Why is $\overline{z^{n}} = (\bar{z})^{n}$ true only for $n \in \mathbb{Z}$? What about a real or complex exponent in general?
Let $$\log z=\{\ln \rho + i (\theta + 2k\pi)\mid k\in\Bbb Z\}$$
then $$ \overline{\log z} = \overline{\ln \rho + i (\theta + 2k\pi)} = \ln \rho - i (\theta + 2k\pi) = \ln \rho + i (-\theta + 2k\pi) = \log \bar{z} $$ So, let $\alpha \in \mathbb{C}$, then $$ \overline{z^{\alpha}} = \overline{e^{\alpha \log z}} = e^{\bar{\alpha} \log \bar{z}} = (\bar{z})^{\bar{\alpha}} $$ It seems than the integer exponent is a particular case of a more general property. Did I make some mistake in my proof?
Edit: My question is twofold. Is the first equality about conjugate of the log correct? And if it is, is the second one about the conjugate of $z^\alpha$ correct? Maybe I am doing something wrong and I'll be glad if somebody could point out where are my mistakes. Thank you.
If we consider $z^\alpha$ to be a multivalued function - in particular "$z^\alpha$" does not represent a complex number, but rather a set of complex numbers, then, yes, $\overline {z^\alpha} = {\overline z}^{\overline \alpha}$ holds. But we have to be careful about what exactly we mean by it.
If $z = \rho e^{i\theta} = \rho e^{i(\theta + 2k\pi)}$ for all $k\in \Bbb Z$, and $\alpha = x + iy$, then $$z^\alpha = \{\rho^xe^{-y(\theta + 2k\pi)}e^{i(y\ln \rho + x(\theta + 2k\pi))} : k \in \Bbb Z\}$$ $$\overline{z^\alpha} = \{\rho^xe^{-y(\theta + 2k\pi)}e^{-i(y\ln \rho + x(\theta + 2k\pi))} : k \in \Bbb Z\}$$ while $${\overline z}^{\overline\alpha} = \{\rho^xe^{-(-y)(-\theta - 2k\pi)}e^{i((-y)\ln \rho + x(-\theta - 2k\pi))} : k \in \Bbb Z\}$$ Since $(-y)(-\theta - 2k\pi) = y(\theta + 2k\pi)$ and $i((-y)\ln \rho + x(-\theta - 2k\pi)) = -i(y\ln \rho + x(\theta + 2k\pi))$. The two sets are the same. Everything that is in one is in the other. I.e., $\overline{z^\alpha} = {\overline z}^{\overline\alpha}$ is actually a statement about set equality, not value equality.
But the reason most sources will tell you that this holds for general $z$ only when $\alpha \in \Bbb Z$ is that they are not defining $z^\alpha$ as a multi-valued function. Instead they are using a "branch-cut" definition, making it a single-valued expression. For each $z, \alpha$ you establish some means (the "branch-cut") of assigning a specific member of the set $\{\rho^xe^{-y(\theta + 2k\pi)}e^{i(y\ln \rho + x(\theta + 2k\pi))} : k \in \Bbb Z\}$ to be the value of $z^\alpha$. It can be proven that only when $\alpha \in \Bbb Z$ is it possible to establish a branch cut which gives the same value for $\overline{z^\alpha}$ and ${\overline z}^{\overline\alpha}$ for all $z$.