\(\displaystyle f(z)=\begin{cases}-\frac{1}{z-2}\sum_{n=0}^\infty \left(\frac{z-2}{2}\right)^2, & \text{for}~ |z-2|<2

\\ \frac{2}{(z-2)^2} \sum_{n=0}^\infty \left(-\frac{2}{z-2}\right)^2, &\text{for}~|z-2|>2

\end{cases}\)

Is it possible?

- Thread starter scorpion007
- Start date