Is the residue of $\cos(\frac{z}{1-z})$ at z=1 : sin(1)?
i.e $\frac{1}{2\pi i}\int_c\cos(\frac{z}{1-z})dz=Res(\frac{1}{z^2}f(\frac{1}{z}),0)$=$\frac{1}{2\pi i}\int_c\frac{\cos(\frac{1}{z-1})}{z^2}dz$
Just wanted to clarify the method.
Edit
Using the following theorem:
Note that\begin{align}\cos\left(\frac z{1-z}\right)&=\cos\left(-1-\frac1{1-z}\right)\\&=\cos\left(1-\frac1{z-1}\right)\\&=\cos(1)\cos\left(\frac1{z-1}\right)+\sin(1)\sin\left(\frac1{z-1}\right)\\&=\cos(1)\left(1-\frac1{2!}\frac1{(z-1)^2}+\cdots\right)+\sin(1)\left(\frac1{z-1}-\frac1{3!}\frac1{(z-1)^3}+\cdots\right)\end{align}and therefore the residue is $\sin(1)$, since that's the coefficient of $\frac1{z-1}$.