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My question regards the composition of two functions:

Let f be an analytic function on a subset U of the complex plane and g be a function that's analytic on f(U){s₀}. If g has a pole at s₀ and f(z₀) = s₀ can we assume that the residue of g(f(z₀)) is equal to the residue of g(s₀) and if so, how would somebody show that?

For example: g(s) = 1/s which has by definition a residue of 1 at s=0

if we now assume that f(z₀) = 0, can we say that the residue of g(f(z)) in z₀ is equal to 1 because g(f(z₀)) 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 :)

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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.

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