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,

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

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

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

Now, since

$latex -1 =e^{i \pi}$, raising both sides by $latex \frac{1}{2}$ gives us

$latex (-1)^{\frac{1}{2}} = e^{(i \pi) \frac{1}{2}}$

Since, $latex (-1)^{\frac{1}{2}} = \sqrt{-1}$,

$latex \sqrt{-1} = e^{\frac{\pi}{2}i}$, we have

$latex \sqrt{-1} = e^{\frac{\pi}{2}i}$

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

$latex i^i = (e^{\frac{\pi}{2}i})^i$

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

$latex i^i = e^{\frac{\pi}{2}(-1)}$

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

$latex i^i =0.20787957…$

This shows that $latex i^i$ is a real number.