Laurent Series expansion of $f(z)=\frac{z+8}{z^2\cdot(z-2-2i)}$

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Determine the principal part of the Laurent series expansion for $$f(z)=\frac{z+8}{z^2\cdot(z-2-2i)}$$ on $\{z\in\mathbb{C}:1<\vert z\vert < 2\}$.

I think that this is not a difficult on, but I can't find the right beginning. First of all I want get this function in partial fractions, but the didn't help me a lot. I know that I have to rearrange this term into $$\frac{1}{1-...}$$ to use the geometric series.

Any hints for me to do so? Thank you a lot!

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Hint: Rewrite as $$-\frac1{2(1+i)z^2}\cdot\frac{8+z}{1-\cfrac z{2(1+i)}}=-\frac{1-i}{4z^2}\cdot\frac{8+z}{1-\cfrac{(1-i)z}{4}}.$$