A 'complex' question: exact value of i^i

[ChaoticLlama]

What is the exact answer to i^i.

The google calculator says the numerical solution is approximately i^i = 0.207879576...

how does one arrive at this? I heard that you are supposed to use the identity e^(i * pi) = -1

 

[pka]

If each of u & w is a complex number then the definition of complex exponentiation:
\(\displaystyle \L
w^u = e^{u\log (w)}\).

\(\displaystyle \L
i^i = e^{i\log (i)} = e^{i\left( {\ln (1) + i(\arg (i)} \right)} = e^{i\left( {i\left( {\frac{{(2n + 1)\pi }}{2}} \right)} \right)} = e^{\left( { - \left( {\frac{{(2n + 1)\pi }}{2}} \right)} \right)}\)

Let n=0 to evaluate.

 

[soroban]

Hello, ChaoticLlama!

What is the exact answer to \(\displaystyle i^i\) ?

The google calculator says the numerical solution is approximately \(\displaystyle i^i\:=\:0.207879576...\)

How does one arrive at this?
I heard that you are supposed to use the identity \(\displaystyle e^{^{i\pi}}\:=\:-1\)

\(\displaystyle \text{Start with: }\:-1\;=\;e^{^{i\pi}}\)

\(\displaystyle \text{Take the square root of both sides: }\:\sqrt{-1}\;=\;\left(e^{i\pi}\right)^{\frac{1}{2}}\)

\(\displaystyle \text{And we have: }\:i\;= \;e^{i\frac{\pi}{2}}\)

\(\displaystyle \text{Raise both sides to the power }i:\;\;i^i\;=\;\left(e^{i\frac{\pi}{2}}\right)^i \;=\;e^{(i^2)\frac{\pi}{2}}\;=\;e^{-\frac{\pi}{2}}\)

\(\displaystyle \text{Therefore: }\:i^i\;=\;e^{-\frac{\pi}{2}} \;= \; 0.207879576...\)

 

[ChaoticLlama]

Thanks for the responses.