Complex function derivation
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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
x^i=i answer is x=exp(PI/2)
problem 2, in i^x=x find x
problem 2 correct up to here
approximate answer m=t=0.7
better numerical solution
<a name=index02>
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

<a name=index03>
problem 1 in x^i = i -----(A00) find x
answer: x=exp(PI/2)=4.810477380965351

problem 2 in i^x = x -----(B02) find x
answer: x=0.4382829367270321+0.36059247187138543i

problem 3 in x^x = i -----(C00) find x
answer: x=1.3606248702911176+1.1194391662423495i
Problem 3 is not in this page, you can 
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 
ask a math expert near-by. Please put 
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

<a name="prob0a02">
2009-03-02-08-29 begin the following 
calculation. All "a20319" lines are 
2013-03-19 LiuHH added general equation. 
//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
<a name="powc1c2Ans">
[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 
//function, (A13) is answer.

<a name="prob0a05">

Explain above derivation as following.

<a name="EulerFormula">
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 
<a name="prob0a06b">
2013-03-19-18-53 start 
After First step, we move toward what 
direction? The answer is clear, 
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).

<a name="prob0a10">

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 〕 

<a name="prob0a15">

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

<a name="add2npi1">
2009-03-14-09-58 start
Answer is trigonometric function, they 
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)
Answers involve trigonometric function 
these answers have many solution.
after include 2*n*PI final equation is
(a+b*i)^(p+q*i)=(c*exp(i*d))^(p+q*i)
={(c^p)*exp(-d*q)} ----(A13add2npi)
*{  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

<a name="add2npi2">
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 
trigonometric function, then add or 
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 
answer is one answer.
〔 cos(x+2*n*PI) and cos(x) have same 
 numerical value. 〕
That is, in answer no need to consider 
2*n*PI .

<a name="add2npi3"> index
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

<a name="prob0a17">
2009-03-03-22-22
x^i=i
find x
Answer is x=exp(pi/2)
[[
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)

<a name="prob0a21">
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
 c=exp(2*n*PI+PI/2) //9803071933 add 2*n*PI
 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">
Please go to
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
 c=exp(2*n*PI+PI/2) //9803071933 add 2*n*PI
 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 
ask a math expert near-by. Please put 
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)
answer is x=exp(PI/2)
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

<a name="a20321b02">
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 analytic answer is x=exp(PI/2)=4.810477380965351 <a name="prob0b02"> 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 numerical answer is 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)]} <a name="prob0b07"> 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] <a name="prob0b12"> 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) <a name="prob0b17"> 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"> please access http://freeman2.com/graph09e.htm [[ 9803062052 please pay attention later part of this file has drawing tool no need go to graph09e.htm 9803071023 please pay attention 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" button read user equation "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 ]] <a name="prob0b22"> 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 2009-03-05-16-30 get i^x=x numerical answer 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 ]] <a name="prob0b27"> 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 answer is numerical answer. 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 ask a math expert near-by. Please put 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 <a name="9803062032"> 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 Earlier said please access 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 Please goto a name=graph01 base complex: c [2] *exp(i*d [0] )    power index complex: p [0] +i*q [1] 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 x^i=i has infinite many answers. 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 <a name="980307h"> 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 〔 Alert, above 6.123031769111886e-17,1 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 <a name="cexp01"> 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 is acceptable answer. 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! + ..... //please review calculus textbook. 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 <a name="csin02"> 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">  exp(iz)-exp(-iz)=2i*sin(z) ---(AD0) 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 <a name="csin07"> 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 <a name="ccos02"> 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 9803162138 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 and too complicate, please read cvch1.pdf 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 ask arg(z) be principal value. [ 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 <a name="980318d"> 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"> Alert ! 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, get different answer. 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 <a name="980515"> 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 θ=π/3 (radian) =180/3=60 degree 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 level rader understand better. 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 answer? or different answer? (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) <a name="980318i"> 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*z2) some q value answer. Complex log addition rule Log z1 + Log z2 = Log(z1*z2) Complex log addition VICE rule 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 and too complicate, please read cvch1.pdf which is concise and professional. 2013-03-24-13-14 translation stop <a name="a20324a1"> 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. Answer is in the form 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 Reader please remember: 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 <a name="key4Arc"> 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 all arc-function() has multiple answer. 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 Subject: reply: complex number asin/acos/atan <a name="980319b"> 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 Equation (AD1)≡(d04) a203241850 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 quadratic equation, because there is general solution for quadratic equation. (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) which is a quadratic equation 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"> To simplify quadratic equation, let p = e^(iw) then (d08) become p^2 - (2iz)p - 1 = 0 -----(d09) solve p = e^(iw) //w is unknown quadratic 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) final answer is 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 //result (AD1). reader please derive cosine //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"> quadratic equation, which has general 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) Compare with general quadratic coefficients 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) <a name="980320g"> arccos(z) final answer is 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 equation. Quadratic has general solution. <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)]} <a name="980320n"> 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 answer and compare with qccalc.exe 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 2009-03-20-21-38 above is qccalc.exe answer. below is complex1.htm answer Next is answer for n=0 which is principal answer. 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 quadratic equation which has general 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 quadratic solution is 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)) <a name="980321f"> Final answer for asinh(z)=w 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)] convert to a quadratic equation. 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 quadratic solution is 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)) <a name="980321l"> arc-hyperbolic cosine acosh(z)=w answer is 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) Link (k02) with (k04) get (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) <a name="980321p"> 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 Add index [a name=index04] 2009-03-25-09-22 stop 2013-03-25-22-41 translate stop 2013-03-27-16-21 1st proofread stop 2013-03-29-15-38 2nd proofread stop <a name="a20402a"> 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. <a name="a20402f"> 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) <a name="a20402k"> 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) <a name="a20402p"> 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) a4=caddf(a1,a3); //cos(c0)+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 b3=cnewf(b1,b2+'i')//(CGr) answer 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] see answer at box 4. 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 Example a20402 exp(i*'1+8i') two answers 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 complex4.htm box4 report answer [[ 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
First upload Chinese tutc0006.htm 2009-03-04
First upload English tute0006.htm 2013-03-30

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




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Λ Μ Ν Ξ Ο Π Ρ ΢ Σ Τ
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