Laurent series of 1/(z²-3z+2), 1<|z|<2

Taylor's theorem

If f(z) is analytic inside and on the circle C: |z − a| = R, then for every z inside C,

f(z) = Σ_{n=0}^{∞} aₙ (z − a)ⁿ, where aₙ = f⁽ⁿ⁾(a)/n! = (1/2πi) ∮_C f(ζ)/(ζ − a)ⁿ⁺¹ dζ.

Laurent series in 1 < |z| < 2

f(z) = 1/(z² − 3z + 2) = 1/((z − 1)(z − 2)).

Partial fractions: 1/((z − 1)(z − 2)) = 1/(z − 2) − 1/(z − 1).

The poles are at z = 1 and z = 2; in the ring 1 < |z| < 2 we have |z| > 1 and |z| < 2, so expand each fraction in the matching way.

Term 1: 1/(z − 2). Since |z/2| < 1:

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

Term 2: 1/(z − 1). Since |1/z| < 1:

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

Result

f(z) = 1/(z − 2) − 1/(z − 1)

f(z) = − Σ_{n=0}^{∞} zⁿ/2ⁿ⁺¹ − Σ_{n=1}^{∞} 1/zⁿ, for 1 < |z| < 2.

Written out:

f(z) = … − 1/z³ − 1/z² − 1/z − 1/2 − z/4 − z²/8 − …

The principal part (negative powers) comes from the pole at z = 1, and the analytic part (non-negative powers) from the pole at z = 2.