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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?

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    What's $\gamma$? – 2017-01-21
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    @MyGlasses Think you need your glasses – 2017-01-21
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    @Jahambo99 I have. I ask dimvolt to go ahead.! – 2017-01-21
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    The winding numbers are really easy to find. Do you know what your curve looks like? Open GeoGebra, create a number $t\in\left[0,2\pi\right]$ with a slider and type $\cos(t)+3i*\sin(t)$ in the input line, then move your slider. You'll see the complex number move. Activate its trace if necessary. – 2017-01-21
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    The curve traces an ellipse with winding number is $1$. – 2017-01-21
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    You 're right, I thought it would be more difficult and that I should compute the winding numbers with the definition. Thank you all ! – 2017-01-21
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    You can use Cauchy's Integral theorem and focus only on the singularity $z_0=0$, the other singularity is outside the region, so it is not of interest. – 2017-01-21

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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.