# A Mathematical Proof that i^i is a Real Number

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

In this post, we are going to show that $i^i$ 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: $i^i$ is a real number.

Proof

In the proof, we are going to use the Euler’s Formula. Using the Euler’s Formula, $e^{i \pi} = \cos \pi + i \sin \pi$. $e^{i \pi} = -1 + 0 i$ $e^{i \pi} = -1$

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

Since, $(-1)^{\frac{1}{2}} = \sqrt{-1}$, $\sqrt{-1} = e^{\frac{\pi}{2}i}$, we have $\sqrt{-1} = e^{\frac{\pi}{2}i}$ $i = e^{\frac{\pi}{2}i}$ $i^i = (e^{\frac{\pi}{2}i})^i$ $i^i = e^{\frac{\pi}{2}i^2}$ $i^i = e^{\frac{\pi}{2}(-1)}$ $i^i = e^{- \frac{\pi}{2}}$ $i^i =0.20787957...$

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

## 2 thoughts on “A Mathematical Proof that i^i is a Real Number”

1. Vinícius on said:

The theorem says i**i=1 and the proof states that i**i=2.0787…

• Guillermo Bautista on said: