Prove that $\overline{\ln(\frac{c}{\overline{z}}+d)}=\ln(\frac{c}{z}+d)$

Question

Prove that $\overline{\ln(\frac{c}{\overline{z}}+d)}=\ln(\frac{c}{z}+d)$

My attempt; first note that $z=re^{i\theta}$ thus $\frac{1}{\overline{z}}=z$ now substituting this into the lhs of what we are trying to prove we get the desired result, noting that since $c,d \in \Bbb R $ there signs do not change.

Here I'm talking about the principal branch of the logarithm. Could anyone tell me if this is correct ?

Answer 1

Simply note that $\overline{\ln z}=\ln\overline z$ then

$$\overline{\ln\left(\frac{c}{\overline{z}}+d\right)}=\ln\left(\overline{\frac{c}{\bar z}+d}\right)=\ln\left(\frac{c}{z}+d\right)$$