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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 $f_b(z)$?

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4 Answers 4

up vote 8 down vote accepted

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 decide to spit out depends on your definition.

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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.

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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 $e^z:=\exp(z)$.

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

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