# g3popstar's question at Yahoo! Answers (Principal value of i^i)

MHB Math Helper

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Hello g3popstar,

For $z,w$ complex nombers and $z\neq 0$ we define $z^w=e^{w\log z}$. In our case, $i^i=e^{i\log i}$. On the other hand, $\log z=log |z|+i\arg z$. The principal argument of $i$ is $\frac{\pi}{2}$, so the principal value of $\log i$ is $\log i=\log 1+i\frac{\pi}{2}=i\frac{\pi}{2}$. As a consequence $$i^{i}=e^{i\log i}=e^{i\cdot i\frac{\pi}{2}}=e^{-\frac{\pi}{2}}$$ If you have further questions, you can post them in the Analysis section.