# Exotic imaginaries

1. Jun 24, 2008

### The_ArtofScience

Imaginary numbers have always intrigued me from the very beginning I was taught to believe that some negative root number could have an "i" factored out!

(1+i)^(1+i) = e^ln(root 2)-pi(1/4+2k)e^i(ln(root2)+pi(1/4+2k))

I'm not quite sure as to how 1+i = root 2e^(pi/4+2pi k). I do know that i^i = e^-pi/2 from the identity e^ipi=-1 which gives i = ln(-1)/pi. So my question is why is 1+i = root 2e^(pi/4+2pi k) true? This statement does not seem to make sense to me because it describes both cosine and sine giving 1 as an answer which does not seem to be the case on the unit circle and neither from Euler's identity

Last edited: Jun 24, 2008
2. Jun 24, 2008

### dx

1 + i is not root 2.

3. Jun 24, 2008

### The_ArtofScience

Ha! Sorry found out my own answer

1+i is really just a case of a+ib on the complex xy plane. The modulus can be calculated using (a^2+b^2)^1/2 and the argument can be done using the fact that tangent = b/a

edit:Yes, dx I'm aware of that. I changed it after you made your post

4. Jun 26, 2008

### the1ceman

Complex powers are not what they seem. You stated that i^i=e^-pi/2, in fact i^i has infinitely many values:
$$i^i=e^{ilogi}=e^{i(i(\frac{\pi}{2}+2k\pi))}=e^{-\frac{\pi}{2}}e^{-2k\pi}$$
Where k is an integer. This 'multi-valuedness' arises from the fact that the argument of a complex number is 'multi-valued', adding an integer multiple of 2pi to the angle a complex number makes with the positive x-axis (real axis) gives you the same number.
Moral of story: Beware of complex powers!

Edit: woops in my haste i forgot to re-read your 3rd line which shows you have taken into account the multivaluedness.