# Why does $i^i$ not equal $e^{-\frac{5\pi}{2}}$?

It can be shown that $i^i = e^{-\frac{\pi}{2}}$:

\begin{align} i^i &= e^{i\ln i} \\ &= e^{i\ln\left(e^\frac{i\pi}{2}\right)} \\ &= e^{i\times\frac{i\pi}{2}} \\ &= e^{-\frac{\pi}{2}} \end{align}

However, it could just as easily been:

\begin{align} i^i &= e^{i\ln i} \\ &= e^{i\ln\left(e^\frac{5i\pi}{2}\right)} \\ &= e^{i\times\frac{5i\pi}{2}} \\ &= e^{-\frac{5\pi}{2}} \end{align}

since $i = e^\frac{i\pi}{2} = e^\frac{5i\pi}{2} \left(= e^\frac{\left(4n+1\right)i\pi}{2}\right)$.

This solution is obviously false, since $e^{-\frac{\pi}{2}} \neq e^{-\frac{5\pi}{2}}$.

Is there some other method of evaluating $i^i$ that eliminates this solution (and the infinite number of other similar solutions)?

• Exponential with complex base requires the choice of the branch of logarithm, so it is defined only up to such a choice. Moreover, none of those branches are a priori preferred. So depending on the choice, any of your answer can be made correct. – Sangchul Lee Sep 13 '17 at 8:14

The problem with complex numbers is that $e^{i\theta} = e^{i(\theta+2\pi k)}$ for all $k\in \mathbb Z$ (but obviously $\theta \neq \theta +2 \pi k$ for $k\neq 0$!) So you have to choose one branch of the logarithm to actually work with it. This means you have to choose one branch to assign $i^i$ some value.
The following image shows a few of the possible branches of the imaginary part of the logarithm. As you can see the imaginary part is only defined up to a multiple of $2\pi$. No. Complex exponentiation to any other base than $e$ is inherently multivalued. If the base is real and positive, there is one distinguished value (since the complex logarithm of a positive real number has one real value), but otherwise there is no consistent way to say that one result is objectively better / more correct / more useful than any other.