Prove that $\overline{\ln z}=\ln\overline z$ 
I was wondering if anyone could help to show that $$\overline{\ln z}=\ln\overline z.$$

My attempt:
Recall that $z=x+iy$, thus $\overline{\ln z}=\overline{\ln(x+iy)}=\ln\overline{x+iy}=\ln(x-iy)=\ln\overline z$ as required. I was just wondering if anyone could confirm if this is correct, and if not give me a hint on how to proceed.
 A: Your reasoning is circular and thus incorrect; you assume what you're trying to prove across the second equals sign.
Instead, use the polar representation $z=re^{i\theta}$, assuming principal value is used for $\theta$:
$$\overline{\ln z}=\overline{\ln r+i\theta}=\ln r-i\theta$$
$$\ln\overline z=\ln re^{-i\theta}=\ln r-i\theta$$
A: No.  Rewriting $z$ as $x+iy$ has nothing to do with whether you can pull conjugation inside the $\ln$.
This is somewhat complicated because $\ln$ is not uniquely defined (unless you do something like barto's comment and take a particular branch of the function).  However, you can prove something like "if $w$ is a logarithm of $z$, then $\overline w$ is a logarithm of $\overline z$" by applying the definition of logarithm as "inverse of exponential" and working from there.
A: Note that $$\text{Log}(x+iy)=\ln\sqrt{x^2+y^2}+i\text{Arg}(x+iy)$$ so $$\overline{\text{Log}(x+iy)}=\ln\sqrt{x^2+y^2}-i\text{Arg}(x+iy)$$ $$\text{Log}(x-iy)=\ln\sqrt{x^2+(-y)^2}+i\text{Arg}(x-iy)=\ln\sqrt{x^2+y^2}-i\text{Arg}(x+iy)$$ Hence $$\overline{\text{Log}(z)}=\text{Log}(\overline z)$$
A: The complex logarithm has infinitely many values because the argument of a complex number is only determined up to multiplication by an integral multiple of $2{\pi}$. So for $z = x+iy$, by the definition of the complex logarithm which is, $${\rm ln}(z)= {\rm ln}|z|+i({\rm arg}(z)+2n{\pi})$$ where $n$ runs through the set ${\bf Z}=\{0,{\pm{1}},{\pm{2}},\cdots\},$ since ${\rm arg}(\bar{z}) = -{\rm arg}(z),$ if we apply the definition of ${\rm ln}(z)$ to work out ${\rm ln}({\bar z}),$ we get $${\rm ln}({\bar{z}})= {\rm ln}|{\bar{z}}|+i({\rm arg}({\bar{z}})+2n{\pi})={\rm ln}|z|-i({\rm arg}(z)+2n{\pi})={\overline{{\rm ln}(z)}}$$ since $|\bar{z}|=|z|.$ The above identity should be understood as an identity between two infinite sets indexed by the integers $n$. So the assertion ${\rm ln}(z) = {\rm ln}({\bar z})$ is a statement that two (countably) infinite sets contain precisely the same elements.
