If you have been introduced to Complex Numbers, then you know that $latex i = \sqrt{-1}$. Operations on $latex i$ give both real and complex results. For instance, $latex (i)(i) = -1$ and $latex i + i = 2i$. But one of the surprising results is the value of $latex i^i$.

In this post, we are going to show that $latex i^i$ is a real number. The proof is credited to Nick Benallo’s blog MathyNick. Nick has permitted me to include this beautiful proof in The Book.

**Theorem**: $latex i^i$ is a real number.

**Proof**

In the proof, we are going to use the Euler’s Formula. Using the Euler’s Formula,

$latex e^{i \pi} = \cos \pi + i \sin \pi$.

$latex e^{i \pi} = -1 + 0 i$

$latex e^{i \pi} = -1$