I have a question as follows. Let
$$f(z)=\frac{\sin z}{(z-1-i)^2}$$
and
$$a_n=\frac{f^{(n)}(0)}{n!}$$
Determine the radius of convergence of
$$\sum_{n=0}^{\infty}a_nz^n$$
In my class we have talked extensively about Cauchy's Integral Formula as well as power series, and we've often used
$$\frac{1}{R}=\limsup_{n \to 0}\left\lvert \frac{a_{n+1}}{a_n} \right\rvert$$
to determine radius of convergence. The main issue I'm having is determining a closed form for $f^{(n)}(0)$. I have taken multiple derivatives of $f(z)$ and have not noticed any sort of pattern. I suspect I'm supposed to utilize Cauchy's Integral Formula in some way in order to do this (I could be wrong); however, I'm unsure on how to do so. Any help would be appreciated. Thanks!
$f$ has a singularity at $z=1+i$.
The radius of convergence of the Taylor series of $f$ around the origin is the distance from the origin to the singularity, $\sqrt 2$ in this case.
The reason is simple:
If $R < \sqrt2$, then $f$ is holomorphic in the disk of radius $R$ centered at the origin.
If $R \ge \sqrt2$, then $f$ is not holomorphic in the disk of radius $R$ centered at the origin because it is not even defined at the singularity.