Holomorphic domain of $ \Phi\left(z\right)=\ln\left(\frac{z-a}{z+a}\right)$

Question

I have to find a domain $D$ of the complex plane where the function ( $a \in \mathbb{R}^{*+}$ ) defined by $$ \Phi\left(z\right)=\ln\left(\frac{z-a}{z+a}\right) $$ is holomorphic. Hence i search for the set of points where it is not holomorphic.

I guess the point $z=-a$ is one of them and that $$ \frac{z-a}{z+a} \in \mathbb{R}^{-} $$ I've chosen $z=x+iy$ and i've found that $y$ must be $0$ ( dont know if i'm correct ) and that $x$ must be $$ \frac{x+a}{x-a}=K \in \mathbb{R}^{*-}$$ What should I do ?

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Answer 1

For the nature of the logarithm, after choosing an angle for the $Arg$ function, you have discontinuity on the negative line $\{x+iy:x\leq 0\}$.

We can look for such values: for $z$ complex $$ \frac{z-a}{z+a} = \frac{(x-a)+iy}{(x+a)+iy} \frac{(x+a)-iy}{(x+a)-iy} = \frac{(x-a)(x+a)-iy(x-a)+iy(x+a)+y^2}{(x+a)^2+y^2} = \frac{x^2+y^2-a + i2ya}{(x+a)^2+y^2} \in \mathbb{R}^-_0 \quad \iff \quad \begin{cases} x^2+y^2-a^2 \leq 0 \\ 2ya = 0, \\ \end{cases} $$ with the further condition that in $(x,y)= (a,0)$ the function is not even defined, as given by the denominator.

At this point we have three cases to analyze:

1. $a=0$, the second equation is automatically satisfies, therefore we get $x^2+y^2-a^2 = x^2+y^2 \leq 0$ if and only if $x+iy=0$, that is, the function is not analytic in the origin.
2. $a<0$, then the second equation gives us $y=0$, so that we need $x^2+y^2-a = x^2-a\leq 0$ if and only if $x^2\leq a < 0$, impossible, which means that the function is everywhere holomorphic but in $(a,0)$ as discussed before.
3. $a>0$, then $y=0$ and $x^2+y^2-a =x^2-a\leq 0$ if and only if $|x|\leq \sqrt{a}$, that is, $\phi$ is holomorphic everywhere but in $\{x+iy:-\sqrt{a}\leq x\leq \sqrt{a}\} \cup \{(a,0)\}$.