(x+yi)^i how?!

[Jenkar]

Just i as far as i have read.

[YawnChainHow]

I probably know less about the subject at hand than anyone else, but seem to have been luckier with my searching:
(Don't mind the redundancy)
Complex Numbers to Complex Powers
Understanding Imaginary Exponents
Imaginary and Complex Exponents
How to raise a complex number to the power of another complex number

Just i as far as i have read.

Seems to be a+bi as far as a+bi have read 

[ralouf]

Did it in school last year it's something that deal with the definition of the ln of a complex number, and I forgot what it's.

[Nelde95]

I was really wanting to do (a+bi)^(c+di), but I thought that it could be written as: ((a+bi)^c)*(((a+bi)^d)^i). However I have some doubts now,
but if this is true, I really just nead to know how to do (a+bi)^i. Can anybody comfirm wether this^ is true for complex numbers too?

Spoiler for thinking...:

e^(i*ln(a+bi)), real(ln(a+bi))=log_e^2(a^2+b^2),imag(ln(a+bi))=... What is with that blue graph that just poped up on my screan?!

[Nelde95]

Finally figured out something helpfull: xⁱ=i^log_epi/2(x)

[Nelde95]

(a+b*i)^(c+di)
=
cos(((ln(a^(2)+b^(2))*d)/(2))-(((2*arctan(((a)/(b)))-sign(b)*pi)*c)/(2)))*(a^(2)+b^(2))^(((c)/(2)))*e^(arctan(((a)/(b)))*d-((sign(b)*d*pi)/(2)))
+
sin(((ln(a^(2)+b^(2))*d)/(2))-(((2*arctan(((a)/(b)))-sign(b)*pi)*c)/(2)))*(a^(2)+b^(2))^(((c)/(2)))*e^(arctan(((a)/(b)))*d-((sign(b)*d*pi)/(2)))*i

or

cos(u)+sin(u)*i where
u=((ln(a^(2)+b^(2))*d)/(2))-(((2*arctan(((a)/(b)))-sign(b)*pi)*c)/(2)))*(a^(2)+b^(2))^(((c)/(2)))*e^(arctan(((a)/(b)))*d-((sign(b)*d*pi)/(2))

Thanks to mathprograms... why did I not do this sooner? oh well...

But now it has no reason to remain active... D:

[andretimpa]

[troll]
exp(-pi/2) = exp(i*pi/2)^i = i^i = exp(5*i*pi/2)^i = exp(-5*pi/2)
exp(-pi/2) = exp(-5*pi/2)
exp(2*pi) = 1
pi = 0
[/troll]

now this thread is complete

[Nelde95]

[troll]
exp(-pi/2) = exp(i*pi/2)^i = i^i = exp(5*i*pi/2)^i = exp(-5*pi/2)
exp(-pi/2) = exp(-5*pi/2)
exp(2*pi) = 1
pi = 0
[/troll]

now this thread is complete

Huh... Did not know about that one.
Tested it in my math program and it gave "exp(5*i*pi/2)^i = exp(-5*pi/2) -> False", but I see how it was done.
Did someone divide by 0 somewhere or is it allowed to say that exp(5*i*pi/2)^i = exp(-5*pi/2)?

[andretimpa]

This is the reason essentially:
http://en.wikipedia.org/wiki/Complex_logarithm#Problems_with_inverting_the_complex_exponential_function

You basically switch the branch you are using for your calculations in the middle of the "demonstration" (much more subtle than dividing by zero)