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.
The theorem says i**i=1 and the proof states that i**i=2.0787…
Yes, changed it already.