Is complex conjugate distributive over exponentiation The complex conjugate is distributive over addition, subtraction, multiplication and division:
$$ \overline{z+w} = \bar z + \bar w, $$
$$ \overline{z*w} = \bar z * \bar w, $$
etc.
Is it also distributive over exponentiation, i.e., is ($\overline{z^w} = \bar z ^ \bar w$ ?)
I was able to prove this for real $w$:
$$ z^w = (a*e^{i\theta})^w = a^w*e^{iw\theta} = a^w(\cos w\theta + i\sin w\theta) $$
Thus,
$$ \overline {z^w} = \overline {a^w(\cos w\theta + i\sin w\theta)} = a^w(\cos w\theta - i\sin w\theta) = a^w*e^{iw(-\theta)} = a^w*e^{(-iw\theta)} = a^w*(e^{-i\theta})^w. $$
Then, as $e^{\overline {i\phi}} = \overline {e^{i\phi}}$ for any real $\phi$,
$$ \overline {z^w} = a^w*{\overline {(e^{i\theta})}}^w = \overline {a^w * {e^{i\theta}}^w} = \overline {(a*e^{i\theta})}^w. $$
Therefore
$$ \overline {z^w} = (\bar z) ^ w $$ 
which for real $w$ equals $(\bar z) ^ {\bar w}$. However, I could not prove the theorem for complex $w$. 
Does the property still hold? If it does, how can it be proved?
 A: In general, $z^w$ is a multivalued function, unless you specify a particular branch.  By definition,
$$ z^w = \exp(w \log(z))$$
so
$$ \overline{z^w} = \overline{\exp(w \log(z)} = \exp(\overline{w}\; \overline{\log(z)})$$
Now $\overline{\log(z)}$ will be one of the branches of $\log(\overline{z})$, i.e. $$\exp(\overline{\log(z)}) = \overline{\exp(\log(z))} = \overline{z}$$
so that $\overline{z^w}$ is one of the branches of $\overline{z}^\overline{w}$,
but it will not necessarily be your favourite branch, whichever that is.
And it is impossible to choose a branch consistently that will make 
$\overline{z^w} = \overline{z}^{\overline{w}}$ always true.
For example, consider 
$ (-1)^{1/2}$.  Depending on which branch you choose, it is either $i$ or $-i$.  But  in either case
$$ (\overline{-1})^{\overline{1/2}} = (-1)^{1/2} \ne \overline{(-1)^{1/2}}$$
A: The answer is in the affirmative provided $\mathrm{Arg}(z) \neq -\pi$.  For $z,w \in \mathbb{C}$ we write $z^w = e^{w\mathrm{Log}(z)}$. Also, recall that:


*

*$\mathrm{Arg}(\overline{z})=-\mathrm{Arg}(z)$, for $\mathrm{Arg}(z)\neq-\pi$ and 

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


Now, \begin{equation}
\overline{z^w}=\overline{e^{w\mathrm{Log}(z)}}=e^{\overline{w(\mathrm{ln}|z|+i\mathrm{Arg}(z))}}=e^{\overline{w}(\mathrm{ln}|z|-i\mathrm{Arg}(z))}=e^{\overline{w}(\mathrm{ln}|\overline{z}|+i\mathrm{Arg}(\overline{z}))}=\overline{z}^{\overline{w}}
\end{equation}
However, if  $\mathrm{Arg}(z)= -\pi$, the above argument fails. Consider
$-2^i=e^{\pi}e^{i\mathrm{ln}2}$. Then $\overline{-2^i}=e^{\pi}e^{-i\mathrm{ln}2}$ but $-2^{-i}=e^{-\pi}e^{-i\mathrm{ln}2}$
A: Is there anything wrong with this approach? Start by proving that $\ln(\bar{z}) = \overline{\ln(z)}$. Setting $z = re^{i\theta}$:
\begin{align*}
\ln(\bar{z}) &= \ln(\overline{re^{i\theta}})\\
&= \ln(re^{-i\theta})\\
& = \ln(r) - i \theta \\
&= \overline{\ln(r) + i \theta} \\
&= \overline{\ln(re^{i\theta})} \\
&= \overline{\ln(z)}
\end{align*}
So taking the natural log of the conjugate of $z^w$:
\begin{align*} 
\ln(\overline{z^w}) &= \overline{\ln(z^w)} && \text{\# Using result above} \\
&= \overline{w\ln(z)}  && \text{\# Logarithm rules}\\
&= \bar{w}\overline{\ln(z)}  && \text{\# Conjugation distributes across complex multiplication}\\
&= \bar{w}\ln(\bar{z})  && \text{\#  Using result above}\\
&= \ln(\bar{z}^{\bar{w}})  && \text{\# Logarithm rules}
\end{align*}
And exponentiating both sides: $\overline{z^w} = \bar{z}^{\bar{w}}$
