Question

Find series expansion of 1/ z^2-3z+2 in region |z| <2

Answer 1

Expand the given expression into partial fractions.

$\dfrac1{z^2 - 3z + 2} = \dfrac1{(z-2)(z-1)} = \dfrac1{(2 - z) (1 - z)} = \dfrac1{1 - z} - \dfrac1{2 - z}$

Recall the infinite geometric series,

$\displaystyle \sum_{n=0}^\infty z^n = \dfrac1{1-z}$

which converges in the unit disk |z| < 1.

The unit disk is a subset of the given region |z| < 2, so there are no issues with the convergence of series expansion of the first expression. For the second expression, we rearrange terms as

$\dfrac1{2 - z} = \dfrac12 \times \dfrac1{1 - \frac z2}$

Then if |z/2| < 1, or equivalently |z| < 2, the series expansion for this term is

$\displaystyle \frac1{2-z} = \frac12 \sum_{n=0}^\infty \left(\frac z2\right)^n = \sum_{n=0}^\infty \frac{z^n}{2^{n+1}}$

Putting everything together, the series expansion of the given expression over |z| < 2 is

$\displaystyle \frac1{z^2-3z+2} = \sum_{n=0}^\infty \left(1 + \frac1{2^{n+1}}\right) z^n$