Evaluation of complex integral

Question

$$\int_{\text{c}}\frac{\sin \pi z^2 + \cos \pi z^2}{(z+1)(z+2)}$$

Where $\text{C}$ is the circle $|z| =3$

I'm a little confused about how to do this. Should this be done the normal way ? How do I use the concept of what C is and the concepts of complex analysis ?

Before looking at the answer below I feel obliged to ask if that squared is on the trig function or their argument — Triatticus, Jun 13 '16 at 02:00

Mark Viola · Accepted Answer · 2016-06-14T13:47:04.103

3

First, use partial fraction expansion to write

$$\frac{1}{(z+1)(z+2)}=\frac{1}{z+1}-\frac{1}{z+2}$$

Then, applying Cauchy's Integral Formula, we obtain

$$\begin{align} \oint_{|z|=3}\frac{\sin(\pi z^2)+\cos(\pi z^2)}{(z+1)(z+2)}\,dz&=\oint_{|z|=3}\frac{\sin(\pi z^2)+\cos(\pi z^2)}{z+1}\,dz-\oint_{|z|=3}\frac{\sin(\pi z^2)+\cos(\pi z^2)}{z+2}\,dz\\\\ &=2\pi i\left(\sin(\pi)+\cos(\pi)-\sin(4\pi)-\cos(4\pi)\right)\\\\ & =-4\pi i \end{align}$$

edited Jun 14 '16 at 13:47

answered Jun 13 '16 at 01:51

Mark Viola

179,405

Thanks a lot for your help – Saikat Jun 13 '16 at 02:07
You're welcome. My pleasure. -Mark – Mark Viola Jun 13 '16 at 02:08
I just have a doubt. Doesn't Cauchy have another theorem which says that the integral around any closed path is zero ? – Saikat Jun 14 '16 at 02:09
@user230452 You are correct. Cauchy's Integral Theorem states precisely that. Cauchy's Integral Formula is the one used here. – Mark Viola Jun 14 '16 at 13:46
If you used, Cauchy's theorem instead, the integral would be zero ? – Saikat Jun 15 '16 at 03:00
Cauchy's Integral Theorem does not apply since the integrand is not analytic in the region $|z|\le 3$ – Mark Viola Jun 15 '16 at 04:21

Evaluation of complex integral

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