I do not know how to solve the following exercise:
Let $f(z) := (z-1-i)^{-2}\cdot sin(z)$ and let $c_n := \frac{1}{n!} f^{(n)}(0)$. Find the radius of convergence of $\sum_{n=0}^{\infty}c_nz^n$.
I tried to think about how $\sum_{n=0}^{\infty}c_nz^n$ looks, so I used the product rule to calculate the first derivative of $f$:
$$f^\prime(z) = -2\cdot (z-1-i)^{-3} \sin(z)+(z-1-i)^{-2}\cos(z)$$
This looks quite complicated to me and I suppose the higher derivatives will be even worse. I do not see how to proceed here. Could you help me?
The function $\sin z$ is entire and its zeros are first order. The function $(z-\alpha)^{-2}$ has a pole of second order at $\alpha$. Thus, $\forall \alpha \ne 0$, the product $f_\alpha(z)=(z-\alpha)^{-2}\cdot \sin z$ has a pole at $\alpha$ and has a convergent MacLauren series inside the circle $|z|=\alpha.$
Therefore, the radius of convergence of the series $\sum c_n z^n \sim f(z)=f_{1+i}(z)$ is $$R=|i+i|=\sqrt{2}.$$