Find the residue of $(z^2-1)\cos\frac {1}{z-1}$ at $z=1$.

Question

Question:

Find the residue of $(z^2-1)\cos\frac{1}{z-1} $ at $z=1$.

Attempt:

I tried to expand the series of $\cos\frac{1}{z-1}$ about $z=1$ and multiply through by $z^2-1$ but I couldn't isolate the $\frac{1}{z-1}$ term.

Answer 1

First of all, make a change of variables $w=z-1$ to reduce typing. Thus we want the residue of $$ (w^2+2w)\cos \frac1w $$ at $w=0$. The Laurent series of $\cos \frac1w$ is $$ \cos \frac1w = 1 - \frac{1}{2!w^2} + \frac{1}{4!w^4} + \cdots $$ Since this is an even function and so is $w^2$, the $w^2\cos \frac1w$ won't pick up any $w^{-1}$-term, and we are left with just a $-1$ as the residue after we multiply by $2w$.

Answer 2

$$(1) \quad (z^2-1) \cdot \cos\left(\cfrac{1}{z-1}\right)$$

You can expand $(1)$ about the point $z=1$ by substituting the Taylor series of $\cos(x)$. We get,

$$(2) \quad \cfrac{-1}{z-1}+\cfrac{1}{24 \cdot (z-1)^2}+o(z^{-3})$$

So $(1)$ actually has an essential singularity rather than a pole at $z=1$. You could still say the residue is $-1$.

On the real line, we have,

From the complex line, we just have an infinite mess.