I need to compute the integral $\displaystyle I= \int _\gamma \frac{1}{z^5(2z-3)^6}$, where $\gamma (t) = \cos{t} + 3i \sin{t} , t\in[0,2\pi]$ .
My main problem is to find the winding number of the curve $\gamma$ around $0$ and $3/2$, which are the poles of the function $\displaystyle f(z)=\frac{1}{z^5(2z-3)^6}$. Is there an easier way to compute this integral, avoiding the residue theorem?
Let $f\colon z\mapsto\dfrac 1{(2z-3)^6}$ and $U=\left\{x+iy\ \colon\ (x,y)\in\mathbb{R}^2\wedge\left(x^2+\left(\dfrac y3\right)^2 < 1\right)\right\}$.
Then $f$ is holomorphic on the open set $U$ and continuous on the closure of $U$, the path $\gamma$ being the boundary of $U$. The winding number of $\gamma$ about $0$ is $1$.
By Cauchy's integral formula, for all $n\geqslant 0$, we have: $$f^{(n)}(0)=\dfrac{n!}{2i\pi}\oint_\gamma\dfrac{dz}{z^{n+1}(2z-3)^6}$$
By choosing $n=4$, and computing the fourth derivative of $f$, you can find the value of your integral.