Find the Laurent series of $f(z) = \frac{1}{z-2} + \frac{1}{z-3}$ for $2 < |z| < 3$ and for $|z| > 3$

Question

Find the Laurent series of $f(z) = \frac{1}{z-2} + \frac{1}{z-3}$ for $2 < |z| < 3$ and for $|z| > 3$.

Is the first step here to notice that $$ \frac{1}{z-2} + \frac{1}{z-3} = \frac{2z-5}{(z-2)(z-3)}$$ and compute the Laurent series as a single term rather than in two parts?

Answer 1

Case I. $2<\lvert z\rvert<3$. $$ \frac{1}{z-2}+\frac{1}{z-3}=\frac{1}{z}\cdot\frac{1}{1-\frac{2}{z}}-\frac{1}{3}\frac{1}{1-\frac{z}{3}}=\sum_{n=1}^\infty\frac{2^{n-1}}{z^n}-\sum_{n=0}^\infty \frac{z^n}{3^{n+1}}. $$ Case II. $\lvert z\rvert>3$. $$ \frac{1}{z-2}+\frac{1}{z-3}=\frac{1}{z}\cdot\frac{1}{1-\frac{2}{z}}+ \frac{1}{z}\cdot\frac{1}{1-\frac{3}{z}}=\sum_{n=1}^\infty\frac{2^{n-1}+3^{n-1}}{z^n}. $$

Answer 2

Hint: Instead of combining the fractions, think geometric series.

Answer 3

Hint: This is the first step $$\frac{1}{z-2} =\frac{1}{z}\Big(\frac{1}{1-\frac{2}{z}}\Big) \ \text{and}\ \ \frac{1}{3} \Big(\frac{1}{1-\frac{z}{3}}\Big)\\ \frac{1}{z-3} = \frac{1}{z} \Big(\frac{1}{1-\frac{3}{z}}\Big)\ \ \text{and}\ \ \frac{1}{2} \Big(\frac{1}{1-\frac{z}{2}}\Big)$$