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I am struggling with how to show the following:

Suppose two holomorphic functions f(z),g(z) have poles with order k1 > 0 and k2 > 0 respectively, show that the order of the pole of

f(z)g(z) = k1 + k2

Any help appreciated, Thanks!

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    Let this pole be located at $z=0$ for ease of notation. Then you have $f(z)=z^{-k_1}*F(z)$, where $F$ doesn't vanish at zero and similarly $g(z)=z^{-k_2}*G(z)$. Then $f(z)g(z)=z^{-k_1-k_2}F(z)G(z)$. – 2017-02-22

1 Answers 1

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Suppose that $f$ and $g$ are holomorphic on $G:=D \setminus \{z_0\}$ ($D$ an open set in $\mathbb C$) with a poles at $z_0$ of order $k_1$ and $k_2$. Then there are holomorphic fuctions $a,b :D \to \mathbb C$ such that

$f(z)=\frac{a(z)}{(z-z_0)^{k_1}}$ for all $z \in G$ and $a(z_0) \ne 0$

and

$g(z)=\frac{b(z)}{(z-z_0)^{k_2}}$ for all $z \in G$ and $b(z_0) \ne 0$.

Then:

$f(z)g(z)=\frac{a(z)b(z)}{(z-z_0)^{k_1+k_2}}$ for all $z \in G$ and $a(z_0)b(z_0) \ne 0$.

This gives $fg$ has a pole of order $k_1+k_2$ at $z_0$