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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.

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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$$\frac1{5-2(z+1/z)}=\frac1{3(2z-1)}-\frac2{3(z-2)}.$$

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  • $\begingroup$ Yeah, I see. I had extra $-$. Thanks for the answer! $\endgroup$ Commented Feb 11, 2021 at 12:39

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