Find $\int_{\gamma}\frac{(\sin z)^2}{(z-z_0)^2}dz$

Question

Find $\int_{\gamma}\frac{(\sin z)^2}{(z-z_0)^2}dz$ where $\gamma$ is a linear pieces parameterization of closed polygonal chain $[w_0,w_1,w_2,w_3,w_0]$ with vertices $w_0=1+i, w_1=1-i, w_2=-1+i,w_3=2$. Consider the cases $z_0=\frac 12, z_0=-1$.

I know that such task I should do by using the Cauchy's integral theorem. However, on the basis of the theorem itself, I have no intuition yet how to go about such tasks. Anyone would like to guide me how to think about a task so that I can do the next task one myself?

Answer 1

Since $-1$ is outside the region of $\Bbb C$ bounded by the range of the loop (see the picture below),$$\int_\gamma\frac{\sin^2z}{z+1}\,\mathrm dz=0.$$

Concerning $\frac12$, we have\begin{align}\int_\gamma\frac{\sin^2z}{z-1/2}\,\mathrm dz&=2\pi i\operatorname{Ind}_\gamma\left(\frac12\right)\frac{\sin'(1/2)}{1!}\\&=-2\pi i\cos\left(\frac12\right).\end{align}

Answer 2

The case $z_0=-1$. There is an open and connected set $D$ with the following properties:

the convex hull of $[w_0,w_1,w_2,w_3,w_0]$ is contained in $D$ and $z_0 \notin D.$ Hence the function $\frac{(\sin z)^2}{(z-z_0)^2}$ is holomorphic on $D$. Cauchy's theorem gives then:

$$\int_{\gamma}\frac{(\sin z)^2}{(z-z_0)^2}dz=0.$$

The case $z_0=1/2.$ Let $f(z)= \sin^2 z$. Since $z_0$ is an element of the interior of the convex hull of $[w_0,w_1,w_2,w_3,w_0]$, we get by Cauchy's integral formula for the first derivative, that

$$\int_{\gamma}\frac{(\sin z)^2}{(z-z_0)^2}dz= 2 \pi i f'(1/2).$$