I'm trying to find the integral along $\partial D$ of $$ \mathrm{f}(z) = {z^4 + 1 \over z^2(z - 3)(3z - 1)} \quad\text{where } D \text{ is the open unit disc centred at } z=0. $$
It's clear that I have to use the Cauchy's theorem, but when the pole is $z=0$ is where I have problems.
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By the residue theorem, $$ \int_\gamma\!\frac{z^4 + 1}{z^2(z - 3)(3z - 1)}\,dz = 2\pi i\left(\operatorname{Res}(f,0) + \operatorname{Res}(f,1/3)\right). $$ Now, because $z = 1/3$ is a simple pole of $f$, \begin{align*} \operatorname{Res}(f,1/3) &= \lim_{z\to 1/3}(z - 1/3)f(z)\\ &=\lim_{z\to 1/3}\frac{z^4 + 1}{3z^2(z - 3)}\\ &=\frac{(1/3)^4 + 1}{3(1/3)^2((1/3) - 3)}\\ &=-\frac{41}{36}. \end{align*} Similarly, $z = 0$ is a pole of order $2$ of $f$, so \begin{align*} \operatorname{Res}(f,0) &= \lim_{z\to 0}\frac{d}{dz}\left[z^2f(z)\right]\\ &=\lim_{z\to 0}\frac{d}{dz}\left[\frac{z^4 + 1}{(z - 3)(3z - 1)}\right]\\ &=\lim_{z\to 0}\frac{2 (5 - 3 z + 6 z^3 - 15 z^4 + 3 z^5)}{(-3 + z)^2 (-1 + 3 z)^2}\\ &= \frac{10}{9}. \end{align*} Thus, $$ \int_\gamma\!\frac{z^4 + 1}{z^2(z - 3)(3z - 1)}\,dz = 2\pi i\left(-\frac{41}{36} + \frac{10}{9}\right) = \frac{-\pi i}{18}. $$
Without the $z^2$ in the denominator, $g(z)=z^2f(z)=1/3+(10/9)z+(91/27)z^2+\cdots.$ So after division by $z^2,$ the residue at $z=0$ of $f$ is $10/9.$ The residue theorem can now be applied.