- 2069 Poush
Singular points, poles, residue of z²/((z-2)(z²+1))
Short answer
A singular point is where f is not analytic; a pole is an isolated singularity where f(z)=g(z)/(z−a)^m with g(a)≠0. f(z)=z²/(( z−2)(z²+1)) has simple poles at z=2 (residue 4/5), z=i (residue (1−2i)/10), z=−i (residue (1+2i)/10).
Define singular points and poles. Compute the residue of f(z) = z²/((z - 2)(z² + 1)) at its pole(s).
Definitions
- Singular point: a point z = a where f(z) is not analytic (it fails to be differentiable there), although every neighbourhood of a contains points where f is analytic. An isolated singular point has a punctured neighbourhood in which f is analytic.
- Pole: an isolated singular point z = a where f(z) = g(z)/(z − a)ᵐ with g analytic and g(a) ≠ 0. The integer m is the order of the pole (m = 1: simple pole). Near a pole, |f(z)| → ∞.
Residues of f(z) = z²/((z − 2)(z² + 1))
Denominator = (z − 2)(z − i)(z + i). The numerator z² does not vanish at 2, i or −i, so there are three simple poles: z = 2, z = i, z = −i.
For a simple pole, Res at a = lim (z → a) (z − a) f(z).
- At z = 2: Res = z²/(z² + 1) = 4/5
- At z = i: Res = z²/((z − 2)(z + i)) = i²/((i − 2)(2i)) = −1/(−2 − 4i) = 1/(2 + 4i) = (2 − 4i)/20 = (1 − 2i)/10
- At z = −i: Res = z²/((z − 2)(z − i)) = (−1)/((−i − 2)(−2i)) = −1/(−2 + 4i) = 1/(2 − 4i) = (1 + 2i)/10
Check: sum = 4/5 + (1 − 2i)/10 + (1 + 2i)/10 = 4/5 + 1/5 = 1, which equals the coefficient of 1/z in f at infinity (f ~ 1/z), as it should.
Answer: Res(2) = 4/5, Res(i) = (1 − 2i)/10, Res(−i) = (1 + 2i)/10.
✓ Answer checked by Claude Opus 5.5 (high effort)
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