I am trying to find the domain of analyticity for $\text{Log}(f(z))$ where $$f(z) = \frac{1+iz}{1-iz}$$

My attempt: the domain of analyticity is the set of points $z\in \mathbb{C}$ where $f(z)$ is defined and $f(z)$ is not in the set $\{z\in \mathbb{C} : \Re(z)\leq 0\}$.

So clearly $z\neq -i$ from the first condition. For the second condition we want $f(z)$ not to be on the negative real axis.

$$f(z) = f(x+iy) = \frac{(1-y)^2-x^2+2ix(1-y)}{(1+y)^2+x^2}$$

When $y = 1$, $f(x+iy)$ lies on the real axis, so we the points we want to exlcude are where $-x^2\leq 0$. So we exclude all $z = x+i, x\in \mathbb{R}$.

Also if $x = 0$, $f(z)$ lies on the real axis, so we want to exclude points such that $(1-y)^2\leq 0$ which is only $y =1$ (but we have excluded this already).

Hence the domain of analyticity is given by $z = x+iy, y \neq 1$. Is this correct?


I think your computation is possibly off (have not checked how you got the expression), but you are doing the right thing otherwise. Let's try: $$ \frac{1+iz}{1-iz} = \frac{(1+iz)(1+i\bar{z})}{|1-iz|^2} = \frac{1-|z|^2 + i(z+\bar{z})}{|1-iz|^2} = \frac{1-|z|^2 + i2\Re z}{|1-iz|^2} $$ So $f(z)$ is on the negative real axis (or zero) if $|z|^2 \geq 1$ and $\Re z = 0$. Work it out from there.

The places where $f(z)$ is zero or where it has a pole are the two "endpoints" of the negative real axis. The function you are looking at, $f(z)$, is linear fractional, and therefore it is a nice transformation of the Riemann sphere to itself, so it is best understood there. The negative reals together with 0 and infinity is a nice closed "line segment" on the Riemann sphere and the set that you are looking for is the preimage of this line segment under $f$. As $f$ is linear fractional it will be another such "line segment" on the Riemann sphere. But note that in this case it will "go through infinity". You should also check that $f$ "takes" $\infty$ to $-1$, to see this replace $z$ with $\frac{1}{z}$ and plug in zero.


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