I can't really think of any way to prove this. I think it starts with the fact that I'm not really sure what my desired outcome is. I was thinking there might be some way to get it in the form $\displaystyle \sum_{n=0}^{\infty } a_n(z-z_0)^n$. Also, it is very plain to see that $\displaystyle \frac{1}{L}$ is radius of convergence $\displaystyle \sum_{n=0}^{\infty } a_n(z-z_0)^n$, which, it seems, is somehow related since $\displaystyle

|z-z_0| < \frac{1}{L} \implies \frac{1}{|z-z_0|} > L$. But...that can't be sufficient...