# Laurent series of a cosine function

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.

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