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

f(z) = 1/((1 + z²)(z + 2)), region 1 < |z| < 2.

Partial fractions: 1/((z² + 1)(z + 2)) = A/(z + 2) + (Bz + C)/(z² + 1).

1 = A(z² + 1) + (Bz + C)(z + 2). Putting z = −2 gives A = 1/5. Comparing z²: A + B = 0, so B = −1/5. Comparing constants: A + 2C = 1, so C = 2/5.

f(z) = (1/5) [ 1/(z + 2) + (2 − z)/(z² + 1) ]

In the ring 1 < |z| < 2: |z| > 1 so expand in 1/z for the z² + 1 factor, and |z| < 2 so expand in z/2 for the z + 2 factor.

1/(z + 2) (|z/2| < 1):

1/(z + 2) = (1/2) · 1/(1 + z/2) = (1/2) Σ (n = 0 to ∞) (−1)ⁿ zⁿ/2ⁿ = 1/2 − z/4 + z²/8 − …

(2 − z)/(z² + 1) (|1/z²| < 1):

1/(z² + 1) = (1/z²) · 1/(1 + 1/z²) = Σ (n = 0 to ∞) (−1)ⁿ z^(−2n−2) = 1/z² − 1/z⁴ + 1/z⁶ − …

so (2 − z)/(z² + 1) = −1/z + 2/z² + 1/z³ − 2/z⁴ − 1/z⁵ + 2/z⁶ + … = Σ (n = 0 to ∞) (−1)ⁿ [ 2 z^(−2n−2) − z^(−2n−1) ]

Answer:

f(z) = (1/5) { Σ (n = 0 to ∞) (−1)ⁿ zⁿ / 2ⁿ⁺¹ + Σ (n = 0 to ∞) (−1)ⁿ [ 2 z^(−2n−2) − z^(−2n−1) ] }

= (1/5) [ … + 2/z² − 1/z + 1/2 − z/4 + z²/8 − … ]