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)?

  • $\begingroup$ 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. $\endgroup$ – Sangchul Lee Sep 13 '17 at 8:14

Note that the logarithm has many branches, see e.g. here: https://en.wikipedia.org/wiki/Complex_logarithm

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.