I want to understand the function: $f(z) = \frac{1}{\sqrt{z^2+11}} dz$.
This functions seems to have two branch points $\sqrt{11}i$ and $-\sqrt{11}i$. Does it also have a brunch at $\infty$? Usually $\infty$ is a branch point of $Log(z)$ and I don't think it cancels here, but I am not sure.
Another thing, it seems $\sqrt{11}i$ and $-\sqrt{11}i$ are also poles of $f(z)$. So are those points both poles and branch points? How should I consider them?
Also something that I would like to understand is: since we define $\sqrt{z^2+11} = exp(\frac{1}{2} Log(z^2+11))$ and the exponent never reaches $0$, are those points really a singularity?
After all this, where is the function analytic? Is $f(z)$ analytic in $\mathbb{C} \setminus [-\sqrt{11}i,\sqrt{11}i]$? What about the branch point of $\infty$?
Can I prove that the function is analytic in $\{|z| > \sqrt{11} \}$?
Finally, is how do I expand $f(z)$ to a power series (Taylor or Laurent)?
Help would be appreciated.
You can choose a branch so that $\infty$ is a removable singularity, the branch cut running, say, on the imaginary axis from $-\sqrt{11} i$ to $+\sqrt{11} i$. Namely, write your function as $$ \frac{1}{z \sqrt{1 + 11/z^2}}$$ with the principal branch of the square root. Note that $1 + 11/z^2 \in (-\infty, 0]$ if and only if $z$ is on that branch cut.