First four terms of Taylor series about z = 2

Taylor series: f(z) = Σ aₙ (z − a)ⁿ about the centre z = a = 2, valid for |z − 2| < R, R = distance to the nearest singularity.

f(z) = (z + 1)/((z − 3)(z − 4)) has poles at 3 and 4. The nearest is z = 3, at distance 1, so R = 1.

Partial fractions

(z + 1)/((z − 3)(z − 4)) = A/(z − 3) + B/(z − 4)

A = (3 + 1)/(3 − 4) = −4, B = (4 + 1)/(4 − 3) = 5.

f(z) = −4/(z − 3) + 5/(z − 4).

Expand about z = 2

Let t = z − 2. Then z − 3 = t − 1 and z − 4 = t − 2.

−4/(t − 1) = 4/(1 − t)          = 4 Σ tⁿ                (|t| < 1)
 5/(t − 2) = −(5/2)/(1 − t/2)   = −(5/2) Σ (t/2)ⁿ       (|t| < 2)

f(z) = Σ_{n=0}^{∞} [ 4 − 5/2ⁿ⁺¹ ] (z − 2)ⁿ

First four terms

n = 0 : 4 − 5/2  = 3/2
n = 1 : 4 − 5/4  = 11/4
n = 2 : 4 − 5/8  = 27/8
n = 3 : 4 − 5/16 = 59/16

f(z) = 3/2 + (11/4)(z − 2) + (27/8)(z − 2)² + (59/16)(z − 2)³ + …, |z − 2| < 1.

Check: f(2) = 3/((−1)(−2)) = 3/2 ✓.