Expand the given expression into partial fractions.
![\frac1{z^2 - 3z + 2} = \frac1{(z-2)(z-1)} = \frac1{(2 - z) (1 - z)} = \frac1{1 - z} - \frac1{2 - z}](https://img.qammunity.org/2023/formulas/physics/college/fqo2oskisnk9szo1blmoeuo14h5khdlpzl.png)
Recall the infinite geometric series,
![\displaystyle \sum_(n=0)^\infty z^n = \frac1{1-z}](https://img.qammunity.org/2023/formulas/physics/college/x0hb0gdhqslppdgdxqj4gt7v866s9lp1g2.png)
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
![\frac1{2 - z} = \frac12 * \frac1{1 - \frac z2}](https://img.qammunity.org/2023/formulas/physics/college/7jeyehf11hlmedwgmkg0qxamaxb7343jx1.png)
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 (z^n)/(2^(n+1))](https://img.qammunity.org/2023/formulas/physics/college/zdyshrrkuzoz6qokfultodajpl44m198zv.png)
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](https://img.qammunity.org/2023/formulas/physics/college/659fcm7zfxal90uwsf8eeqibkk2q1lyee3.png)