# Solving I to the I power

1. Jul 14, 2011

### SeanofWar

Solving "I" to the "I" power

What is the proper formula to calculate "I" to the "I" power? I have seen numerous formulas, however, I was curious if someone could provide me with the simplest solution.

2. Jul 14, 2011

### Hurkyl

Staff Emeritus
Re: Solving "I" to the "I" power

In complex analysis, the (multi-valued) function $x^y$ is defined to be $\exp(y \log x)$....

3. Jul 14, 2011

### hunt_mat

Re: Solving "I" to the "I" power

Note that $i=e^{\frac{\pi i}{2}}$, so $i^{i}=e^{i\frac{\pi i}{2}}$

4. Jul 14, 2011

### dextercioby

Re: Solving "I" to the "I" power

Hunt_mat, that's only half the story. As Hurkyl noted, the complex exponential is multiple-valued, so we must not forget that.

5. Jul 14, 2011

### micromass

Staff Emeritus
Re: Solving "I" to the "I" power

A typo: it's the logarithm that's multiple valued but you knew that of course.

Anyway, while it is certainly true that $a^b$ has multiple values for complex numbers, mathematicians sometimes pick one value as a principal value. That is, they define

$$a^b=e^{bLog(a)}$$

where Log is the principal branch of the logarithm, which is not multivalued (since we restricted it).

We consider the principal values of $a^b$ in the Riemann-zeta function, for example, where

$$\zeta (z)=\sum{\frac{1}{n^z}}$$

there we take the exponentiation to be the principal value, and not the multi-valued one. The value

$$i^i=e^{-\frac{\pi}{2}}$$

that hunt_mat gave was the principal value. If you type i^i in google, you will see that they also return the principal value.

I'm not saying that anybody did anything wrong here. But I just wanted to tell the OP that there are multiple values of ab, but that we often restrict these multiple values to get a principal value.

6. Jul 14, 2011

### hunt_mat

Re: Solving "I" to the "I" power

$$i^{i}=e^{i\left(\frac{i\pi}{2}+2n\pi i \right)}$$