Complex powers question, raising i to the power of i

Question

I am confused as to what $$i^{i}$$ looks like. I followed the procedure our prof described and got $$e^{i\log i}=e^{\log i+2ki\pi }=e^{-2k\pi (\cos(\log i)+i\sin(\log i))}.$$

However, $$i^{i}=e^{-\pi/2}$$

Is it the same thing or am I doing something wrong?

Thanks!

Use $i = e^{i \pi / 2}$ to evaluate the logarithm in the middle. — Sean Lake, Feb 25 '17 at 17:39
You might wish to define $a^b=e^{b\operatorname{Log}(a)}, where $\operatorname{Log}$ denotes the principal log. — Simply Beautiful Art, Aug 25 '17 at 20:36

Tig la Pomme · Answer 1 · 2017-02-26T12:27:10.097

2

There's no unambiguous definition for complex logarithm, hence several answers are legitimate.

edited Feb 26 '17 at 12:27

answered Feb 25 '17 at 16:54

Tig la Pomme

641

Complex powers question, raising i to the power of i

1 Answers1