- 2073 Bhadra · 1+4 marks
Zero of order m; poles and residues
Short answer
Zero of order m: f(z)=(z−a)^m g(z) with g analytic and g(a)≠0. f(z)=(1+z)/((z+2)(1−z)²) has a simple pole at z=−2 with residue −1/9 and a double pole at z=1 with residue 1/9.
Define zero of order m of function of complex variable. Determine the poles and residue at poles of the function f(z) = (1 + z)/((z + 2)(1 - z)²).
Zero of order m
A point z = a is a zero of order m of an analytic function f(z) if f(a) = f′(a) = … = f⁽ᵐ⁻¹⁾(a) = 0 but f⁽ᵐ⁾(a) ≠ 0. Equivalently f(z) = (z − a)ᵐ g(z) where g is analytic and g(a) ≠ 0. A zero of order 1 is a simple zero.
Poles and residues of f(z) = (1 + z)/((z + 2)(1 − z)²)
The denominator vanishes at z = −2 and z = 1; the numerator is non-zero at both (−1 and 2). So:
- z = −2 is a simple pole.
- z = 1 is a pole of order 2 (since (1 − z)² = (z − 1)²).
Residue at z = −2 (simple pole):
Res = lim (z → −2) (z + 2) f(z) = (1 + z)/(1 − z)² at z = −2 = (−1)/9 = −1/9
Residue at z = 1 (double pole):
Res = d/dz [ (z − 1)² f(z) ] at z = 1 = d/dz [ (1 + z)/(z + 2) ] at z = 1 = [ (z + 2) − (1 + z) ] / (z + 2)² = 1/(z + 2)² = 1/9
Answer: simple pole at z = −2 with residue −1/9; double pole at z = 1 with residue 1/9. (Their sum is 0, as expected since f(z) ~ 1/z² at infinity.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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