A Mathematical Proof that i^i is a Real Number

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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.

 

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