# Complex power series expansion for $f(z) = \frac{e^z}{a-z}$

I have the following homework problem in my complex analysis class: Find the complex power series expansion for the function $$f(z) = \frac{e^z}{a-z}$$ where $a \in \mathbb{C}$, and $z \ne 0$.
I know the power series expansions for the functions: $$g(z) = e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$$ valid for each $z \in \mathbb{C}$, and $$h(z) = \frac{1}{a-z} = \sum_{n=0}^{\infty} \frac{z^n}{a^{n+1}}$$ valid for $\mid \frac{z}{a} \mid < 1$.
However, I do not see how to combine these two expressions to yield the desired expression. More specifically, even though $f(z) = g(z)h(z)$, we do not obtain the correct power series expansion for $f$ by multiplying the power series expansions for $g(z)$ and $h(z)$ term-wise. It doesn't seem like the Cauchy Product of the two series will be of much help either, although I could be mistaken.
Any help would me most appreciated!

• The Cauchy product is what you need. On a different note, you lost an $1/a$ in the series for $h$. Oct 10 '13 at 21:34
• @DanielFischer Thanks for noticing the error, I edited to fix it. Would you mind saying a little bit about how we use the Cauchy product for this example? Oct 10 '13 at 21:38

The Cauchy product says that the coefficient of $z^n$ will be $\sum_{k=0}^n 1/(a^{k+1}(n-k)!)$ in your series, and it will be valid if $|z| < |a|$.