# Evaluating a contour integral given its Laurent series and singularity.

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 $$\int_{c}\frac{1}{f(z)}dz=2\pi i\cdot (-50)=-100\pi i$$

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