
All calculation must bring BLUE i to
RED i as next expression
sin(x+iy)=
sin(x)*[exp(-y)+exp(+y)]/2
+i*cos(x)*[exp(+y)-exp(-y)]/2
Above expression x and y must be given.
real1 + i*real2 -----(AC3)
<a name="a20324a4">
There are how many road (channel, bus)
bring us from func(x+iy)
to answer real1 + i*real2 ?
Liu,Hsinhan's observation is that
one and only one road bring blue i
func(x+iy) to red i: real1 + i*real2
<a name="a20324a5">
This is EULER ! formula
exp(i*t)=cos(t)+i*sin(t) ----(A91)
Because Euler's formula is the only way,
LiuHH observed that all calculation first
step link to Euler's formula, next use
Euler's formula to get
real1 + i*real2
<a name="a20324a6">
This observation explain why solve
w = Log z                ---(LC2)
first write
w = u + iv               ---(LC3)
and
z = r(cos θ + i sin θ)   ---(LC4)
z is given, w is target to find.
<a name="a20324a7">
(LC3) and (LC4) allow us re-write
log() problem as an exp() problem
z = e^w                  ---(LC1)
//red to red, blue to blue
//red z is given, blue w is unknown.
become
r(cos θ + i sin θ) = e^(u + iv) --(LC6)
<a name="a20324a8"> <key4all> summary //begin
Key for ALL complex calculation
bring blue i in func(x+iy)
to red i: real1 + i*real2
Inside func(), blue i can not be answer.
Outside func(), red  i can be answer.
what you can do, what you can NOT
EULER ! ONLY EULER ! // ---(A91)
exp(i*t)=cos(t)+i*sin(t)
Convert all functions to
exp(z). sin() to exp() 2013-03-24-15-52 stop
2013-03-25-19-03 start
Change arc-function(given_complex)=unknown_complex
to regular-function(unknown_complex)=given_complex
Convert regular-function(unknown_complex) to
exp(unknown_complex), apply Euler's formula
find exp(unknown_complex)=f(given_complex)
find unknown_complex=Log[f(given_complex)]
all arc-function() involve Log(), Log() has
multiple value, therefore
2013-03-25-19-08 stop
2013-03-24-18-05 start
<a name="980319a">　index
2009-03-19-17-26 start
Next discuss casin(), arc-sine(complex)
This is also study notes, source come
from mathforum.org Dr.Math page
2009-03-12-21-21 access
http://mathforum.org/library/drmath/view/52235.html
save as complex_power_980312.txt

Date: 7/14/96 at 21:49:33
From: Doctor Jerry

Define z=x+iy -----(d01)
use the function name
casin(z)  tute0006.htm use this name
arcsin(z) Dr.Math  use this name
We solve the next equation
sin(w) = z -----(d02)
in which unknown w=u+iv -----(d03)
(remind: z=x+iy is given)
<a name="980319btip">
We are solving complex arc-sine
w=casin(z)     -----(d00)
z=x+iy is given, data in hand.
w=u+iv is unknown.
write arc-sin(z)=w as sin(w) = z
allow us use sin() to exp() relation (AD1)
arc-sine has no direct link to exp().
sin() has link to exp() and remember
EULER ! ONLY EULER
Euler change exp(i*t) to cos(t)+i*sin(t)
2013-03-24-18-59 stop
2013-03-24-22-08 start
<a name="980319c">
Because
sin(w) = (e^(iw)-e^(-iw))/(2i) -----(d04)
sin(w) = z -----(d02)
How to get (d04)? See csin01 to csin06
From next equation (d05) solve for w
(e^(iw)-e^(-iw))/(2i) = z  -----(d05)
// w is unknown, z is given.
that is to find
w = f(z) -----(d06)
Equation (d05) multiply by [(2i)*e^(iw)]
<a name="980319d">
change from e^(iw)-e^(-iw) to
(e^(iw)*[(2i)*e^(iw)]
- e^(-iw)*[(2i)*e^(iw)] ) /(2i)
= z*[(2i)*e^(iw)] -----(d07)
Left side cancel (2i) get
e^(iw)*e^(iw) - e^(-iw)*e^(iw)
= z*(2i)*e^(iw)
<a name="980319e">　index
Cancel e^(-iw)*e^(iw) to one, get
e^(iw)*e^(iw) - 1 = 2i*z*e^(iw)
e^(iw)*e^(iw) - 2i*z*e^(iw) - 1 = 0 -----(d08)
in (d08) w is unknown,
z is given, e and i are constants.

<a name="980319f">
let p = e^(iw) then (d08) become
p^2 - (2iz)p - 1 = 0 -----(d09)

solve p = e^(iw) //w is unknown

a*x*x + b*x + c = 0 -----(d10)
general solution is
x = [-b +/- sqrt(b*b-4*a*c)]/(2*a) -----(d11)

<a name="980319g">
Compare (d09) and (d10) find
a = 1
b = - (2iz)
c = -1
unknown x is p = e^(iw) in (d09)

<a name="980319h">
x = p = e^(iw)
= [-b +/- sqrt(b*b-4*a*c)]/(2*a)
= {-(-2iz) +/- sqrt[(-2iz)*(-2iz)-4*1*(-1)]}/(2*1)
= { 2iz +/- 2*sqrt[(iz)*(iz)-(-1)]}/(2*1)
= {  iz +/-   sqrt[(iz)*(iz)+1]}/(1)
= {  iz +/-   sqrt[i*i*z*z+1]}
= {  iz +/-   sqrt[1-z*z]}

<a name="980319i">
e^(iw) = { iz +/- sqrt[1-z*z]} -----(d12)
w is unknown, //a203271545 next 5 lines
log[e^(iw)] = log{ iz +/- sqrt[1-z*z]}
(iw)*log[e] = log{ iz +/- sqrt[1-z*z]}
(iw)*1 = log{ iz +/- sqrt[1-z*z]}
i*(iw) = i*log{ iz +/- sqrt[1-z*z]}
-1*w = i*log{ iz +/- sqrt[1-z*z]}
write w explicitly
w = -i*ln(iz + (1-z^2)^(1/2)) -----(d13)
Note: take log for exp(x)=a get
log(exp(x)) = log(a)
because log(exp(x))=x
get x = log(a)
here  x=iw   a=iz +/- sqrt[1-z*z]
z is given, w is unknown to find
<a name="980319j">　index
therefore iw=log(iz +/- sqrt[1-z*z])
whole equation multiply by i and take +
in +/- selection, get (d13)
note: w = arcsin(z) = arcsin(x+iy)
arcsin(z) = -i*log[iz + (1-z^2)^(1/2)] ---(d14)
take - in +/- selection, get second solution
arcsin(z) = -i*log[iz - (1-z^2)^(1/2)] ---(d15)
Above is solution from Dr.Math Doctor Jerry .
2009-03-19-18-17 stop
2013-03-24-22-40 stop
2013-03-25-16-47 start
<a name="980320a">　index
2009-03-20-12-42 start
Find arc-cosine of a complex number z.
z=x+iy -----(e01)
cacos(z)  tute0006.htm use this name
arccos(z) Dr.Math use this name
we solve the following equation
cos(w) = z -----(e02)
unknown is w=u+iv -----(e03)
given data z=x+iy
Because //see (CC1)
cos(w) = (e^(iw)+e^(-iw))/(2) -----(e04)
//sine has derivation from csin01 to csin06
//equation (e04). Main point is Euler's eq.
//Convert cos(w) to exp(w) is reader homework.
//a203251705

<a name="980320b">
Solve w from next equation.
(e^(iw)+e^(-iw))/(2) = z  -----(e05)
w is unknown, z is given data.
That is need find
w = f(z) -----(e06)
Multiply [(2)*e^(iw)] to (e05) both side.
let e^(iw) + e^(-iw) change to
<a name="980320c">
solution. The result is
(e^(iw) *[(2)*e^(iw)]
+ e^(-iw)*[(2)*e^(iw)]) /(2)
= z*[(2)*e^(iw)] -----(e07)
Left hand side cancel (2), get
e^(iw)*e^(iw) + e^(-iw)*e^(iw)
= z*(2)*e^(iw)

<a name="980320d">
e^(-iw)*e^(iw) cancel to one, get
e^(iw)*e^(iw) + 1 = 2*z*e^(iw)
Write as quadratic = 0 next
e^(iw)*e^(iw) - 2*z*e^(iw) + 1 = 0 -----(e08)
in which, w is unknown, z is given,
e and i are constants.
let p = e^(iw) then (e08) become
p^2 - (2*z)p + 1 = 0 -----(e09)
a = 1
b = - (2*z)
c = +1

<a name="980320e">　index
Solve p = e^(iw)
= [-b +/- sqrt(b*b-4*a*c)]/(2*a)
= {-(-2z) +/- sqrt[(-2z)*(-2z)-4*1*(+1)]}/(2*1)
= { 2*z +/- 2*sqrt[(z)*(z)-1]}/2
= {   z +/-   sqrt[z*z-1]}/(1)
= {   z +/-   sqrt[z*z-1]}

<a name="980320f">
e^(iw) = {  z +/- sqrt[z*z-1]} -----(e12)
w is unknown, write w explicitly get
iw =    log[z +/- sqrt(z*z-1)]
w = -i*log[z +/- sqrt(z*z-1)] -----(e13)

arccos(z) = -i*log[z + sqrt(z*z-1)] ---(e14)
In +/- take - get second answer
arccos(z) = -i*log[z - sqrt(z*z-1)] ---(e15)
2009-03-20-13-01 stop
a203251810 here
<a name="980320h">
2009-03-20-20-29 start
Find arc-tangent of a complex number z
z=x+iy -----(f01)
catan(z)   tute0006.htm use this name
arctan(z)  Dr.Math use this name
We solve the following
tan(w) = z -----(f02)
unknown is w=u+iv -----(f03)
given data z=x+iy

<a name="980320i">
tan(w) = sin(w)/cos(w)
={[e^(iw)-e^(-iw)]/(2i)}
/{[e^(iw)+e^(-iw)]/(2)}
-----(f04)

<a name="980320j">
We solve for w from next equation
{[e^(iw)-e^(-iw)]/(2i)}
/{[e^(iw)+e^(-iw)]/(2)} = z  -----(f05)
//unknown is w, given data z
That is to find w = f(z) -----(f06)
in (f05) multiply [(2i)*e^(iw)] to both
numerator and denominator.
Let e^(iw) - e^(-iw) change to quadratic
<a name="980320k">
{[e^(iw) *(2i)*e^(iw)
-e^(-iw)*(2i)*e^(iw)]/(2i)}
/{[e^(iw) *(2i)*e^(iw)
+e^(-iw)*(2i)*e^(iw)]/(2)} = z

{[e^(iw) *e^(iw)
-e^(-iw)*e^(iw)]}
/{[e^(iw) *i*e^(iw)
+e^(-iw)*i*e^(iw)]} = z -----(f07)

<a name="980320l">
{[e^(iw)*e^(iw)
-1]}
/{[e^(iw)*i*e^(iw)
+i]} = z

e^(iw)*e^(iw)-1
= z*{[e^(iw)*i*e^(iw)+i]}
= z*e^(iw)*i*e^(iw)+z*i -----(f08)

<a name="980320m">
e^(iw)*e^(iw)-1
-z*i*e^(iw)*e^(iw)-z*i = 0

e^(iw)*e^(iw)*(1-z*i)-1-z*i = 0

e^(iw)*e^(iw)=(1+z*i)/(1-z*i)

e^(iw)= +/- sqrt[(1+z*i)/(1-z*i)] -----(f09)

iw= Log{+/- sqrt[(1+z*i)/(1-z*i)]}

w= -i*Log{+/- sqrt[(1+z*i)/(1-z*i)]}

w= -i*Log{+/- sqrt[(i+z*i*i)/(i-z*i*i)]}

Final step, take square root positive
or take square root negative for two
solutions.
arctan(z)=w=
w= -i*Log{+sqrt[(i-z)/(i+z)]} -----(f10)
w= -i*Log{-sqrt[(i-z)/(i+z)]} -----(f11)
2009-03-20-20-50 stop

<a name="980320o">
2009-03-20-21-53 start
on
2009-03-17-10-38-57 access
http://www.ece.ucsb.edu/bears/class/engr5a/complex.pdf
page bottom has equation
arctan(z)=Log((i-z)/(i+z))/(2*i)

Another reference is
ISBN 0-12-059820-5 page 359, exercise 6.1.16(c)
arctan(z)=i*Log((i+z)/(i-z))/2

<a name="980320p">
Both are different from LiuHH solution
w= -i*Log{+/- sqrt[(i-z)/(i+z)]}
LiuHH review about five times, can not
find where is wrong. Only way to check
is write program look for numerical
output. Result is that LiuHH answer
and qccalc.exe answer are the same.

atan(2.1+1.2*i)
1.2139472868623191665689789249293418302375913380213300807070034965023543E0
+1.8292986383654624516275986589694639600525463857579863036271574884480371E-1i
//+1.82929863... E-1i
Next is answer for n=0 which is principal
1.2139472868623191+0.18292986383654638i
2009-03-20-21-59 stop

<a name="980321a">
2009-03-21-08-42 start
Next find arc-hyperbolic sine asinh(z)
z=x+iy -----(g01)
asinh(z)   tute0006.htm use this name
arcsinh(z)
Hope find w=asinh(z) -----(g00) //a203251830
Reverse (g00) to (g02)
sinh(w) = z -----(g02)
unknown is w=u+iv -----(g03)
given data z=x+iy -----(g01)
<a name="980321b">
hyperbolic sine sinh(w) is defined as
sinh(w) = 0.5*(exp(w)-exp(-w)) -----(g04)
See cvch1.pdf page 14 (page 17/18) line 1,2

relate (g02) and (g04) get
0.5*(exp(w)-exp(-w)) = z -----(g05)
Solve for w from (g05)
That is to find w = f(z) = asinh(z) -----(g06)
z is given data, w is unknown.
<a name="980321c">
multiply [2*exp(w)] to (g05)
Let exp(w) - exp(-w) change to a
solution.

0.5*[exp(w)*2*exp(w)-exp(-w)*2*exp(w)] = z*2*exp(w)

exp(w)*exp(w)-exp(-w)*exp(w) = z*2*exp(w)

<a name="980321d">
exp(w)*exp(w) -2*z*exp(w) -1 = 0 -----(g07)

This is quadratic in exp(w) .
let exp(w) be x
compare (g07) with
a*x*x +b*x +c = 0
get
a = 1
b = -2*z
c = -1
x=[-b +/- sqrt(b*b-4*a*c)]/2
<a name="980321e">
then (g07) has solution
exp(w) = [-(-2*z) +/-
sqrt((-2*z)*(-2*z)-4*1*(-1))]/2

exp(w) = [ 2*z +/-2*sqrt(z*z+1)]/2

exp(w) = z +/- sqrt(z*z+1)
Take log() to get explicit expression
for unknown w
w = Log(z +/- sqrt(z*z+1))

w = Log(z + sqrt(z*z+1)) -----(g08)
w = Log(z - sqrt(z*z+1)) -----(g09)
2009-03-21-09-08 stop
2013-03-25-18-46 stop
2013-03-25-22-01 start
<a name="980321g">
2009-03-21-13-46 start
Next find arc-hyperbolic cosine acosh(z)
Given z=x+iy -----(h01)
acosh(z)   tute0006.htm use this name
arccosh(z)
Hope find w=acosh(z) -----(h00) //a203252205
Reverse (h00) to (h02)
cosh(w) = z -----(h02)
unknown is w=u+iv -----(h03)

<a name="980321h">
hyperbolic cosine cosh(w) is defined
cosh(w) = 0.5*(exp(w)+exp(-w)) -----(h04)
Combine (h02) and (h04) get
0.5*(exp(w)+exp(-w)) = z -----(h05)
where z is given, w is unknown.
Solve unknown w , that is to find
w = f(z) = acosh(z) -----(h06)
w = some other function involve z

<a name="980321i">
Multiply [2*exp(w)] to (h05) such that
[exp(w)+exp(-w)]*[2*exp(w)]
0.5*[exp(w)*2*exp(w)+exp(-w)*2*exp(w)] = z*2*exp(w)

exp(w)*exp(w)+exp(-w)*exp(w) = z*2*exp(w)

<a name="980321j">
exp(w)*exp(w) -2*z*exp(w) +1 = 0 -----(h07)

(h07) is quadratic in exp(w), let x=exp(w)
Compare (h07) with
a*x*x +b*x +c = 0
find
a = 1
b = -2*z
c = +1
x=[-b +/- sqrt(b*b-4*a*c)]/2
<a name="980321k">
Apply solution to (h07) get
exp(w) = [-(-2*z) +/-
sqrt((-2*z)*(-2*z)-4*1*(+1))]/2

exp(w) = [ 2*z +/-2*sqrt(z*z-1)]/2

exp(w) = z +/- sqrt(z*z-1)
Whole equation take log() solve for w

w = Log(z +/- sqrt(z*z-1))

arc-hyperbolic cosine acosh(z)=w
w = Log(z + sqrt(z*z-1)) -----(h08)
w = Log(z - sqrt(z*z-1)) -----(h09)
2009-03-21-13-54 stop
//a203252227 here

<a name="980321m">
2009-03-21-14-21 start
Next find arc-hyperbolic tangent atanh(z)
Given z=x+iy -----(k01)
atanh(z)   tute0006.htm use this name
arctanh(z)
Hope find w=atanh(z) -----(k00) //a203252230
Reverse (k00) to (k02)
tanh(w) = z -----(k02)
unknown is w=u+iv -----(k03)

<a name="980321n">
hyperbolic tangent tanh(w) is defined as
tanh(w) = (exp(w)-exp(-w))
/(exp(w)+exp(-w)) -----(k04)

(exp(w)-exp(-w)) = z*(exp(w)+exp(-w)) -----(k05)
in (k05), z is given, w is unknown.
In (k05) solve for w [solve for exp(w)]
Hope to find
w = f(z) = atanh(z) -----(k06)
w = some other function of z

<a name="980321o">
Multiply exp(w) to (k05) let
exp(w) - exp(-w) change to quadratic
equation, which has given formula.
exp(w)*exp(w)-exp(-w)*exp(w)
= z*[exp(w)*exp(w)+exp(-w)*exp(w)]

exp(w)*exp(w)-1
= z*[exp(w)*exp(w)+1] -----(k07)

exp(w)*exp(w)-1 - z*exp(w)*exp(w)-z = 0

exp(w)*exp(w)*(1-z) = 1+z

exp(w)*exp(w) = (1+z)/(1-z) -----(k08)

exp(w+w) = (1+z)/(1-z)

Take log() to express w explicitly
arc-hyperbolic tangent atanh(z)=w
solution is
w = 0.5*Log((1+z)/(1-z)) -----(k09)

2009-03-21-14-42 stop

<a name="980325a">
2009-03-25-09-16 start
Above moved derivation notes from
http://freeman2.com/polyroot.htm
to tute0006.htm main derivation is
complex power　　complex sine
complex cosine　　complex tangent
hyperbolic sine　　hyperbolic cosine
hyperbolic tangent

2009-03-25-09-22 stop
2013-03-25-22-41 translate stop

2013-04-02-12-21 start
Earlier notes say "EULER ! ONLY EULER !"
exp(i*t)=cos(t)+i*sin(t) ---(A91)
Convert all functions to
exp(z). Here, it has a trap. If
apply exp(i*t)=cos(t)+i*sin(t) ---(A91)
with t be real. This is correct.
<a name="a20402b">
If change real t to complex c0=c+i*d ---{CG1)
apply exp(i*c0)=cos(c0)+i*sin(c0) ---{CG2)
is wrong. Not only wrong, it is in fact
go backward ! This is easy to understand.
We want final answer be real1 + i*real2
exp(i*t)=cos(t)+i*sin(t) ---(A91)
satisfy real1 + i*real2, because t is
real, both cos(t) and sin(t) are real.
You can use (A91), t be real
<a name="a20402c">
You can NOT use (CG2), c0=complex
But in exp(i*c0)=cos(c0)+i*sin(c0) ---{CG2)
c0 is complex, both cos(c0) and sin(c0)
are complex. Then we still have blue i
in cos(c+i*d)+i*sin(c+i*d) This blue i
in function can not be our answer.
Convert cos(c0) to real3+i*real4 that
is go backward. We should go from
exp(i*complex) to exp(i*real) directly.
2013-04-02-13-05 here

<a name="a20402d">
Above suggest reader to review the book
George B. Thomas, Jr. and Ross L. Finney
ISBN 0-201-16290-3 for Euler's formula
proof by Taylor series expansion.

<a name="a20402e">
Next is another proof Euler's formula
by differentiation. Textbook is written
by Tom M. Apostol, Calculus volume 1.
Second ed. 1967 page 366.
Assume y be real number, assume i=sqrt(-1)
assume exp(iy) can be express by
exp(iy) = A(y) + i*B(y) ---(CG3)
Here A(y), B(y) are unknown functions.
Differentiate d/dy (CG3) once get
i*exp'(iy) = A'(y) + i*B'(y) ---(CG4)
Because exp'(iy)=exp(iy) ---(CG5)
is exponential function property, get
i*exp(iy) = A'(y) + i*B'(y) ---(CG6)
Differentiate d/dy (CG6) once get
i*i*exp(iy) = A''(y) + i*B''(y) ---(CG7)
We know i*i=-1  ---(CG8)
<a name="a20402g">
(CG7) become
-exp(iy) = A''(y) + i*B''(y) ---(CG9)
compare with negative of start equation
-exp(iy) =  -A(y) - i*B(y)   ---(CGa)
compare (CG9) with (CGa) we have
real number equality A''(y)=-A(y) ---(CGb)
imag number equality B''(y)=-B(y) ---(CGc)

<a name="a20402h">
In second order differential equation
f''(x) + f(x) = 0  ---(CGd)
we need initial condition to solve f(x)

In (CG3) set y=0 to get
exp(i*0) = 1 = A(0) + i*B(0) ---(CGe)
For real equality, A(0)=1 ---(CGf)
For imag equality, B(0)=0 ---(CGg)

<a name="a20402i">
In (CG6) set y=0 to get
i*exp(i*0) = i = A'(0) + i*B'(0) ---(CGh)
For real equality, A'(0)=0 ---(CGi)
For imag equality, B'(0)=1 ---(CGj)

<a name="a20402j">
Now
A''(y)=-A(y) ---(CGb)
A(0)=1 ---(CGf)
A'(0)=0 ---(CGi)
and
B''(y)=-B(y) ---(CGc)
B(0)=0 ---(CGg)
B'(0)=1 ---(CGj)

Second order differential equation
with constant coefficient tell us that
the solution for A(y) is
A(y)=cos(y) ---(CGk)
and
B(y)=sin(y) ---(CGl)
<a name="a20402l">
Differential equation theory tell us
that the solution is unique. Assume
second solution exist,
first solution - second solution = 0
identically. Then the initial problem
exp(iy) = A(y) + i*B(y) ---(CG3)
has solution
exp(iy) = cos(y) + i*sin(y) ---(CGm)
Here y be REAL.
Above is study notes. textbook is
Tom M. Apostol, Calculus volume 1.
Second ed. 1967 page 366.
2013-04-02-13-40 here

<a name="a20402m">
Assume z=x+i*y be complex and x,y both
be real. Equation (AC2) give us
exp(z)= exp(x+iy)
=exp(x)*(cos(y) + i*sin(y)) -----(AC2)
Express in symbolic get
exp(realOne+i*realTwo)
=  exp(realOne)*cos(realTwo)
+i*exp(realOne)*sin(realTwo) -----(CGn)

<a name="a20402n">
Now a little bit twist
exp(i*z)= exp(i*x+i*iy)
= exp(i*x-y)  -----(CGo)
set realOne=-y, realTwo=x  -----(CGp)
exp(i*z)= exp(i*x+i*iy)
= exp(realOne+i*realTwo)
=  exp(-y)*cos(x)
+i*exp(-y)*sin(x) -----(CGq)
<a name="a20402o">
That is
exp(i*z)= exp(i*(x+iy))=
=  exp(-y)*cos(x)
+i*exp(-y)*sin(x) -----(CGr)
compare with
exp(i*c0)=cos(c0)+i*sin(c0) ---{CG2)

If z==c0, then (CGr),(CG2) are the
same. Use (CGr), drop (CG2) because
exp(-y),sin(x),cos(x) are real GOOD
cos(c0), sin(c0) are complex. NO !
2013-04-02-14-00 stop

<a name="a20402q">
Please try next code you can
change only first line c0=...
[[
//Example a20402
//exp(i*c0)=cos(c0)+i*sin(c0)
c0='1+8i';    //you can change number
a1=ccosf(c0); //eqn. (CC3)
a2=csinf(c0); //eqn. (CS4)
a3=cmulf('i',a2); //i*sin(c0)
//above is what you should NOT do
//below is what you should do
x=cgetr(c0); //x=1, if c0='1+8i';
y=cgeti(c0); //y=8, if c0='1+8i';
b1=exp(-y)*cos(x); //(CGr) real
b2=exp(-y)*sin(x); //(CGr) imag
a4 //correct ans. but walk long way
b3 //better ans. less twist.
]]
<a name="a20402r">
copy and paste to complex4.htm Box3
click [test box3 command output to box4]
a4 and b3 both be exp(i*c0) You should
see a4 = b3. If not, LiuHH code wrong.
a4 is cos(i*c0)+i*sin(i*c0)
b3 is exp(-y)*cos(x)+i*exp(-y)*sin(x)
a4 is (CG2) , b3 is (CGr) .
here c0=x+i*y be given by user.
2013-04-02-14-21 stop

<a name="a20403a">
2013-04-03-13-47 start
a4=0.00018125123131085274,0.00028228206770108954
b3=0.00018125123138831292,0.00028228206786737157
both correct, which one is better answer?
LiuHH mark b3 be better answer. Why it
is so? because b1,b2,b3 path has less
calculation, less error involved.
On the other hand, a1,a2,a3,a4 path
walk long way, more calculation and
more error introduced.
<a name="a20403b">
Below check whether b3 is better answer
than a4. The problem is
given c0='1+8i'; find exp(i*c0)
The input complex is i*c0 that is
i*(1+8i) or 1*i+8*i*i .
The input complex is -8+i
[[
a4 //correct ans. but walk long way
0.00018125123131085274,0.00028228206770108954
b3 //better ans. less twist.
0.00018125123138831292,0.00028228206786737157
]]
<a name="a20403c">
Run the reverse. take log of exp(i*'1+8i')
should recover i*c0. Put next two lines
clogf('0.00018125123131085274,0.00028228206770108954i')
clogf('0.00018125123138831292,0.00028228206786737157i')
to complex4.htm box3 click
[test box3 command output to box4]
box4 output next
[[
clogf('0.00018125123131085274,0.00028228206770108954i')
-8.000000000541858,0.9999999999264834
clogf('0.00018125123138831292,0.00028228206786737157i')
-8,0.9999999999999999
]]
<a name="a20403d">
Input is : -8                +i
clogf(a4): -8.000000000541858,0.9999999999264834
|| error           || error
clogf(b3): -8                ,0.9999999999999999
<a name="a20403e">
eqn.(CG2) is clogf(a4),
(CG2) walk long way , (CG2) get poor answer.
eqn.(CGr) is clogf(b3),
(CGr) walk short way, (CGr) get better answer.
Use (CGr), drop (CG2) !!
2013-04-03-14-01 stop
```

<a name=graph01>　index Draw complex1^complex2 curve

complex base rise to complex power [c*exp(i*d)]^(p+i*q)
In c, d, p, q allow one be variable, other three must assign value.
step size 　 start 　 end
base complex﹕c *exp(i*d ) 　 default (1+2i)
complex power﹕p +i*q 　 　 default (3+4i)
x-axis min , max 　 　min/max determine
y-axis min , max 　 　curve aspect ratio
draw method

Above is [draw 601], no [modify 601] because [draw 601]
draw default data. User can not control 9803070919
All curve squeeze to one small corner? Please
reduce x-axis min , max range ; reduce y-axis min , max range.
All curve fly to outer space? (not see one) Please
increace x-axis min , max range ; increace y-axis min , max range.

Box4, curve equation

Box5, curve data

<a name=graph02>　index
in i^x=x find x one middle step curve. Click "draw 602" view graph.

<a name=graph03>　index
in i^x=x find x one middle step curve.
"draw 602" see one zero point, "draw 603" see all.
"draw 602" and "draw 603" draw same equation, but different range.

2009-03-06-18-17 start draw code
2009-03-06-20-06 done draw code

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

Chinese version tutc0006.htm created on 2009-03-04
English version tute0006.htm created on 2013-03-18

URL of this English page
http://freeman2.com/tute0006.htm
URL of this Chinese page
http://freeman2.com/tutc0006.htm

Thank you for visiting freeman2.com
Freeman　　2009-03-04-13-14
Freeman　Liu,HsinHan　劉鑫漢　2013-03-27-16-28

2009-01-27-10-08-05
http://www.sftw.umac.mo/~fstitl/10mmo/inequality.html

2013-03-27-16-24 decide to keep next equation,
to display one method to write math equation in
web page. Similar code learned from sftw.umac.mo
See freeman2.com/tute0055.htm for more equation.
Choose text size=normal for correct table width.

2009-01-27-13-30
 Γ(z) = t=∞ ∫ t=0 e(-t) *t(z) *dt

2009-01-27-13-49

Α Β Γ Δ Ε Ζ Η Θ Ι Κ
Λ Μ Ν Ξ Ο Π Ρ ΢ Σ Τ
Υ Φ Χ Ψ Ω Ϊ Ϋ ά έ ή
ί ΰ α β γ δ ε ζ η θ
ι κ λ μ ν ξ ο π ρ ς
σ τ υ φ χ ψ ω