The function $f(z) :=\frac{ z−1}{z+ 1}$ maps the domain $\mathbb{C}\setminus[−1,1]$ into $\Bbb{C}\setminus(−\infty,0]$. Using this fact, re-express the function $$g(z) =\int_{-1}^1\frac{dt}{t−z}$$ $\forall z\in \Bbb{C}\setminus [−1,1]$ as a formula involving the principal holomorphic branch of the logarithm.
I am trying to show that the denominator is missing $(-\infty,0]$. But that is not the case. But if we just think that branch of log exists, then we can see the function $g(z)=\log\circ f(z). \forall z\in \Bbb{C}\setminus (-\infty,0] $
Best Answer
Let $\operatorname{Log}$ denote the principal branch of the logarithm on $\Bbb{C}\setminus(−\infty,0]$, and $D = \Bbb{C}\setminus[−1,1]$. Note that $1/f(z) = (z+1)/(z-1)$ maps $D$ onto $\Bbb{C}\setminus(−\infty,0]$ as well.
Then $G(z) = \operatorname{Log}\left(\frac{z+1}{z-1}\right)$ is holomorphic in $D$ with $$ G'(z) = \frac{1}{z+1} - \frac{1}{z-1} = \int_{-1}^1 \frac{dt}{(t-z)^2} = g'(z) $$ and therefore $$ g(z) = \operatorname{Log}\left(\frac{z+1}{z-1}\right) + C $$ for some constant $C$. Considering the limit $z \to \infty$ shows that $C=0$.