Contour integral $\int_{|z|=1}\exp(1/z)\sin(1/z)dz$

Question

Evaluate the contour integral $$\int_{|z|=1}\exp(1/z)\sin(1/z)\,dz$$ along the circle $|z|=1$ counterclockwise once.

The singularities are $\dfrac1{\pi k},k\in\mathbb{Z}$ plus the limit point $0$. So I can't apply the residue theorem. Any other alternative?

Answer 1

Note that $\vert z \vert = 1 \implies \vert 1/z \vert = 1$. Hence, setting $w=1/z$, we get that \begin{align} \int_{\vert z \vert = 1} \exp(1/z) \sin(1/z) dz & = \overbrace{\underbrace{\int_{\vert w \vert = 1} \exp(w) \sin(w) \dfrac{dw}{w^2}}_{\text{by change in orientation of integral}}}^{\text{Negative sign gets cancelled}}\\ & = \int_{\vert w \vert = 1} \dfrac{\left(1+ \mathcal{O}(w)\right)(1+\mathcal{O}(w^2))}w dw = 2 \pi i \end{align}