How to integrate $\int_{|z|=4} \frac {z} {\sin z}\,\mathrm dz$?

Question

How can I integrate $\displaystyle\int_{|z|=4} \frac {z} {\sin z}\,\mathrm dz$?

The problem is that $\sin z$ has three zeros in $|z|=4$ and hence the integrand is not analytic there. How can I overcome it? Please help me.

Thank you in advance.

Answer 1

Hint:

Use series expansion of $\sin$ : $$\dfrac{w}{\sin w}=\dfrac{w}{w-\frac{1}{3!}w^3+\frac{1}{5!}w^5-\cdots}=\dfrac{1}{1-\frac{1}{3!}w^2+\frac{1}{5!}w^4-\cdots}=1+\frac{1}{3!}w^2+\frac{7}{3\cdot5!}w^4+\cdots$$ so $$\dfrac{z}{\sin z}=\dfrac{-\frac{z(z-\pi)}{\sin{(z-\pi)}}}{z-\pi}=\dfrac{-z(1+\frac{1}{3!}(z-\pi)^2+\frac{7}{3.5!}(z-\pi)^4+\cdots)}{z-\pi}$$ also apply this for $-\pi$ and $0$ inside $|z|<4$.

Answer 2

Your integral is $0$. In order to prove that you don't need the residue theorem or even Cauchy's theorem. All you need is to notice that you are integrating an even function along a path which is symmetric with respect to the origin. Let $\gamma\colon[0,\pi]\longrightarrow\mathbb C$ be the path defined by $\gamma(t)=4e^{it}$. Let $f(z)=\frac{\sin z}z$. Then\begin{align}\int_{|z|=4}f(z)\,\mathrm dz&=\int_\gamma f(z)\,\mathrm dz+\int_{-\gamma}f(z)\,\mathrm dz\\&=\int_0^\pi f\bigl(\gamma(t)\bigr)\gamma'(t)\,\mathrm dt+\int_0^\pi f\bigl(-\gamma(t)\bigr)\bigl(-\gamma'(t)\bigr)\,\mathrm dt\\&=\int_0^\pi f\bigl(\gamma(t)\bigr)\,\mathrm dt-\int_0^\pi f\bigl(\gamma(t)\bigr)\gamma'(t)\mathrm dt\text{ (because $f$ is even)}\\&=0.\end{align}