Does complex conjugation commute with exponentiation? For integer exponents, complex conjugation commutes with exponentiation. Is the same true for an arbitrary real exponent? Does it depend on the choice of branch cut?
 A: First of all recall that for $z,w\in\mathbb{C}$, the most common definition for complex powers is
\begin{align}
z^w:=\exp{(w\log(z))}:=\exp{(w(\log(|z|)+\imath\mathrm{arg}(z)))},
\end{align}
where we take the principal branch of the logarithm that is $\mathrm{arg}(z)\in[-\pi,\pi)$. Note, this is a nice definition because this agrees with the real exponential function. So now, what can we say for $x\in\mathbb{R}$. Well, then we know that this is the same as saying
\begin{align}
z^x=|z|^xe^{\imath x\mathrm{arg}(z)}.
\end{align}
So from here we want to check: is $(\bar{z})^x=\overline{(z^x)}$? To answer this, recall the following facts about the complex conjugate:


*

*$|z|=|\bar{z}|$,

*$\mathrm{arg}(\bar{z})=-\mathrm{arg}(z)$, for $\mathrm{arg}(z)\neq-\pi$

*$e^{\bar{z}}=\overline{e^z}$.


So using our formula, we find that for $\mathrm{arg}(z)\neq-\pi$:
\begin{align}
(\bar{z})^x=|\bar{z}|^xe^{\imath x\mathrm{arg}(\bar{z})}=|z|^xe^{-\imath x\mathrm{arg}(z)}=\overline{(z^x)}.
\end{align}
When $\mathrm{arg}(z)=-\pi$, we have $\bar{z}=z$ so this is trivially true here, therefore it is true for all cases. This answer is of course dependent on the choice of branch of the logarithm. See if you can derive other results based on different choices for the logarithm (bear in mind that you want the function $z^w$ to be well defined!!).
