Inverse z of (z-4)/((z-1)(z-2)²), inversion integral

Inversion integral method. x(k) = (1/2πj) ∮ X(z) z^{k−1} dz = sum of residues of X(z) z^{k−1} at all poles inside the contour (which encloses every pole).

Here X(z) = (z − 4)/[(z − 1)(z − 2)²], so

X(z) z^{k−1} = (z − 4) z^{k−1} / [(z − 1)(z − 2)²]

Poles: z = 1 (simple), z = 2 (double), and z = 0 for k = 0 only.

Residue at z = 1 (simple pole)

R₁ = (z − 4) z^{k−1} / (z − 2)²  at z = 1
   = (−3)(1)/(1) = −3

Residue at z = 2 (double pole)

R₂ = d/dz [ (z − 4) z^{k−1} / (z − 1) ] at z = 2

d/dz = z^{k−1}/(z−1) + (k−1)(z−4) z^{k−2}/(z−1) − (z−4) z^{k−1}/(z−1)²
at z = 2:  2^{k−1} − 2(k−1) 2^{k−2} + 2 · 2^{k−1}
         = 2^{k−1} [ 1 − (k − 1) + 2 ]
         = (4 − k) 2^{k−1}

Result

For k ≥ 1:

x(k) = −3 + (4 − k) 2^{k−1}

For k = 0 there is an extra pole at z = 0 with residue (−4)/[(−1)(4)] = 1, giving x(0) = −3 + 2 + 1 = 0 (as expected, because X(z) → 0 when z → ∞).

Check: x(0), x(1), x(2), … = 0, 0, 1, 1, −3, −19, −67, … Expanding X(z) in powers of z⁻¹ gives the same sequence.

Answer: x(k) = (4 − k) 2^{k−1} − 3 for k ≥ 1, and x(0) = 0.