$\displaystyle \frac{Log z}{sin(\pi z)}$ Find convergences radius at z = 1 and develop a Taylor series.

Just by looking at the function I see that deviser and numerator has zeros in z = 1, so theres a limit in z = 1. The numerator isn't defined in z = 0. So the convergence radius is |z-1| < 1. (See deviser also, zeros in z = 2, 3, 4 ...)

First the log function, I think I've got it right. The Taylor expansion is also pretty straight forward.

$\displaystyle \log z$ = $\displaystyle \log (z - 1 + 1)$ = $\displaystyle \log 1 + (z - 1)$ =

$\displaystyle (z-1) - \frac{(z - 1)^2}{2} + \frac{(z - 1)^3}{3} - \frac{(z - 1)^4}{4}+O((z-1)^5)$

In this part I'm starting to get shaky.

$\displaystyle \sin \pi z$ = $\displaystyle - \sin \pi (z - 1)$

$\displaystyle \frac {1}{\sin \pi z}$ = $\displaystyle -\frac{1}{\pi(z-1)} \frac{1}{1 - \frac{\pi^2}{6}(z-1)^2 - O((z-1)^4)}$ - This is where I need some pointers. I can't simplify this enough to multiply it with the log z part to get the whole series.