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i need to resolve this problem:

Given $f(z)=\frac{1}{\big(g(z)\big)^2}$, show that if $g(z_0)=0$ and $g'(z_0)\neq 0$ then $f$ has a second order pole at $z_0$ and $\text{Res}_{z=z_0}f(z)=-\frac{g''(z_0)}{\big(g'(z_0)\big)^3}$

It's easy to show that $f$ has a second order pole at $z_0$ however i have some difficulties finding de residue, since $z_0$ is a pole of order $2$ then we have that

$$\text{Res}_{z=z_0}f(z)=\lim_{z\to z_0}\frac{d}{dz}\big((z-z_0)^2f(z)\big)\\ =\lim_{z\to z_0}\frac{d}{dz}\Big(\frac{z-z_0}{g(z)}\Big)^2=\lim_{z\to z_0}2\Big(\frac{z-z_0}{g(z)}\Big)\Big(\frac{g(z)-(z-z_0)g'(z)}{g(z)^2}\Big)$$

By L'Hôpital's rule I get the correct answer, but i would like to compute this limit without the aid of L'Hôpital's rule, just with the properties/theorems of residue calculus. Thanks.

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    $\begingroup$ Have you considered the Laurent series of $f$ around $z_0$? $\endgroup$ Nov 12, 2020 at 19:02

1 Answer 1

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Assuming that $g$ is analytic with $g(z_0)=0$ and $g'(z_0)\ne0$, we can write

$$\begin{align} \lim_{z\to z_0}\frac{d}{dz}\left(\frac{z-z_0}{g(z)-g(z_0)}\right)^2&=2\lim_{z\to z_0}\left(\frac{z-z_0}{g(z)-g(z_0)}\right)\frac{d}{dz}\left(\frac{z-z_0}{g(z)-g(z_0)}\right)\\\\ &=\frac2{g'(z_0)}\lim_{z\to z_0}\left(\frac{g(z)-g(z_0)-g'(z)(z-z_0)}{(g(z)-g(z_0))^2}\right)\\\\ &=\frac2{g'(z_0)}\lim_{z\to z_0}\left(\frac{\color{red}{\frac{g(z)-g(z_0)-g'(z_0)(z-z_0)}{(z-z_0)^2}}-\color{blue}{\frac{(g'(z)-g'(z_0))(z-z_0)}{(z-z_0)^2}}}{\frac{(g(z)-g(z_0))^2}{(z-z_0)^2}}\right)\\\\ &=\frac2{(g'(z_0))^3}\left(\color{red}{\frac12 g''(z_0)}-\color{blue}{g''(z_0)}\right)\\\\ &=-\frac{g''(z_0)}{(g'(z_0))^3} \end{align}$$

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  • $\begingroup$ Thanks!! I didn't realized that way of rewriting the numerator $\endgroup$ Nov 12, 2020 at 21:20
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$
    – Mark Viola
    Nov 12, 2020 at 21:23

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