Complex Power Functions

# Complex Power Functions

Recall from The Complex Natural Logarithm Function page that if $z \in \mathbb{C} \setminus \{ 0 \}$ then we defined:

(1)
\begin{align} \quad \log (z) = \log \mid z \mid + i \arg (z) \end{align}

Where $\arg(z)$ is specified in some branch of the complex natural logarithm function.

With this definition, we are able to define complex power functions. Let $a, b \in \mathbb{C}$ with $a \neq 0$. We would like a meaningful definition for $a^b$. If $a \neq 0$ then we know that $a = e^{\log a}$ for some branch of the logarithm function. Defining $a^b$ follows naturally.

 Definition: If $a, b \in \mathbb{C}$ and $a \neq 0$ then the Complex Power Function $a^b$ is defined as $a^b = (e^{\log a})^b = e^{b \log a}$ with some branch for the complex logarithm function.

For example, suppose that we want to compute $i^i$. Then from our definition:

(2)
\begin{align} \quad i^i = e^{i \log (i)} = e^{i[\log \mid i \mid + i \arg(i)]} = e^{i \left [0 + \left ( \frac{\pi}{2} + 2k\pi \right )i \right ]} = e^0 \cdot e^{i^2 \left ( 2k + \frac{1}{2} \right )\pi} = e^{-\left ( 2k + \frac{1}{2} \right )\pi} \end{align}

In general, since the complex natural logarithm function is a multi-valued function, the power function $a^b$ may take on multiple values. The following theorem tells us when multiple values of $a^b$ occur.

 Theorem 1: Let $a, b \in \mathbb{C}$, $a \neq 0$. a) $a^b$ has a unique value if and only if $b \in \mathbb{Z}$. b) If $b \in \mathbb{Q}$, i.e., $\displaystyle{b = \frac{p}{q}}$ where $p, q \in \mathbb{Z}$ ($q \neq 0$), and $p$ and $q$ are in lowest terms, i.e., $\gcd (p, q) = 1$, then $a^b$ has $q$ distinct values which are the $q^{\mathrm{th}}$ roots of $a^p$. c) If $b \in \mathbb{R} \setminus \mathbb{Q}$ ($b$ is irrational) or if $b \in \mathbb{C} \setminus \mathbb{R}$ ($b$ is a nonreal complex number), then $a^b$ takes on infinitely many values.

As we will see in the proof below, the various values of $a^b$ differ only by factors of $e^{2kb\pi i}$ where $k \in \mathbb{Z}$ is determined by the choice of branch for the logarithm function.

• Proof) Consider the complex natural logarithm function $\log z$ with a chosen branch, say $[0, 2\pi)$. Then:
(3)
\begin{align} \quad a^b = e^{b \log a} = e^{b [\log \mid a \mid + i \mathrm{Arg} (a)]} = e^{b \log \mid a \mid} \cdot e^{i b \mathrm{Arg} (a)} \end{align}
• For any other branch of the complex natural logarithm function we would have for some $k \in \mathbb{Z}$:
(4)
\begin{align} \quad a^b = e^{b \log a} = e^{b [\log \mid a \mid + i (\mathrm{Arg} (a) + 2k\pi)]} = e^{b \log \mid a \mid} \cdot e^{ib[\mathrm{Arg} (a) + 2k\pi]} = e^{b \log \mid a \mid} \cdot e^{ib\mathrm{Arg} (a)} \cdot e^{2kb\pi i} \end{align}
• These values different only by the factor $e^{2kb \pi i}$ where $k \in \mathbb{Z}$ is determined by the choice of branch for the logarithm function. If this value remains the same as $k$ varies then $a^b$ has a single value. If this value differs as $k$ varies then $a^b$ will have multiple (possibly infinite) values. We cover these cases below:
• Proof of a) Suppose that $b \in \mathbb{Z}$. Then $kb \in \mathbb{Z}$. So $2kb \pi i$ is an integer multiple of $2\pi i$. But from one of the theorems on the Properties of the Complex Exponential Function page we know that $2kb \pi i = 1$ if and only if $kb \in \mathbb{Z}$ (and $kb \in \mathbb{Z}$ if and only if $b \in \mathbb{Z}$. Therefore $a^b$ has a unique value if and only if $b \in \mathbb{Z}$. $\blacksquare$
• Proof of b) If $b \in \mathbb{Q}$ and $\displaystyle{b = \frac{p}{q}}$ in lowest terms, then we know by the theorem on the nth Roots of Complex Numbers page that $a^p$ has $q$ many $q^{\mathrm{th}}$ roots, i.e., there exists $q$ values for $a^b = a^{p/q}$. $\blacksquare$
• Proof of c) Suppose that $b$ is irrational and that:
(5)
• But from the result mentioned in part (a), this happens if and only if $b(k - j) \in \mathbb{Z}$. Since $b$ is irrational, $b(k - j)$ is never an integer - a contradiction. Therefore the equality at $(*)$ never happens. In other words, $a^b$ takes on infinitely many values.
• Now suppose that $b$ is a nonreal complex number. Say $b = x + yi$ with $y \neq 0$. Then:
• $e^{-2ky \pi}$ takes on multiple values as $k$ varies (by choice of the branch for the logarithm function), so $a^b$ takes on infinitely many values.