# Is the power of complex number defined yet?

Let $z$, and $b$ be two complex numbers. What is $$f_b(z)=z^b.$$

If I write it like this:
$$\left(re^{i\theta}\right)^{b}=r^{b}e^{ib\theta}.$$

Would this even make sense?

Wolframalpha gives me $(-i)^i=e^{\pi/2}$ using the formula above. How to calculate $f_b(z)$?

#### Solutions Collecting From Web of "Is the power of complex number defined yet?"

It makes sense, but it is ambiguous unless you’re more careful.

On the one hand, we can say that $i = e^{\pi i/2}$, so that
$$i^i = (e^{\pi i/2})^i = e^{- \pi /2}$$
On the other hand, we can equally claim that $i = e^{(2 \pi + \pi / 2)i }$, so that
$$i^i = (e^{(2 \pi + \pi / 2)i})^i = e^{-5 \pi/2}$$
That is, your $f_b$ will generally, at its heart, be a “multi-valued” function. Exactly which number you assign to $z^b$ depends on your definition whenever $b$ is not a real number.

The “definition” of $z^b$ for $z=r\mathrm e^{\mathrm i\theta}$ as $r^{b}\mathrm e^{\mathrm ib\theta}$ actually does not make sense when $r\ne0$ and $b$ is not an integer because $z$ is also $z=r\mathrm e^{\mathrm i\theta+2\mathrm i\pi}$, say, hence $z^b$ should also be $r^{b}\mathrm e^{\mathrm ib\theta+2b\mathrm i\pi}$, which is not equal to $r^{b}\mathrm e^{\mathrm ib\theta}$ when $r\ne0$ and $b$ not an integer.

Once you define a complex log; $\log z$ , then $z^b:=e^{b \log z}$ , wherever $\log z$ is defined.

Here $\log z$ is a local inverse of $e^z:=\exp(z)$.

The complex power of a complex number is commonly defined using logarithms as $$x^y=e^{y\ln x},$$ where some branch of the logarithm function is used.

(Recall that $\ln z=ln| z|+i\arg z=\ln r+i\theta$, and $e^z=e^{\Re_z}(\cos\Im_z+i\sin\Im_z)$).

So,
$$(re^{i\theta})^y=e^{\Re_y\ln r-\Im y\theta}[\cos(\Im_y\ln r+\Re_y\theta)+i\sin(\Im_y\ln r+\Re_y\theta)].$$