Calculate the complex integral $\oint_{|z|=1}\sin{\frac{1}{z}} dz$

Question

How do I calculate this complex integral?

$$\displaystyle\oint_{|z|=1}\sin\left ({\displaystyle\frac{1}{z}}\right ) dz$$

I made the Taylor series for this: $$\displaystyle\sum_{n=0}^\infty \displaystyle\frac{(-1)^n*(1/z)^{2n+1}}{(2n+1)!}$$

But now I don't know what to do.

Answer 1

You can either apply the Residue Theorem or use the power series of $\sin$. Using the former we have \begin{eqnarray} \int_{|z|=1}\sin\left(\frac{1}{z}\right)\,dz&=&\int_{|z|=1}\frac{1}{z}\,dz+\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)!}\int_{|z|=1}\frac{1}{z^{2n+1}}\,dz\\ &=&\int_{|z|=1}\frac{1}{z}\,dz=\int_0^{2\pi}\frac{ie^{it}}{e^{it}}\,dt=2i\pi. \end{eqnarray}

Answer 2

Here is a trick. Let $z'=\frac{1}{z}$. Then $$dz=d(\frac{1}{z'})=\frac{-dz'}{z'^{2}}$$

So all you need is to integrate $$-\int_{|z'|=1}\frac{\sin[z']}{z'^{2}}dz'$$

But you can evaluate it by Cauchy's integral formula by noticing this is $\sin[z']'$ at $z'=0$ times some constant.