complex-analysis
The question states: Where is the function $(z^2+4)^{-1}\text{Log}(z+2i)$ analytic (where Log denotes the principal complex logarithm).
My attempt:
First note that if $z=\pm 2i$, then $(z^2+4)^{-1}$ is not analytic.
Also, we know that Log$(g(z))$ is not analytic if $g(z)\in(-\infty,0]$. So $\text{Log}(z+2i)$ is not analytic iff $$z+2i\in(-\infty,0].$$
But I'm unsure of how to simplify this expression. Do we equate real and imaginary components to determine a condition for $x$ and $y$?
If $2xy=0$ then either (a) $x=0$, in which case the other inequality becomes $-y^2-1\leq 0$ which is satisfied by all $y\in\mathbb{R}$, or (b) $y=0$, where the other inequality becomes $x^2 - 1 \leq 0$ which is satisfied by all $|x| \leq 1$.
These inequalities must both be satisfied together. You are describing the union of the sets they are satisfied on individually, where what you really want is the intersection.
The way we define the Logarithm is the inverse of the exponential.
If $z= re^{i\theta}$,then $ln(z)=ln(r)+i\theta$.
Now , a problem that is there is, with this equation alone , logarithm is multi-valued i.e for every z there are infinitely many values for the logarithm . For $\theta$ differing by a value of $2\pi$ , one will get the same value. It isn't injective. This leads us restricting its domain so that each point(z) corresponds to only one value. This is referred to as a branch cut. Check the image from Wiki,
So, for log(z), simply remove any ray joining 0 and infinity.(Restricting complete rotation(0 to $2\pi$!). Also, see in the graph what happens. The principle log removes one particular ray, the negative real axis, I guess. This is only a convention, I guess.
Now, coming back to your question, you want log(iz)(PS:-The $iz$ only changes the labels on the axes) to be analytic in some given region. All you need to do is to remove a ray(not in the given region) to make it analytic in the given region.
Best Answer
$$ \begin{align} &z+2i\in(-\infty,0] \\ &\implies\Re(z+2i)\in(-\infty,0] \text{ and } \Im(z+2i)=0 \\ &\implies\Re(z)\in(-\infty,0] \text{ and } \Im(z)=-2\\ \end{align} $$
Therefore, $\frac{\operatorname{Log}(z+2i)}{z^2+4}$ is analytic on $$\mathbb C\setminus\{2i\}\setminus\{z~:~ \Re(z)\in(-\infty,0] , \Im(z)=-2\} $$