Let $\Lambda$ be a circle centered at zero on $\mathbb{C}\text{\ }\{0\}$. The value of $$\int_{\Lambda}\left(\sin\frac1z+\sin^2\frac 1z\right)dz\ \ \ \ \ \text{is equal to}$$
$(a) \ 0 \ \ \ \ \ \ (b)\ \ \pi i\ \ \ (c) \ \ 2\pi i \ \ \ \ \ \ (d)\ \ 4\pi i$
I am really confused how to find the poles & then apply residue theorem.
Someone please help me solve this integral.
On
Instead of Laurent expansion, another way to evaluate the integral is compute the residue at "infinity".
To achieve this, we change variable to $w = \frac1z$.
For any $r > 0$, let $\Lambda(r)$ be a circular contour of radius $r$ in counterclockwise orientation. Let $-\Lambda(r)$ be the same contour but with orientation reversed. In terms of $w$, we have.
$$\oint_{\Lambda(r)}\left(\sin\frac1z + \sin^2\frac1z\right) dz = \oint_{\color{red}{-}\Lambda(r^{-1})}\left(\sin w + \sin^2 w\right)\left(\color{red}{-}\frac{dw}{w^2}\right)\\ = \oint_{\Lambda(r^{-1})} \left[\frac{1}{w}\left(\frac{\sin w}{w}\right)+ \left(\frac{\sin w}{w}\right)^2\right] dw$$ Since $\displaystyle\;\frac{\sin w}{w}$ has a removable singularity at $w = 0$ with limiting value $1$, the residue of what inside the square bracket at $w = 0$ is $1$. This means the original contour integral over $z$ is $2\pi i$.
For all $z\ne0$ one has $$\sin{1\over z}={1\over z}-{1\over 6z^3}+{1\over 120 z^5}-\ldots\ .$$ whereby the series converges uniformly on any circle of radius $\rho>0$ around the origin. It follows that $$\int_\Lambda\sin{1\over z}\>dz=2\pi i\ .$$ The function $z\mapsto\sin^2{1\over z}$ is even, hence the integral along $\Lambda$ of this function is $0$. It follows that option c) is correct.