$$\oint_{|z|=2}\frac{1}{(z)^2(z-1)^3}.d\underline{z} = 0$$
I wanted to ask if there is a faster way rather than the Cauchy Integral Theorem(?)
The problem is partial fraction is taking too much time; I just need to know if I am missing a short cut?
Any advice will be helpful. Thanks
By the residue theorem, the integral $$\oint_{|z|=R}\frac{dz}{z^2(z-1)^3}$$ is constant for any $R>1$, and by taking a very large $R$ we get
$$\left|\oint_{|z|=R}\frac{dz}{z^2(z-1)^3}\right|\leq \oint_{|z|=R}\frac{dz}{|z|^2 |z-1|^3}\leq 2\pi R\cdot \frac{1}{R^2(R-1)^3} $$ hence the previous constant is zero.