Why is $\frac1{2i}\ln \frac{i-z}{i+z}=\frac i{2}\ln \frac{i+z}{i-z}$ possible with complex logarithm?

Question

I stumbled upon something today that I couldn't calculate myself. On a problem set, I managed to get the answer on the left side of this equation, but for some reason it could be simplified even further.

How do I simplifiy this?

$$\frac1{2i}\ln \frac{i-z}{i+z}=\frac i{2}\ln \frac{i+z}{i-z}$$

Answer 1

Recall that

$$\frac 1i=\frac1i\frac ii =-i$$

and $$-\log A=\log \frac1A$$