Find the integral $\int_{\lambda}\frac{\sin(z)}{z(z-1)}$ where $\lambda(t) = 10e^{it},t\in[0,2\pi]$
We notice that there are poles at $z = 0$ and $z=1$. So we can use residue theorem but I am confused as to how to handle the coefficient 10 in $10e^{it}$.
The contour if integration is a circle of radius 10 centered at the origin. Both singularities at $z=0$ and $z=1$ lie inside the circle. The singularity at the origin is a removable singularity, as $\lim_{z\rightarrow 0}\frac{\sin(z)}{z}=1 $, so the residue is zero. The residue at $z=1$ is $\lim_{z\rightarrow 1}(z-1)\frac{\sin(z)}{z(z-1)}=\frac{\sin(1)}{1}$. Therefore the integral is just $2i\pi\sin(1)$.