# Equality between $\ln(1-iz)-\ln(1+iz)$ and $\ln\left(\frac{1-iz}{1+iz}\right)$

In this question, I only consider the principal determination of the logarithm, that is $\arg(z) \in (-\pi,\pi].$ Consider the two functions $f(z) = \ln(1-iz)-\ln(1+iz)$ and $g(z) = \ln\left(\frac{1-iz}{1+iz}\right)$ which are both holomorphic on $\mathbb{C}\backslash ((-\infty,-1]i\cup i [1,+\infty))$. I'm interested in the region of the complex plane where $f=g$.

This condition is satisfied if and only if $$\arg(1-iz)-\arg(1+iz) \in (-\pi,\pi].$$ If $-1<\Im(z)<1$, then the condition is obviously satisfied. But if I consider, for example the region where $\Re(z)>0, \Im(z)>1$, then $\arg(1-iz) \in (-\pi/2,0)$ and $\arg(1+iz) \in (\pi/2,\pi)$. Hence, in this region $$\arg(1-iz)-\arg(1+iz) \in (-3\pi/2,\pi/2).$$ Thus, there are some $z$ that are "good" (the equality $f=g$ does hold) and some that are "bad". The transition is made when $$\arg(1-iz)-\arg(1+iz) = -\pi.$$ So I try now to solve this equation to find a curve in the complex plane. After that I would be able to conclude with some topological argument. Using the definition of the principal argument and writing $z=x+iy,$ I got $$\arctan\left(\frac{-x}{1+y} \right)+\arctan\left(\frac{1-y}{x} \right) = -\frac{\pi}{2}.$$ And if I'm not mistaken, this equation has ... no solutions. I'm pretty sure I'm missing something with this exercice but I don't see what exactly.

Any help will be highly appreciated.

• what happens when the argument of a logarithm is 0? – danimal Nov 23 '16 at 12:04
• Are you sure that your last equation covers all cases? The $\arctan$ values are in $[-\pi/2, \pi/2]$ and $\arg$ is conveniently computed with the two-argument function $\mathrm{arctan2},$ see en.wikipedia.org/wiki/…. – gammatester Nov 23 '16 at 13:36
• If I'm not mistaken, for $z$ with $\Re(z)>0, \Im(z)>1$, we have $\arg(1-iz) = \arctan(\Re(1-iz)/\Im(1-iz))$ and $\arg(1+iz) =\pi-\arctan(\Im(1-iz)/\Re(1-iz))$. – C. Dubussy Nov 23 '16 at 19:18