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My question regards the composition of two functions:

Let $f$ be an analytic function on a subset $\mathcal{U}$ of the complex plane and $g$ be a function that's analytic on $f(\mathcal{U})\setminus \{s_0\}$. If $g$ has a pole at $s_0$, and $f(z_0) = s_0$, can we assume that the residue of $g(f(z_0))$ is equal to the residue of $g(s_0)$ and if so, how would somebody show that?

For example: Let $g(s) := \displaystyle{\frac{1}{s}}$ which has by definition a residue $1$ at $s=0$. If we now assume that $f(z_0) = 0$, can we say that the residue of $g(f(z))$ in $z_0$ is equal to $1$ because $g(f(z_0))$ is equal to $g(0)$ or is some sort of other mathematical trickery going on? :P

Thank you for reading to this part of the question, it's very much appreciated.

Sincerely, Cedric :)

Brass One
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C. Brendel
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1 Answers1

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Suppose $g$ has a simple pole at $s_0$ with residue $r$ there, and $f'(z_0) \ne 0$.

We have $$f(z) = s_0 + f'(z_0) (z - z_0) + O((z-z_0)^2)$$ and $$\eqalign{g(f(z)) &= r (f(z) - s_0)^{-1} + O(1)\cr &= r(f'(z_0) (z-z_0) + O((z-z_0)^2))^{-1} + O(1)\cr &= \frac{r}{f'(z_0)}(z-z_0)^{-1} + O(1)}$$ so the residue of $g(f(z))$ at $z=z_0$ is $r/f'(z_0)$.

If $g$ has a higher-order pole at $s_0$, or if $f'(z_0) = 0$, things can be more complicated.

Robert Israel
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