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

Taylor's theorem

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

f(z) = Σ (n = 0 to ∞) aₙ (z − a)ⁿ, where aₙ = f⁽ⁿ⁾(a)/n! = (1/2πi) ∮_C f(w)/(w − a)ⁿ⁺¹ dw.

Laurent series of f(z) = z/((z + 1)(z + 2)) for 1 < |z| < 2

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

In the region 1 < |z| < 2 we have |1/z| < 1 and |z/2| < 1, so expand each term in the convergent form.

First term (|z| > 1, use powers of 1/z):

−1/(z + 1) = −(1/z) · 1/(1 + 1/z) = −(1/z)[1 − 1/z + 1/z² − …] = −1/z + 1/z² − 1/z³ + … = Σ (n = 1 to ∞) (−1)ⁿ z⁻ⁿ

Second term (|z| < 2, use powers of z/2):

2/(z + 2) = 1/(1 + z/2) = 1 − z/2 + z²/4 − z³/8 + … = Σ (n = 0 to ∞) (−1)ⁿ zⁿ/2ⁿ

Answer:

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

= … − 1/z³ + 1/z² − 1/z + 1 − z/2 + z²/4 − z³/8 + …

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