Let C be the positively oriented circle of radius 2 centered at the origin. Evaluate $\int_C\frac{\sin{\frac{\pi z}{2}}}{z^3+z^2}$

Question

I'm trying to apply the Cauchy integral formula, but since no matter we choose f(z) as $\frac{\sin{\frac{\pi z}{2}}}{z^2}$ or $\frac{\sin{\frac{\pi z}{2}}}{z(z+1)}$, we will always have singularity within the contour... Is there any way to avoid brute force calculation? Thanks

Answer 1

The point is that there is a singularity inside the contour.

Now evaluate the residue at each singularity.

$\lim_\limits{z\to 0} zf(z) = \frac {\pi}{2}$

$\lim_\limits{z\to -1} (z+1) f(z) = -\sin \frac {\pi}2 = -1$

$2\pi i(\frac {\pi}{2}-1) = 0$

Answer 2

By the residue theorem, the integral you have is equal to $2\pi i$ times the residue of the function at any singularities enclosed within the integration contour.

Since the circle of radius $2$ centered at the origins encloses both the pole of order $2$ at the origin and the simple pole at $z=-1$, we have:

\begin{align} \oint_{|z|=2}\frac{\sin\frac{\pi z}{2}}{z^2(z+1)}\,dz &= 2\pi i\left[\operatorname{Res}\left(\frac{\sin\frac{\pi z}{2}}{z^2(z+1)},0\right)+\operatorname{Res}\left(\frac{\sin\frac{\pi z}{2}}{z^2(z+1)},-1\right)\right]\\ &=2\pi i\left[\lim\limits_{z\rightarrow 0}\frac{d}{dz}\frac{\sin\frac{\pi z}{2}}{z+1} +\lim\limits_{z\to -1}\frac{\sin\frac{\pi z}{2}}{z^2}\right]\\ &=2\pi i\left[\lim\limits_{z\to 0}\frac{\pi\cos\frac{\pi z}{2}}{2z+2}+\lim\limits_{z\to -1}\frac{\sin\frac{\pi z}{2}}{z^2}\right]\\ &=2\pi i\left(\frac{\pi}{2} -1\right) \end{align}