# Contour Integral of $\frac{\sin z}{z}$ about the unit circle (counterclockwise)

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.

Concerning the second one, now you use the fact that$$\frac1{(z-1)(z+1)}=\frac{1/2}{z-1}-\frac{1/2}{z+1}.$$
• By your second comment, do you mean that I should do the algebra, or that I should use the fact that $e^{i\theta}$ is never 0? There is a theorem that states the integral of a closed contour over an analytic function is 0, I think. Is this correct? If so, is it what I should be using? – johndisk Oct 9 at 17:19
\begin{aligned} \int_\gamma {\frac{\sin z}{z} \mathrm{d}z} & = \int_{0}^{2\pi} {\frac{\sin(e^{i\theta})}{e^{i\theta}} \mathrm{d}(e^{i\theta})} = i\int_{0}^{2\pi} {\sin(e^{i\theta}) \>\mathrm{d}\theta}\\ & = i\int_{0}^{\pi} {\sin(e^{i\theta}) \>\mathrm{d}\theta} + i\int_{\pi}^{2\pi} {\sin(e^{i\theta}) \>\mathrm{d}\theta}\\ & = i\int_{0}^{\pi} {(\sin(e^{i\theta}) + \sin(e^{i(2\pi-\theta)})) \>\mathrm{d}\theta}\\ & = i\int_{0}^{\pi} {(\sin(e^{i\theta}) + \sin(e^{-i\theta)})) \>\mathrm{d}\theta}\\ & = 2i\int_{0}^{\pi} {\left(\sin\left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right) \cos\left(\frac{e^{i\theta}-e^{-i\theta}}{2}\right)\right) \mathrm{d}\theta}\\ & = 2i\int_{0}^{\pi} {(\sin(\cos\theta) \cos(i\sin\theta)) \>\mathrm{d}\theta}\\ & = 2i\int_{0}^{\pi/2} {(\sin(\cos\theta) \cos(i\sin\theta)) \>\mathrm{d}\theta} + 2i\int_{\pi/2}^{\pi} {(\sin(\cos\theta) \cos(i\sin\theta)) \>\mathrm{d}\theta}\\ & = 2i\int_{0}^{\pi/2} {(\sin(\cos\theta) \cos(i\sin\theta) + (\sin(\cos(\pi-\theta)) \cos(i\sin(\pi-\theta))) \>\mathrm{d}\theta}\\ & = 0 \end{aligned}