How to compute the residue of $(z^2+2z+1)\sin\left(\frac{1}{1+z}\right)$

Question

This was an example given in my notes but all it concluded was with something about an infinite principal part. How do we compute it? we have it equal to $ \left( z + 1 \right)^2 \cdot \sin \left( (z+1)^{-1} \right) = \left( z + 1 \right)^2 \cdot \left[ (z+1)^{-1} + \cdots \right] $. That's all I could get. What now?

Answer 1

$z=-1$ is an essential singularity of the function. So we have $$(z^2+2z+1)\sin \frac{1}{1+z}=(z+1)^2\sin \frac{1}{1+z} $$ therefore $$(z+1)^2\sin \frac{1}{1+z}=(z+1)^2 \left(\frac{1}{1+z}-\frac{1}{3!(1+z)^3}+\frac{1}{5!(1+z)^5}-\cdots\right)$$ this shows that Res$[(z+1)^2\sin \frac{1}{1+z}, -1]=-\frac{1}{3!}$

Answer 2

The function $$ f(z)=(z+1)^{2}\sin(\frac{1}{z+1}) $$ has an essential singularity at $z=-1$, and the residue should be $-\frac{1}{6}$. This can be read from the Taylor expansion explicitly. Sorry for the nonsense wrote earlier.