Question

For 𝑓(𝑧) =1/((𝑧+2)(3−𝑧)), find the Laurent series valid for |𝑧 + 2| < 5.

Answer 1

First split the function into partial fractions:
\[\frac{1}{(z+2)(3-z)}=\frac{1}{5}\left(\frac{1}{z+2}+\frac{1}{3-z}\right)\]
The first term is already a function of \(\dfrac{1}{z+2}\). The second term can be rewritten as a geometric series:
\[\frac{1}{3-z}=\frac{1}{5-(z+2)}=-\frac{1}{z+2}\frac{1}{1-\frac{5}{z+2}}=-\frac{1}{z+2}\sum_{n=0}^\infty\left(\frac{5}{z+2}\right)^n\]It should be clear that the summation only holds for \(\left|\dfrac{5}{z+2}\right|<1\), or \(|z+2|>5\), as required.

Answer 2

Whoops, I meant \(>1\) in that first inequality, so it should be valid for \(|z+2|<5\).