How can I solve $f(a) = \int_C \frac{2}{(2z-1)(z-2)}dz$ where $C$ is the unit circle $|z|=1 $ ?
my approach was:
Since any point on the unit circle is in the parametric form $z = e^{i \theta}$, where $0\leq \theta \leq 2\pi$ and $dz = ie^{i\theta}d\theta$ by substitution in the partial fractions of the main problem find:
$$\int_C \frac{2}{(2z-1)(z-2)}dz = \int_0^{2\pi}(\frac{\frac{2}{3}}{e^{i\theta}-2}-\frac{\frac{4}{3}}{2e^{i\theta}-1})ie^{i\theta}d\theta \\=\frac{2}{3} \int_0^{2\pi}(\frac{ie^{i\theta}}{e^{i\theta}-2}-\frac{ie^{i\theta}}{e^{i\theta}-1/2})d\theta \\=\frac{2}{3}[\ln\frac{e^{i\theta}-2}{e^{i\theta}-1/2}]_0^{2\pi}$$
This evaluates to zero which I'm sure is not the correct answer. What I'm missing here?
No, the integral is not zero. Note that $$ \int_{|z|=1} \frac{2}{(2z-1)(z-2)}dz=\int_{|z|=1} \frac{\frac{1}{z-2}}{z-1/2}dz,$$ now apply the Cauchy's integral formula with $f(z)=\frac{1}{z-2}$.
Can you take it from here?