Laurent series expansion

Resolve into partial fractions: z² − 3z + 2 = (z − 1)(z − 2).

1/((z − 1)(z − 2)) = 1/(z − 2) − 1/(z − 1)

In the annulus 1 < |z| < 2 we have |z/2| < 1 and |1/z| < 1. Expand each term accordingly.

1/(z − 2) = −1/(2 − z) = −(1/2) · 1/(1 − z/2)
          = −(1/2) Σ_{n=0}^∞ (z/2)ⁿ = − Σ_{n=0}^∞ zⁿ/2^(n+1)

1/(z − 1) = (1/z) · 1/(1 − 1/z) = Σ_{n=0}^∞ 1/z^(n+1)

Subtracting:

f(z) = − Σ_{n=0}^∞ zⁿ/2^(n+1) − Σ_{n=1}^∞ 1/zⁿ
     = ... − 1/z³ − 1/z² − 1/z − 1/2 − z/4 − z²/8 − ...

Answer: f(z) = − Σ_{n=0}^∞ zⁿ/2^(n+1) − Σ_{n=1}^∞ z^(−n), valid for 1 < |z| < 2. The principal part has infinitely many negative powers because z = 1 is a singularity inside the inner circle.