# Finding residue of composite function.

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}

• 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 $z_0$. – Kabo Murphy Oct 26 '18 at 9:52
• Hold on. I'll upload a picture. thank you guys! – TheLast Cipher Oct 26 '18 at 9:54
• Don't upload a picture! Use MathJax. – José Carlos Santos Oct 26 '18 at 9:56
• You are reading the product $fg$ as $f\circ g$. @TheLastCipher – Kabo Murphy Oct 26 '18 at 9:58
• As stated it is the product of functions not the composite function. – Shashi Oct 26 '18 at 9:58

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$$g(z)=\frac{b_{-k}}{(z-z_0)^k}+\frac{b_{-k+1}}{(z-z_0)^{k-1}}+\cdots=\frac1{(z-z_0)^k}\left(b_{-k}+b_{-k+1}(z-z_0)+\cdots\right),$$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 $$a_0b_{-1}+a_1b_{-2}+\cdots+a_{k-1}b_{-k}$$.
• 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 $\frac{d^{k-1}}{dz^{k-1}}[h(z)]=(k-1)!a_{-1}+\cdots$. – TheLast Cipher Oct 26 '18 at 10:51