Poles and residues of z²/((z-1)²(z+2))

f(z) = z²/((z − 1)²(z + 2)).

The numerator is non-zero at z = 1 and z = −2, so:

• z = 1 is a pole of order 2 (double pole).
• z = −2 is a simple pole.

Residue at z = −2 (simple):

Res = lim (z → −2) (z + 2) f(z) = z²/(z − 1)² at z = −2 = 4/9

Residue at z = 1 (order 2):

Res = d/dz [ (z − 1)² f(z) ] at z = 1 = d/dz [ z²/(z + 2) ] at z = 1

d/dz [ z²/(z + 2) ] = [ 2z(z + 2) − z² ] / (z + 2)² = (z² + 4z)/(z + 2)²

At z = 1: (1 + 4)/9 = 5/9.

Answer: Double pole at z = 1 with residue 5/9; simple pole at z = −2 with residue 4/9.

Check: f(z) ~ 1/z for large z, so the sum of all residues must be 1: 5/9 + 4/9 = 1.