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let $G$ be an open disc centered around $z_0$ of radius $r$. Let $f(z),g(z)$ be holomorphic functions on $G$. such that $f(z)$ has a simple zero at $z_0$.

Find an expression for the residue of $\frac{g(z)}{f(z)}$ at $z_0$.

When I looked this up, I found that the answer is actually $\frac{g(z_0)}{f'(z_0)}$. I'm currently stuck at this step:

$$\int_{\gamma} \frac{g(z)}{f(z)}\, dz = 2{\pi}i \text{Res}\left(\frac{g(z)}{f(z)}, z_0\right)$$

where ${\gamma}$ is a closed curve in $G$ that wraps around $z_0$ I'm not too sure how to continue. I tried to pull out a $g(z_0)$ by using the cauchy integral formula but I don't think it will work.

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Here's a (in my opinion simpler) way to deduce the formula without power series: $$ \operatorname*{Res}_{z=z_0} g/f = \lim_{z\to z_0} (z-z_0) \cdot \frac{g(z)}{f(z)} = \lim_{z\to z_0} g(z) \frac{z-z_0}{f(z)-f(z_0)} = \frac{g(z_0)}{f'(z_0)} $$ since $f(z_0) = 0$.

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Since both $f$ and $g$ are analytic on $G$, and $f$ has a simple zero at $z_0$, we can write

$$f(z)=\sum_{n=1}^\infty \frac{f^{(n)}(z_0)\,(z-z_0)^n}{n!}$$

and

$$g(z)=\sum_{n=0}^\infty \frac{g^{(n)}(0)\,(z-z_0)^n}{n!}$$

Then, the ratio $g(z)/f(z)$ has a simple pole at $z_0$ with

$$\begin{align} \frac{g(z)}{f(z)}&=\frac{\sum_{n=0}^\infty \frac{g^{(n)}(0)\,(z-z_0)^n}{n!}}{\sum_{n=1}^\infty \frac{f^{(n)}(z_0)\,(z-z_0)^n}{n!}}\\\\ &=\frac{g(z_0)+\sum_{n=1}^\infty \frac{g^{(n)}(0)\,(z-z_0)^n}{n!}}{f'(z_0)(z-z_0)+\sum_{n=2}^\infty \frac{f^{(n)}(z_0)\,(z-z_0)^n}{n!}} \end{align}$$

The residue is simply the limit

$$\begin{align} \lim_{z\to z_0}\left(\frac{(z-z_0)g(z)}{f(z)}\right)&=\lim_{z\to z_0}\left(\frac{g(z_0)+\sum_{n=1}^\infty \frac{g^{(n)}(0)\,(z-z_0)^n}{n!}}{f'(z_0)+\sum_{n=2}^\infty \frac{f^{(n)}(z_0)\,(z-z_0)^{n-1}}{n!}}\right)\\\\ &=\frac{g(z_0)}{f'(z_0)} \end{align}$$

as was to be shown!

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  • $\begingroup$ How do you expose the constant $a_{-1}$ in the Laurent expansion of the quotient? That is, can you derive this expression without using the limit? @Dr.MV $\endgroup$ – sequence Mar 25 '17 at 22:15
  • $\begingroup$ @sequence The residue is equal to $a_{-1}$. What do you mean? $\endgroup$ – Mark Viola Mar 25 '17 at 22:16
  • $\begingroup$ $a_{-1}$ in the Laurent expansion $\sum\limits_{-\infty}^\infty a_n(z-z_0)^n$. $\endgroup$ – sequence Mar 25 '17 at 22:18
  • $\begingroup$ @phantom Please let me know how I can improve my answer. I really want to give you the best answer I can. If the answer was not useful, then I am happy to delete it. -Mark $\endgroup$ – Mark Viola Mar 25 '17 at 22:18
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    $\begingroup$ And yet, it is correct. For example, $$\frac{1}{1+x}=1-x+x^2-x^3\cdots=1+O(x)$$ $\endgroup$ – Mark Viola Mar 25 '17 at 22:47

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