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Given $f(z)$ is a continuous function with zeroes at the origin and outside $\vert z\vert=5$ and $\frac{1}{f(z)}=\frac{10}{z^3}-\frac{25}{z}+3z-2z^2+\cdots$. Evaluate $\int_{c}\frac{1}{f(z)}dz$ such that C is any simple closed contour lying within $\vert z\vert=5$.

Attempt:

Since the principal part of the Laurent series has only $2$ terms and $0$ is its only singularity within $\vert z\vert=5$, then $0$ is a pole of $\frac{1}{f(z)} $order $2$. Before using the Cauchy Residue Theorem, we first evaluate \begin{align} \operatorname{Res}(f(z_0),0)=\lim_{z \to 0}\frac{d}{dz}[z^2\frac{1}{z^2 g(z)}] \quad \text{s.t.}\quad f(z)=z^{2} g(z) \\ \text{but}\quad \frac{1}{g(z)}=\frac{10}{z}-25z+3z^3 -2z^4+ \cdots\\ \text{and}\quad \frac{d}{dz}(\frac{1}{g(z)})=\frac{-10}{z^2}-25+9z^2-8z^3 \end{align} So as $z \to 0$, $\operatorname{Res}(f(z),0)\to -\infty$. How should I go about this?

EDIT: for future reference.

Note that we must instead have \begin{align} \operatorname{Res}(f(z),0)=\frac{1}{2!}\lim_{z \to 0}\frac{d^2}{dz^2}[\frac{1}{g(z)}] \end{align} because $\frac{1}{f(z)}$ has a pole of order 3. Then it can be verified that $\lim_{z\to 0}g^{(2)}(0)=-50$. Then by the Cauchy Residue Theorem $\int_{c}\frac{1}{f(z)}dz=2\pi i\cdot (-50)=-100\pi i$

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    $\begingroup$ The principal parte has 3 components, of whose one has coefficient 0. $\endgroup$ – N74 Oct 24 '18 at 11:35
  • $\begingroup$ Thanks! I think that is what's missing. $\endgroup$ – TheLast Cipher Oct 24 '18 at 11:42

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