Is complex conjugate distributive over exponentiation

Question

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?

Answer 1

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 $-2^{-i}=e^{-\pi}e^{-i\mathrm{ln}2}$

Answer 2

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 $$ (\overline{-1})^{\overline{1/2}} = (-1)^{1/2} \ne \overline{(-1)^{1/2}}$$