# multiplication of complex power series

Let $f(z) = \sum\limits_{n=0}^\infty a_n z^n$ and let $g(z) = \sum\limits_{n=0}^\infty b_n z^n$. I need to show that $\sum\limits_{n=0}^\infty a_n z^n \sum\limits_{n=0}^\infty b_n z^n$ converges to $\sum\limits_{n=0}^\infty c_n z^n$ where each $c_n$ is given by $c_n = \sum\limits_{k=0}^n a_k b_{n-k}$.

This is not homework by the way.

The question says to use the Cauchy integral theorem and look at $f(z)g(z)/z^n = \sum\limits_{k=0}^\infty a_k z^{k-n} g(z)$, but I don't understand why it tells me to proceed this way.

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Actually I got it. Simply following the hint and integrating termwise does the trick. –  nigelvr Dec 10 '12 at 4:04

## 1 Answer

Let $\phi(z) = f(z)g(z)$. $\phi$ is analytic and can be represented as $\phi(z) = \sum_{n=0}^\infty c_n z^n$. Then Cauchy's integral formula gives $c_n = \frac{1}{2 \pi i} \int_\gamma \frac{\phi(z)}{z^{n+1}} dz$. Expanding $f$ gives

\begin{eqnarray} c_n &=& \frac{1}{2 \pi i} \int_\gamma \frac{\sum_{k=0}^\infty a_k z^k g(z)}{z^{n+1}} dz \\ &=& \frac{1}{2 \pi i} \int_\gamma \frac{\sum_{k=0}^{n} a_k z^k g(z)}{z^{n+1}} dz \\ &=& \sum_{k=0}^{n} a_k \frac{1}{2 \pi i} \int_\gamma \frac{g(z)}{z^{n+1-k}} dz \\ &=& \sum_{k=0}^{n} a_k b_{n-k} \end{eqnarray}

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Thanks. I actually got it a split second before you posted this, but I appreciate your reply. –  nigelvr Dec 10 '12 at 4:05
Minor point: Note that the denominator should be $z^{n+1}$. –  copper.hat Dec 10 '12 at 4:05