Find residue of $(z-2) \cos (1/(z-1)) $

Question

I want to find the residue around $z_0 = 1 $ for the following function: $f(z) = (z-2) \cos\left(\frac{1}{z-1}\right)$ I've made $\cos (\frac{1}{z-1})$ like this: $\cos\left(\frac{1}{z-1}\right) = 1 - \frac {1}{2!(z-1)^2} + \frac {1}{4!(z-1)^4}- \frac {1}{6!(z-1)^6} $

So, I have $\cos\left(\frac{1}{z-1}\right) = (z-2) - \frac {(z-2)}{2!(z-1)^2} + \frac {(z-2)}{4!(z-1)^4}- \frac {(z-2)}{6!(z-1)^6} + .....$

I don't know how to go on though. Any ideas ?

Answer 1

Now, you use the fact that $z-2=(z-1)-1$:\begin{align}(z-2)\cos\left(\frac1{z-1}\right)&=\bigl((z-1)-1\bigr)\cos\left(\frac1{z-1}\right)\\&=\bigl((z-1)-1\bigr)\left(1-\frac1{2!(z-1)^2}+\frac1{4!(z-1)^4}+\cdots\right)\\&=(z-1)-1-\frac1{2!(z-1)}-\frac1{2!(z-1)^2}+\frac1{4!(z-1)^3}+\cdots\end{align}Therefore,$$\operatorname{res}_{z=1}\left((z-2)\cos\left(\frac1{z-1}\right)\right)=-\frac12.$$

Answer 2

Just write a few terms...:

$$(z-2)\cos\frac1{z-1}=\left[(z-1)-1\right]\left(1-\frac1{(2(z-1)^2}+\frac1{24(z-1)^4}-\ldots\right)=$$

$$=(z-1)-\frac1{2(z-1)}+\frac1{24(z-1)^3}-\ldots-1+\frac1{(2(z-1)^2}-\frac1{24(z-1)^4}+\ldots=$$

$$=\ldots-\frac1{2(z-1)}+\ldots$$

and thus the residue at $\;z=1\;$ is $\;-\cfrac12\;$ .