Evaluation of Integral via the Residue Theorem

Question

I was trying to solve some complex integrals via Contour Integration and found myself stuck with the following exercise: $$ I = \oint_{\gamma}z\sin\left(\frac{1+z}{1-z}\right)\mathrm{d}z,\,\gamma = 2e^{i\theta},\,\theta \in [0,\,2\pi]$$ There's a singularity inside the integration path, so my first thought was to use the residue theorem but then realized that wouldn't work since we're dealing with an essential singularity in $z=1$. I'd like to know how to properly treat the function inside the integral since I have little to no experience with these kind of functions. Thanks in advance

Answer 1

Yes, it's an essential singularity, but you can still apply the residue theorem. If $z\ne1$,\begin{align}z\sin\left(\frac{1+z}{1-z}\right)&=-z\sin\left(\frac{z+1}{z-1}\right)\\&=-z\sin\left(1+\frac2{z-1}\right)\\&=-((z-1)+1)\left(\sin(1)\cos\left(\frac2{z-1}\right)+\cos(1)\sin\left(\frac2{z-1}\right)\right)\\&=-((z-1)+1)\left(\sin(1)\left(1-\frac2{(z-1)^2}+\cdots\right)\right.+\\&\qquad+\left.\cos(1)\left(\frac2{z-1}-\frac4{3(z-1)^3}+\cdots\right)\right),\end{align}and therefore$$\operatorname{res}_{z=1}\left(z\sin\left(\frac{1+z}{1-z}\right)\right)=2\sin(1)-2\cos(1).$$So,\begin{align}\oint_\gamma z\sin\left(\frac{1+z}{1-z}\right)\,\mathrm dz&=2\pi i\bigl(2\sin(1)-2\cos(1)\bigr)\\&=4\pi i\bigl(\sin(1)-\cos(1)\bigr)\\&\approx3.7846i.\end{align}