What Is i to the Power i?
Let $i = \sqrt{-1}$. Evaluate $i^{\,i}$.
Show that the (principal) value is a real number, give it in closed form, and comment on whether the value is unique.
Hints
- Write $i$ in exponential form. Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ (which follows from the Taylor series of $e^x$, $\cos x$, $\sin x$) gives $i = e^{i\pi/2}$.
- Then $i^{i} = \left(e^{i\pi/2}\right)^{i} = e^{i \cdot i\pi/2}$. Simplify $i \cdot i$.
- Because $e^{i\theta}$ is $2\pi$-periodic, $i = e^{i(\pi/2 + 2\pi k)}$ for every integer $k$; each choice of $k$ gives a different real value. The principal value uses $k = 0$.
Worked Solution
How to Think About It: Complex powers are defined through the exponential and logarithm: $a^{b} = e^{b \ln a}$. So the whole problem is to compute $\ln i$, which Euler's formula gives as $i\pi/2$ (plus multiples of $2\pi i$).
Quick Estimate: $e^{-\pi/2} = e^{-1.571}$. Since $e^{-1.5} \approx 0.223$ and $e^{-0.07} \approx 0.93$, the value is about $0.21$.
Formal Solution:
*Step 1 -- Euler's formula from Taylor series.* Using $e^{x} = \sum x^n/n!$ with $x = i\theta$ and separating even and odd powers of $i$, $$e^{i\theta} = \sum_{n \text{ even}} \frac{(-1)^{n/2}\theta^{n}}{n!} + i\sum_{n \text{ odd}} \frac{(-1)^{(n-1)/2}\theta^{n}}{n!} = \cos\theta + i\sin\theta.$$
*Step 2 -- Polar form of $i$.* Taking $\theta = \pi/2$ gives $e^{i\pi/2} = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = i$. Hence the principal logarithm is $\ln i = i\pi/2$.
*Step 3 -- Exponentiate.* $$i^{\,i} = e^{\,i \ln i} = e^{\,i \cdot i\pi/2} = e^{-\pi/2} \approx 0.2079.$$
*Step 4 -- Non-uniqueness.* Since $e^{i\theta}$ has period $2\pi$, also $i = e^{i(\pi/2 + 2\pi k)}$ for any integer $k$, so $i^{i}$ can equal $e^{-\pi/2 - 2\pi k}$ for any $k \in \mathbb{Z}$: all real, all positive. The principal branch ($k = 0$, argument in $(-\pi, \pi]$) gives $e^{-\pi/2}$.
Answer: $i^{\,i} = e^{-\pi/2} \approx 0.208$ (principal value); in general $e^{-\pi/2 - 2\pi k}$, $k \in \mathbb{Z}$, all of which are real.
Intuition
Euler's formula turns exponentiation of complex numbers into arithmetic on exponents, and $i^i$ is the famous case where an imaginary base and exponent conspire to produce the real number $e^{-\pi/2} \approx 0.208$. The same Taylor-series machinery is what makes characteristic functions $E[e^{itX}]$ work: complex exponentials encode rotations, so multiplying exponents replaces convolving distributions.