# The product $(fg ^ 2)(p)$ is different of zero

Could anyone help me understand what this passing a demonstration is highlighted in yellow?

Let $f, g$ be the analytic functions defined in the complex plane. Assume that $p$ is zero of order $2m$ of $f$ and one pole of order $m$ of $g$, $m\in \mathbb{N}^*$. Show that $(fg ^ 2)(p)\neq 0$.

Here are the pictures below, sorry but the article is in portuguese, but to understand.

I'm finding it strange because if $f(p)=0$ then $(fg^2)(p)=0$.

Think about the Taylor and Laurent expansions about $p$ for $f$ and $g$, respectively. You have (in a punctured neighborhood of $p$) $$f(z) = a_{2m}(z-p)^{2m} h(z) \qquad\text{and}\qquad g(z) = b_{-m}(z-p)^{-m} k(z)$$ for some holomorphic functions $h$ and $k$ nonzero at $p$, with $a_{2m}$ and $b_{-m}$ both nonzero. So $f(z)g(z)^2 = a_{2m}b_{-m}^2 h(z)k(z)^2$ is nonzero at $p$.

• Ted, but is $k$ defined at the pole? Do we make an extension? – Mancala Sep 7 '18 at 0:39
• @Mancala: Once you factor out the highest negative power in the Laurent series, the remaining factor is a (locally convergent) power series with a nonzero constant term. No problem. – Ted Shifrin Sep 7 '18 at 0:49