Find where $f(z)=\mathrm{Log}\left(\frac{z-\alpha}{z-\beta}\right)$ is defined and continuous Let $\alpha,\beta\in\mathbb C$ and let $f(z):\mathbb C \to \mathbb C$ such that:
$$f(z)=\mathrm{Log}\left(\frac{z-\alpha}{z-\beta}\right)$$
We know that the principal branch of $\log(w)$ (denoted as $\mathrm{Log}(w)$) is defined and continuous for every $w\in\mathbb C\setminus A$ when:
$$A=\left\{w\in\mathbb C\mid \Im(w)=0, \Re(w)\leq 0 \right\}$$
My goal is to find where $f$ is defined and continuous. For that, I subtituted $w=\frac{z-\alpha}{z-\beta}$ and tried to calculate the real and imaginary parts of $w$, but it turned out to be an algebraic nightmare. I managed to make things simpler, but the condition I got is an equation which is dependent on $\Re(z),\Im(z),\Re(\alpha),\Im(\alpha),\Re(\beta)$ and $\Im(\beta)$, which makes it harder to define the domain properly.
I would be glad to see your ways of finding the domain of the above function $f$; They must be better than mine.
Thanks!
 A: Two options:


*

*Shift the variable to make things easier, then revert at the end: put $u=z+\beta$. Then
$$ f(u+\beta) = \operatorname{Log}\bigg( \frac{u+\beta-\alpha}{u}\bigg) = \operatorname{Log}\bigg( 1 + \frac{\beta-\alpha}{u}\bigg) $$
and now all you have to do is find when $1+(\beta-\alpha)/u$ is nonpositive.

*We know the set where $\operatorname{Log}(w)$ is defined, with $w=g(z)$, so if we can find the inverse image of this set under $g$, we find out where $\operatorname{Log}(g(z))$ is defined. That is, solve
$$ w = \frac{z-\alpha}{z-\beta} $$
for $z$, and find out where the set of $w \in A$ goes.
Either way, it comes out to be the complement of the line segment joining $\beta$ to $\alpha$ (including the endpoints).
A: Since $\mathrm{Log}$ is discontinuous (and not holomorph) on $D=(-\infty,0]\times \{0\}$, function $f$ is discontinuous (and not holomorph) on $z$ iff 
$$\frac{z-\alpha}{z-\beta}\in D=(-\infty,0]\times \{0\},$$
that is:
$$\mathrm{Re}\left(\frac{z-\alpha}{z-\beta}\right)\leqslant 0$$
and $$\mathrm{Im}\left(\frac{z-\alpha}{z-\beta}\right)= 0.$$
Since $$\frac{z-\alpha}{z-\beta}=\frac{(z-\alpha)(\overline{z}-\overline{\beta})}{|z-\beta|^2}=\frac{z\overline{z}-z\overline{\beta}-\alpha \overline{z}+\alpha \overline{\beta}}{|z-\beta|^2},$$
we get:
$$\mathrm{Re}\left(\frac{z-\alpha}{z-\beta}\right)=\frac{|z|^2-\mathrm{Re}(z\overline{\beta})-\mathrm{Re}(\alpha \overline{z})+\mathrm{Re}(\alpha \overline{\beta})}{|z-\beta|^2},$$
$$\mathrm{Im}\left(\frac{z-\alpha}{z-\beta}\right)=\frac{-\mathrm{Im}(z\overline{\beta})-\mathrm{Im}(\alpha \overline{z})+\mathrm{Im}(\alpha \overline{\beta})}{|z-\beta|^2}.$$
So 
$$\mathrm{Re}\left(\frac{z-\alpha}{z-\beta}\right)\leqslant 0\iff |z|^2-\mathrm{Re}(z\overline{\beta})-\mathrm{Re}(\alpha \overline{z})+\mathrm{Re}(\alpha \overline{\beta})\leqslant 0$$
and:
$$\mathrm{Im}\left(\frac{z-\alpha}{z-\beta}\right)=0\iff -\mathrm{Im}(z\overline{\beta})-\mathrm{Im}(\alpha \overline{z})+\mathrm{Im}(\alpha \overline{\beta})=0.$$
Setting $$A=\{z:-\mathrm{Im}(z\overline{\beta})-\mathrm{Im}(\alpha \overline{z})+\mathrm{Im}(\alpha \overline{\beta})=0 \textrm{  and  } |z|^2-\mathrm{Re}(z\overline{\beta})-\mathrm{Re}(\alpha \overline{z})+\mathrm{Re}(\alpha \overline{\beta})\leqslant 0\},$$
we get that $f$ is continuous and holomorph on $\mathbb{C}\setminus A.$
