In real algebra, $a^c = e^{c\ln a}$ for $a > 0$. The same recipe defines powers of complex numbers, with the complex logarithm in place of $\ln$. Because $\log z$ has many values, so does $z^c$, and the familiar exponent rules need care.
Key idea
For $z \ne 0$ and any complex $c$, $\;z^c = e^{c\log z}$, one value for each value of $\log z$. The principal value is $e^{c\operatorname{Log} z} = |z|^ce^{ic\operatorname{Arg} z}$ (for real $c$).
• $c$ an integer: one value (the usual power).
• $c = \frac{p}{q}$ in lowest terms: exactly $q$ values (the $q$th roots of $z^p$).
• $c$ irrational or not real: infinitely many values. For example, $i^i = e^{i \cdot i(\frac{\pi}{2} + 2\pi k)} = e^{-\frac{\pi}{2} - 2\pi k}$, all of them real.
• $c$ an integer: one value (the usual power).
• $c = \frac{p}{q}$ in lowest terms: exactly $q$ values (the $q$th roots of $z^p$).
• $c$ irrational or not real: infinitely many values. For example, $i^i = e^{i \cdot i(\frac{\pi}{2} + 2\pi k)} = e^{-\frac{\pi}{2} - 2\pi k}$, all of them real.
Why it works
The values of $\log z$ differ by $2\pi ik$, so the values of $z^c$ differ by factors $e^{2\pi ick}$. If $c$ is an integer, these factors are all $1$. If $c = \frac{p}{q}$, they are the $q$th roots of unity $e^{2\pi ipk/q}$, which repeat after $q$ steps. Otherwise $e^{2\pi ick}$ never repeats, giving infinitely many values. For $i^i$: $\log i = i\left(\frac{\pi}{2} + 2\pi k\right)$, and multiplying by $i$ gives the real number $-\frac{\pi}{2} - 2\pi k$.
Worked examples
Worked example: The values of $i^i$
Find all values of $i^i$ and its principal value.
- $\log i = \ln 1 + i\left(\frac{\pi}{2} + 2\pi k\right) = i\left(\frac{\pi}{2} + 2\pi k\right)$.
- $i^i = e^{i\log i} = e^{-\frac{\pi}{2} - 2\pi k}$.
- Principal value ($k = 0$): $e^{-\pi/2} \approx 0.2079$.
Answer: $e^{-\pi/2 - 2\pi k}$; principal $e^{-\pi/2}$
Worked example: A principal cube root
Find the principal value of $(-8)^{1/3}$.
- $|-8| = 8$ and $\operatorname{Arg}(-8) = \pi$.
- $(-8)^{1/3} = 8^{1/3}e^{i\pi/3} = 2\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = 1 + \sqrt{3}\,i$.
- The real cube root $-2$ is one of the three values (with $k = 1$), but it is not the principal one.
Answer: $1 + \sqrt{3}\,i$
Worked example: A rational power
Find the principal value of $(-4)^{3/2}$.
- $4^{3/2} = 8$, and $\frac{3}{2}\operatorname{Arg}(-4) = \frac{3\pi}{2}$.
- $8e^{3\pi i/2} = 8(-i) = -8i$.
Answer: $-8i$
- Find $|z|$ and $\operatorname{Arg} z$ (or, for all values, $\operatorname{Arg} z + 2\pi k$).
- Form $c\log z = c\ln|z| + ic(\operatorname{Arg} z + 2\pi k)$.
- Exponentiate: $z^c = e^{c\log z}$. For real $c$ this is $|z|^ce^{ic(\operatorname{Arg} z + 2\pi k)}$.
- Count the distinct values: one for integer $c$, $q$ for $c = \frac{p}{q}$ in lowest terms, infinitely many otherwise.
The principal power $z^c$ is analytic wherever $\operatorname{Log} z$ is (off the negative real axis), with derivative $cz^{c - 1}$ using the same branch.
Common mistake
Using real exponent rules on principal values. Rules like $(zw)^c = z^cw^c$ and $(z^a)^b = z^{ab}$ can fail. With principal values, $\big((-1)(-1)\big)^{1/2} = 1^{1/2} = 1$, but $(-1)^{1/2}(-1)^{1/2} = i \cdot i = -1$. And $\big((-1)^2\big)^{1/2} = 1$ while $(-1)^{2 \cdot \frac{1}{2}} = -1$. Work from the definition $z^c = e^{c\operatorname{Log} z}$ each time, and check your answer by raising it back to the needed power.