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Suppose that $f$ has a simple pole at $z=a$ and let $g$ be analytic in an open set containing $a$. Show that $Res(fg;a)=g(a)Res(f;a)$.

I know that as $f$ has a simple pole at $z=a$, this means its Laurent series is of the form

$f(z)=\dfrac{Res(f;a)}{z-a}+\displaystyle\sum_{n=0}^{\infty}a_n(z-a)^n$

How can I compute the Laurent series of $fg$ at $z=a$?

Tanius
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2 Answers2

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It shouldn't be too hard to see that $fg$ has at most a simple pole at $z=a$. Then $$ \text{Res}(fg;a)=\lim_{z\to a}(z-a)f(z)g(z)=\left(\lim_{z\to a}(z-a)f(z)\right)\left(\lim_{z\to a}g(z)\right)=\text{Res}(f;a)g(a). $$

For a more convincing argument, write $g(z)=\sum_{n=0}^{\infty}b_nz^n$, and study the product $$ f(z)g(z)=\left(\frac{\text{Res}(f;a)}{(z-a)}+a_0+a_1(z-a)+\cdots\right)\left(b_0+b_1(z-a)+b_2(z-a)^2+\cdots\right) $$ (note $g(a)=b_0$ here).

Blake
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  • Thank you very much! Though I don't see why you say the second one is a more convincing argument. I think both of them are so. – Tanius Dec 11 '15 at 05:00
  • @TRmate I mostly said that because I wasn't sure if you knew the "easy way" to find the residue of a function with a simple pole. – Blake Dec 11 '15 at 05:04
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If $g(a)=0$ then $fg$ has removable singularity at zero so $Res(fg;a)=0=g(a)Res(f;a).$ Again if $g(a)\neq0$ then $fg$ has simple pole at zero. Then $f(z)=\frac{h(z)}{z-a}$ and so $Res(fg;a)=Res(\frac{h(z)g(z)}{z-1};a)=h(a)g(a)=g(a)Res(f;a).$ So in both of the case result is proved.

neelkanth
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