# Laurent series example

I have been trying to understand Laurent series expansion in complex analysis and I need someone's confirmation that what I'm doing is right.

Expand the function $$f(z) = \frac{1}{5-2(z+\frac{1}{z})}$$ on a disk $$\frac{1}{2} < |z| < 2$$.

My approach was to break this fraction onto smaller parts using partial fractions and I got the following: $$\frac{-z}{(z-2)(z+2)} = \frac{-1}{2z-4}-\frac{1}{2z+4}$$
Next thing I think I should do is find Laurent expansion for first fraction on $$\frac{1}{2} < |z|$$ and then $$|z| <2$$. After that I would do same for second fraction and finally I would have two expansions of this function, one on $$\frac{1}{2} < |z|$$ and second one on $$|z| <2$$.

Is this the right approach? This has been confusing me a lot lately. Thanks for any tips in advance.

The global approach is correct and that's how I would do it. But your computations concerning the given function are wrong. In fact, we have$$\frac1{5-2(z+1/z)}=\frac1{3(2z-1)}-\frac2{3(z-2)}.$$
• Yeah, I see. I had extra $-$. Thanks for the answer! Feb 11, 2021 at 12:39