I am doing contour integration, and I'm trying to solve a couple problems. I am not sure if I am doing them correctly, so if anyone can explain the steps to me, it would be much appreciated:
$(a)\int_{|z| = 1}\frac{\sin z}{z}dz$ , counterclockwise
$(b) \int_{|z| = 2}\frac{1}{z^2-1}dz$, counterclockwise
For (a), I think I should use the Cauchy formula, $f(z_0) = \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z – z_0}dz$, so that I could set $z_0 = 0$, which gives me $f(0)2\pi i = \int_\gamma \frac{\sin z}{z-0}dz = \sin(0)2\pi i = 0$. I think that another way to solve it is to parameterize, but I have not gotten past $\int_{|z| = 1}\frac{\sin z}{z}dz = \int_0^{2\pi} \frac{\sin(e^{i\theta})}{e^{i\theta}}\frac{ie^{i\theta}\cos(e^{i\theta}) – \sin(e^{i\theta})}{e^{i\theta}}d\theta$.
For (b), I separated out the denominator to get $\int_{|z| = 2} \frac{1}{(z-1)(z+1)}dz$ and was not sure where to go from there.
As a quick side question, when is it most appropriate to use the Cauchy formula or to parameterize? And when do I need to worry about the points where the function does not exist (i.e. in (a), z = 0 and in (b), z = 1, -1)? And are there other ways to compute these integrals? All help is appreciated.
Best Answer
Your computation of the first integral, through Cauchy's integral formula, is just fine. It's not amazing that you cannot compute it from the definition of contour integral.
Concerning the second one, now you use the fact that$$\frac1{(z-1)(z+1)}=\frac{1/2}{z-1}-\frac{1/2}{z+1}.$$