Math Motives

I have read similar claims on various websites explaining why \(i^i\) is a real number, or how to find \(\sqrt{i}\). So I decided to check whether AI can get it correctly.

Q: Is \(i^i\) a real number?

Gemini: Yes, the value of \(i^i\) is indeed a real number. It’s one of those beautiful mathematical results that feels like a magic trick because you are starting with nothing but imaginary units and ending up with a tangible, real decimal.

Mathematical Proof. To understand why, we use Euler's Formula: \(e^{ix} = \cos(x) + i\sin(x)\).

Step 1. Represent \(i\) in exponential form. On the complex plane, the number \(i\) is located at an angle of 90° (or \(\frac{\pi}{2}\) radians) from the positive real axis with a magnitude of 1. Therefore: \(i = e^{i\frac{\pi}{2}}.\)

Step 2. Raise \(i\) to the power of \(i\). Substitute that exponential form into the expression \(i^i\): \( i^i = (e^{i\frac{\pi}{2}})^i. \)

Step 3. Apply exponent rules: When you raise a power to a power, you multiply the exponents: \(i^i = e^{i \cdot i \cdot \frac{\pi}{2}} = e^{-\frac{\pi}{2}}.\)

Conclusion. The expression \(e^{-\frac{\pi}{2}}\) contains only real constants (\(e\) and \(\pi\)). There is no \(i\) left in the final exponent, making the result entirely real.


A Note on Multi-valuedness. Technically, because the complex plane is circular, you can represent \(i\) using multiple angles (\(\frac{\pi}{2}\), \(\frac{5\pi}{2}\), \(-\frac{3\pi}{2}\), etc.). This means \(i^i\) actually has an infinite number of possible values, but every single one of them is a real number. For any integer \(k\): \(i^i = e^{-(\frac{\pi}{2} + 2k\pi)}.\)

Proving \(i^i\) is real: if we apply the same logic to \(e^{i2\pi} =1\), we would have
\[1 = 1^{1/2} = (e^{i2\pi})^{1/2} = e^{i\pi} = -1.\]
This is absurd.

The mistake happens at the very moment when raising the question: \(a^r\) is a well defined number when either \(a>0\) or \(r \in \mathbb{Z}\), but not simultaneously. Power rules are not applicable to expressions like \(i^i\).

The multivalue justification: Let \(k\) be an integer. Since \(i = e^{i(2k + \frac{1}{2})\pi}\), raising to the power \(i\) would lead to
\[i^i = (e^{i(2k + \frac{1}{2})\pi})^i = e^{-(2k + \frac{1}{2})\pi}.\]
For different \(k\)'s we would get different values that approach \(0\) on one side, and approach \(\infty\) on the other side. We need to treat \(i^i\) as a set of values: \(i^i = \{e^{-(2k + \frac{1}{2})\pi}: k\in \mathbb{Z}\} =e^{(2 \mathbb{Z} - \frac{1}{2})\pi} \) -- all of these values are real. In this sence, one can say \(i^i\) is real, just not a real number.