Laurent series of a cosine function

Question

I want to find Laurent series of the complex function $$f(z) = \cos\left(\frac{z^2-4z}{(z-2)^2}\right)$$ at $z_0=2$. I will be thankful for any hints.

Answer 1

Hint. You can substitute

$$\frac{z^2-4z}{(z-2)^2}=1-\frac{4}{(z-2)^2}$$

into the series for $\cos z$ (which is convergent for all $z\in\mathbf{C}$): $$\sum_{n\geqslant 0}\frac{(-1)^n}{(2n)!}z^{2n}.$$

This will give the Laurent series centered at $z=2$.