# Complex analysis: Evaluating an i^i

• Jul 5th 2010, 02:35 PM
Complex analysis: Evaluating an i^i
"Evaluate $\displaystyle i^{1+i}$. Put the result in the form $\displaystyle u+iv$ when $\displaystyle u,v$ are real."

$\displaystyle i = e^{i(\frac{\pi}{2}+2n\pi)}$
$\displaystyle i^{1+i} = [e^{i(\frac{\pi}{2}+2n\pi)}]^{1+i} = e^{-(2n+\frac{1}{2})\pi}e^{i(2n+\frac{1}{2})\pi}$
$\displaystyle = e^{-(2n+\frac{1}{2})\pi}[\cos(2n+\frac{1}{2})\pi + i\sin(2n+\frac{1}{2})\pi] = e^{-(2n+\frac{1}{2})\pi}$ for $\displaystyle n = 0, \pm1, \pm2,...$

Looks good or did I not need to start with that $\displaystyle 2n\pi$?
• Jul 5th 2010, 03:08 PM
HallsofIvy
Looks good to me. If you want the most general solution it doesn't hurt to have that $\displaystyle 2n\pi$ from the start.
• Jul 5th 2010, 03:09 PM
huram2215
An adjustment. Pick up from ... $\displaystyle e^{-\left( 2n+\frac{1}{2}\right) \pi} e^{i \left( 2n+\frac{1}{2}\right) \pi} = e^{-\left( 2n+\frac{1}{2}\right) \pi} e^{i 2n \pi} e^{i \frac{1}{2} \pi} = e^{-\left( 2n+\frac{\pi}{2}\right) } e^{i \frac{\pi}{2} } = i e^{-\left( 2n+\frac{\pi}{2}\right) }$.