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$$\int_{|z| = 1}\frac{1}{(1-3z)(1-2z)^2}dz$$

z here is a complex number calculus

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    How did you try.? – 2017-01-27
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    Use definitions and maybe some theorems :) – 2017-01-27
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    What poles does the integrand have? What are the values of the residues at those poles? – 2017-01-27
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    To compute residues, there's a [formula](https://en.wikipedia.org/wiki/Residue_(complex_analysis)#Limit_formula_for_higher_order_poles) – 2017-01-27

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Set $\mathcal{D}=\{z\in\mathbb{C}\,:\, |z|\le1\}$. It is clear $z=\frac{1}{2}\in \mathcal{D}$ and $z=\frac{1}{3}\in \mathcal{D}$. We have $$\text{Res}_{z=\frac 12}f(z)=\lim_{z\to\frac12}\frac{d}{dz}\left(\left(z-\frac 12\right)^2\frac{1}{(1-3z)(1-2z)^2}\right)=\frac{1}{4}\lim_{z\to\frac12}\frac{3}{(1-3z)^2}=3$$ and $$\text{Res}_{z=\frac 13}f(z)=\lim_{z\to\frac13}\left(\left(z-\frac 13\right)\frac{1}{(1-3z)(1-2z)^2}\right)=-3$$ By application of Reside theorem, we have $$\oint_{|z|=1}\frac{1}{(1-3z)(1-2z)^2}dz=2\pi\text{i}(-3+3)=0$$