The problem is: Let a and b be complex numbers such that $0<|a|<|b|$. Find a series in positive and negative powers of z that represent the function $\frac{1}{(z-a)(z-b)}$ in the annulus $|a|<|z|<|b|$
I understand these steps:
- Use partial fraction decomposition to write $\frac{1}{(z-a)(z-b)}$ as $\frac{1}{(a-b)(z-a)} - \frac{1}{(a-b)(z-b)}$
- $\frac{1}{(a-b)(z-a)}$ is equivalent to $\frac{1}{z(a-b)}\sum_0^∞\frac{a^n}{z^n}$ when $|a|<z$.
- $\frac{1}{(a-b)(z-b)}$ is equivalent to $\frac{1}{z(a-b)}\sum_1^∞\frac{z^n}{b^n}$ when $z<|b|$.
But where do I go from here? How do I combine these values to get one Laurent series for this function? Could you please explain the steps when you do them? I have seen the answer ($\frac{1}{(z-a)(z-b)} = \sum_{-∞}^∞a_nz^n$ in the annulus $|a|<|z|<|b|$) but do not understand how to get there...
When $|a| <|z| <|b|$ both the series expansions are valid. Hence the required Laurent series is $ \sum\limits_{n=0}^{\infty}\frac {a^{n}} {(a-b)z^{n+1}}-\sum\limits_{n=1}^{\infty} \frac {z^{n-1}} {(a-b)b^{n}}$. (This series is of the form $\sum\limits_{n=-\infty}^{\infty} c_n z^{n}$ and since Laurent series is unique this must be the Laurent series of the given function. Explicitly, $c_n=\frac 1 {(a-b)b^{n+1}}$ for $n \geq 0$ and $c_n=\frac {a^{-n-1}} {(a-b)}$ for $n <0$.