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Let $f$ be analytic at $z_0\in\Bbb{C}$ and $g$ have a simple pole at $z_0$. Find $\operatorname{Res}(f(g(z_0)),z_0)$.

scratchwork (Would like big hints. I don't usually like asking for the full answer, but I have no choice this time.):

If $g$ has a simple pole at $z_0$ with residue $A$, then \begin{align} \operatorname{Res}(g(z),z_0)=\lim_{z\to z_0}(z-z_0)g(z)=\lim_{z\to z_0}(z-z_0)\frac{h(z)}{z-z_0}=\lim_{z\to z_0}h(z)=A \quad \text{, for some function h(z).} \end{align} Since $f$ is analytic at $z_0\in\Bbb{C}$, then it has taylor series representation, say $f(z)=\sum_{k=0}^\infty \frac{f^{(k)}(z_0)}{k!}(z-z_0)^k$.

Original problem statement(verbatim): Let $f$ be analytic at $z=z_0$. calculate $\operatorname{Res}(fg,z_0)$, if

  • (a) $g$ has a simple pole with residue $A$ at z=z_0.
  • (b) $g$ has a pole of order $k$ and principal part given by

\begin{align} \frac{a_{-1}}{z-z_0}+\frac{a_{-2}}{(z-z_0)^2}+\cdots +\frac{a_{-k}}{(z-z_0)^k} \end{align}

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    $\begingroup$ There is something wrong with the way the question is stated. The fact that $f$ is analytic at $z_0$ is of no use because we are interested in $f(g(z))$ for $z$ neat $z_0$. $\endgroup$ Oct 26, 2018 at 9:52
  • $\begingroup$ Hold on. I'll upload a picture. thank you guys! $\endgroup$ Oct 26, 2018 at 9:54
  • $\begingroup$ Don't upload a picture! Use MathJax. $\endgroup$ Oct 26, 2018 at 9:56
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    $\begingroup$ You are reading the product $fg$ as $f\circ g$. @TheLastCipher $\endgroup$ Oct 26, 2018 at 9:58
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    $\begingroup$ As stated it is the product of functions not the composite function. $\endgroup$
    – Shashi
    Oct 26, 2018 at 9:58

1 Answer 1

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The answer is $f(z_0)\operatorname{res}\bigl(g(z),z_0\bigr)$, since$$\lim_{z\to z_0}(z-z_0)f(z)g(z)=\lim_{z\to z_0}f(z)\times\lim_{z\to z_0}(z-z_0)g(z)=f(z_0)\operatorname{res}\bigl(g(z),z_0\bigr).$$


For the second part, note that if$$f(z)=a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots$$and\begin{align}g(z)&=\frac{b_{-k}}{(z-z_0)^k}+\frac{b_{-k+1}}{(z-z_0)^{k-1}}+\cdots\\&=\frac1{(z-z_0)^k}\left(b_{-k}+b_{-k+1}(z-z_0)+\cdots\right),\end{align}then $(z-z_0)^{k-1}$ in$$\left(a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots\right)\times\left(b_{-k}+b_{-k+1}(z-z_0)+\cdots\right),$$which is $a_0b_{-1}+a_1b_{-2}+\cdots+a_{k-1}b_{-k}$.

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  • $\begingroup$ I agree. thanks!! $\endgroup$ Oct 26, 2018 at 10:06
  • $\begingroup$ As for part (b), I am stuck with $\operatorname{Res}(fg,z_0)=\frac{1}{(k-1)!}\lim_{z \to z_0}\frac{d^{k-1}}{dz^{k-1}}[(z-z_0)^{k}fg]=\frac{1}{(k-1)!}\lim_{z\to z_0}\frac{d^{k-1}}{dz^{k-1}}[f(z)h(z)]$ such that $g(z)=h(z)/(z-z_0)^k$ and the fact that $\frac{d^{k-1}}{dz^{k-1}}[h(z)]=(k-1)!a_{-1}+\cdots$. $\endgroup$ Oct 26, 2018 at 10:51
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    $\begingroup$ I've edited my answer. $\endgroup$ Oct 26, 2018 at 11:02
  • $\begingroup$ Wow. How did you deduce the coefficient so quickly by simple expansion? $\endgroup$ Oct 26, 2018 at 11:09
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    $\begingroup$ I computed the product. It's like multiplying polynomials. $\endgroup$ Oct 26, 2018 at 11:11

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