Finding branch cut and choice of logarithm

Question

I've shown that $$\arctan(z)=\frac{1}{2i} \log\left(\frac{1+iz}{1−iz}\right)$$ when $z\ne \pm i.$. But then the question asks, "Which branch cut and choice of branch of the logarithm do you use so that $\arctan$ agrees with its usual definition on the real line?" I am having trouble answering this last part. Would it just be $(0,\pi/2+2\pi)$?

Answer 1

Let $$z = \frac {1 + i x} {1 - i x}.$$ If $x$ is real, then $|1 + i x| = |1 - i x|$, and we have $$w = \frac 1 {2i} \log z = \frac 1 {2i} \log |z| + \frac 1 2 \arg z = \frac 1 2 \arg z,$$ where the choice of $\arg$ isn't specified yet. On the other hand, $$z = \frac {1 - x^2} {1 + x^2} + i\frac {2 x} {1 + x^2}.$$ Consider how the point $z$ moves when $x$ goes from $-\infty$ to $\infty$. $z$ starts at $-1 - 0i$, goes counterclockwise around the unit circle and ends at $-1 + 0i$. Therefore, setting $\arg$ to change from $-\pi$ to $\pi$ makes $w$ coincide with $\arctan x$, and other choices of $\arg$ won't work.

On the negative real axis, we are free to set $\arg z$ to either $+\pi$ or $-\pi$.