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Engineering Mathematics → Complex Variables → Integrals via Residues
Last updated 5 September 2026
Question If
C C C is a circle
∣ z ∣ = 4 |z| = 4 ∣ z ∣ = 4 and
f ( z ) = z 2 ( z 2 − 3 z + 2 ) 2 f(z) = \frac{z^2}{(z^2-3z+2)^2} f ( z ) = ( z 2 − 3 z + 2 ) 2 z 2 , then
∮ C f ( z ) d z \oint_C f(z) dz ∮ C f ( z ) d z is
Solution The function is
f ( z ) = z 2 ( z 2 − 3 z + 2 ) 2 = z 2 ( ( z − 1 ) ( z − 2 ) ) 2 = z 2 ( z − 1 ) 2 ( z − 2 ) 2 f(z) = \frac{z^2}{(z^2-3z+2)^2} = \frac{z^2}{((z-1)(z-2))^2} = \frac{z^2}{(z-1)^2(z-2)^2} f ( z ) = ( z 2 − 3 z + 2 ) 2 z 2 = (( z − 1 ) ( z − 2 ) ) 2 z 2 = ( z − 1 ) 2 ( z − 2 ) 2 z 2 .
The singularities are at
z = 1 z=1 z = 1 and
z = 2 z=2 z = 2 , both are poles of order 2. Both poles lie inside the circle
∣ z ∣ = 4 |z|=4 ∣ z ∣ = 4 .
By Cauchy's Residue Theorem,
∮ C f ( z ) d z = 2 π i [ Res ( f , 1 ) + Res ( f , 2 ) ] \oint_C f(z) dz = 2\pi i [\text{Res}(f, 1) + \text{Res}(f, 2)] ∮ C f ( z ) d z = 2 π i [ Res ( f , 1 ) + Res ( f , 2 )] .
Residue at z = 1 z=1 z = 1 Res ( f , 1 ) = lim z → 1 d d z [ ( z − 1 ) 2 f ( z ) ] = lim z → 1 d d z [ z 2 ( z − 2 ) 2 ] = lim z → 1 2 z ( z − 2 ) 2 − z 2 ⋅ 2 ( z − 2 ) ( z − 2 ) 4 = lim z → 1 2 z ( z − 2 ) − 2 z 2 ( z − 2 ) 3 = 2 ( 1 ) ( − 1 ) − 2 ( 1 ) 2 ( − 1 ) 3 = − 4 − 1 = 4 \text{Res}(f, 1) = \lim_{z \to 1} \frac{d}{dz} \left[ (z-1)^2 f(z) \right] = \lim_{z \to 1} \frac{d}{dz} \left[ \frac{z^2}{(z-2)^2} \right] = \lim_{z \to 1} \frac{2z(z-2)^2 - z^2 \cdot 2(z-2)}{(z-2)^4} = \lim_{z \to 1} \frac{2z(z-2) - 2z^2}{(z-2)^3} = \frac{2(1)(-1) - 2(1)^2}{(-1)^3} = \frac{-4}{-1} = 4 Res ( f , 1 ) = lim z → 1 d z d [ ( z − 1 ) 2 f ( z ) ] = lim z → 1 d z d [ ( z − 2 ) 2 z 2 ] = lim z → 1 ( z − 2 ) 4 2 z ( z − 2 ) 2 − z 2 ⋅ 2 ( z − 2 ) = lim z → 1 ( z − 2 ) 3 2 z ( z − 2 ) − 2 z 2 = ( − 1 ) 3 2 ( 1 ) ( − 1 ) − 2 ( 1 ) 2 = − 1 − 4 = 4 .
Residue at z = 2 z=2 z = 2 Res ( f , 2 ) = lim z → 2 d d z [ ( z − 2 ) 2 f ( z ) ] = lim z → 2 d d z [ z 2 ( z − 1 ) 2 ] = lim z → 2 2 z ( z − 1 ) 2 − z 2 ⋅ 2 ( z − 1 ) ( z − 1 ) 4 = lim z → 2 2 z ( z − 1 ) − 2 z 2 ( z − 1 ) 3 = 2 ( 2 ) ( 1 ) − 2 ( 2 ) 2 ( 1 ) 3 = 4 − 8 1 = − 4 \text{Res}(f, 2) = \lim_{z \to 2} \frac{d}{dz} \left[ (z-2)^2 f(z) \right] = \lim_{z \to 2} \frac{d}{dz} \left[ \frac{z^2}{(z-1)^2} \right] = \lim_{z \to 2} \frac{2z(z-1)^2 - z^2 \cdot 2(z-1)}{(z-1)^4} = \lim_{z \to 2} \frac{2z(z-1) - 2z^2}{(z-1)^3} = \frac{2(2)(1) - 2(2)^2}{(1)^3} = \frac{4-8}{1} = -4 Res ( f , 2 ) = lim z → 2 d z d [ ( z − 2 ) 2 f ( z ) ] = lim z → 2 d z d [ ( z − 1 ) 2 z 2 ] = lim z → 2 ( z − 1 ) 4 2 z ( z − 1 ) 2 − z 2 ⋅ 2 ( z − 1 ) = lim z → 2 ( z − 1 ) 3 2 z ( z − 1 ) − 2 z 2 = ( 1 ) 3 2 ( 2 ) ( 1 ) − 2 ( 2 ) 2 = 1 4 − 8 = − 4 .
Sum of residues =
4 + ( − 4 ) = 0 4 + (-4) = 0 4 + ( − 4 ) = 0 .
Therefore,
∮ C f ( z ) d z = 2 π i ( 0 ) = 0 \oint_C f(z) dz = 2\pi i (0) = 0 ∮ C f ( z ) d z = 2 π i ( 0 ) = 0 .
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Solution The function is
f ( z ) = z 2 ( z 2 − 3 z + 2 ) 2 = z 2 ( ( z − 1 ) ( z − 2 ) ) 2 = z 2 ( z − 1 ) 2 ( z − 2 ) 2 f(z) = \frac{z^2}{(z^2-3z+2)^2} = \frac{z^2}{((z-1)(z-2))^2} = \frac{z^2}{(z-1)^2(z-2)^2} f ( z ) = ( z 2 − 3 z + 2 ) 2 z 2 = (( z − 1 ) ( z − 2 ) ) 2 z 2 = ( z − 1 ) 2 ( z − 2 ) 2 z 2 .
The singularities are at
z = 1 z=1 z = 1 and
z = 2 z=2 z = 2 , both are poles of order 2. Both poles lie inside the circle
∣ z ∣ = 4 |z|=4 ∣ z ∣ = 4 .
By Cauchy's Residue Theorem,
∮ C f ( z ) d z = 2 π i [ Res ( f , 1 ) + Res ( f , 2 ) ] \oint_C f(z) dz = 2\pi i [\text{Res}(f, 1) + \text{Res}(f, 2)] ∮ C f ( z ) d z = 2 π i [ Res ( f , 1 ) + Res ( f , 2 )] .
Residue at z = 1 z=1 z = 1 Res ( f , 1 ) = lim z → 1 d d z [ ( z − 1 ) 2 f ( z ) ] = lim z → 1 d d z [ z 2 ( z − 2 ) 2 ] = lim z → 1 2 z ( z − 2 ) 2 − z 2 ⋅ 2 ( z − 2 ) ( z − 2 ) 4 = lim z → 1 2 z ( z − 2 ) − 2 z 2 ( z − 2 ) 3 = 2 ( 1 ) ( − 1 ) − 2 ( 1 ) 2 ( − 1 ) 3 = − 4 − 1 = 4 \text{Res}(f, 1) = \lim_{z \to 1} \frac{d}{dz} \left[ (z-1)^2 f(z) \right] = \lim_{z \to 1} \frac{d}{dz} \left[ \frac{z^2}{(z-2)^2} \right] = \lim_{z \to 1} \frac{2z(z-2)^2 - z^2 \cdot 2(z-2)}{(z-2)^4} = \lim_{z \to 1} \frac{2z(z-2) - 2z^2}{(z-2)^3} = \frac{2(1)(-1) - 2(1)^2}{(-1)^3} = \frac{-4}{-1} = 4 Res ( f , 1 ) = lim z → 1 d z d [ ( z − 1 ) 2 f ( z ) ] = lim z → 1 d z d [ ( z − 2 ) 2 z 2 ] = lim z → 1 ( z − 2 ) 4 2 z ( z − 2 ) 2 − z 2 ⋅ 2 ( z − 2 ) = lim z → 1 ( z − 2 ) 3 2 z ( z − 2 ) − 2 z 2 = ( − 1 ) 3 2 ( 1 ) ( − 1 ) − 2 ( 1 ) 2 = − 1 − 4 = 4 .
Residue at z = 2 z=2 z = 2 Res ( f , 2 ) = lim z → 2 d d z [ ( z − 2 ) 2 f ( z ) ] = lim z → 2 d d z [ z 2 ( z − 1 ) 2 ] = lim z → 2 2 z ( z − 1 ) 2 − z 2 ⋅ 2 ( z − 1 ) ( z − 1 ) 4 = lim z → 2 2 z ( z − 1 ) − 2 z 2 ( z − 1 ) 3 = 2 ( 2 ) ( 1 ) − 2 ( 2 ) 2 ( 1 ) 3 = 4 − 8 1 = − 4 \text{Res}(f, 2) = \lim_{z \to 2} \frac{d}{dz} \left[ (z-2)^2 f(z) \right] = \lim_{z \to 2} \frac{d}{dz} \left[ \frac{z^2}{(z-1)^2} \right] = \lim_{z \to 2} \frac{2z(z-1)^2 - z^2 \cdot 2(z-1)}{(z-1)^4} = \lim_{z \to 2} \frac{2z(z-1) - 2z^2}{(z-1)^3} = \frac{2(2)(1) - 2(2)^2}{(1)^3} = \frac{4-8}{1} = -4 Res ( f , 2 ) = lim z → 2 d z d [ ( z − 2 ) 2 f ( z ) ] = lim z → 2 d z d [ ( z − 1 ) 2 z 2 ] = lim z → 2 ( z − 1 ) 4 2 z ( z − 1 ) 2 − z 2 ⋅ 2 ( z − 1 ) = lim z → 2 ( z − 1 ) 3 2 z ( z − 1 ) − 2 z 2 = ( 1 ) 3 2 ( 2 ) ( 1 ) − 2 ( 2 ) 2 = 1 4 − 8 = − 4 .
Sum of residues =
4 + ( − 4 ) = 0 4 + (-4) = 0 4 + ( − 4 ) = 0 .
Therefore,
∮ C f ( z ) d z = 2 π i ( 0 ) = 0 \oint_C f(z) dz = 2\pi i (0) = 0 ∮ C f ( z ) d z = 2 π i ( 0 ) = 0 .
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