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

f(z) = (z² − 1)/((z + 2)(z + 3)).

Numerator and denominator have the same degree, so first divide:

(z² − 1)/(z² + 5z + 6) = 1 − (5z + 7)/((z + 2)(z + 3))

Partial fractions: (5z + 7)/((z + 2)(z + 3)) = A/(z + 2) + B/(z + 3)

• A = (5(−2) + 7)/(−2 + 3) = −3
• B = (5(−3) + 7)/(−3 + 2) = 8

so f(z) = 1 + 3/(z + 2) − 8/(z + 3).

In the ring 2 < |z| < 3: |2/z| < 1 and |z/3| < 1.

Term 3/(z + 2) (|z| > 2, powers of 1/z):

3/(z + 2) = (3/z) · 1/(1 + 2/z) = 3 Σ (n = 0 to ∞) (−2)ⁿ / zⁿ⁺¹ = 3/z − 6/z² + 12/z³ − …

Term −8/(z + 3) (|z| < 3, powers of z/3):

−8/(z + 3) = −(8/3) · 1/(1 + z/3) = −(8/3) Σ (n = 0 to ∞) (−1)ⁿ zⁿ/3ⁿ = −8/3 + 8z/9 − 8z²/27 + …

Answer:

f(z) = 1 + 3 Σ (n = 0 to ∞) (−2)ⁿ z^(−n−1) − (8/3) Σ (n = 0 to ∞) (−1)ⁿ zⁿ/3ⁿ

i.e. f(z) = … + 12/z³ − 6/z² + 3/z − 5/3 + 8z/9 − 8z²/27 + …

(constant term: 1 − 8/3 = −5/3).