If f(z) = (8z^2 - 2) / (z(z-1)(z+1)), evaluate ∫c f(z) dz along the countour c where c is the triangle joining?

Question

the points z = 2, z = j, z = -1 - j

Answer 1

The singularities of the integrand are z = 0, 1, -1.

By sketching the region enclosed by C, we find that only z = 0, 1 are inside C.



Deforming C into two closed contours C₁ and C₂ which only enclose exactly z = 0 and z = 1 (but not the other) respectively, we have by Cauchy's Integral Formula

∫c f(z) dz

= ∫c₁ f(z) dz + ∫c₂ f(z) dz

= ∫c₁ [(8z^2 - 2)/((z-1)(z+1))] dz / (z - 0) + ∫c₂ [(8z^2 - 2)/(z(z+1))] dz / (z-1)

= 2πi * [(8z^2 - 2)/((z-1)(z+1))] {at z = 0} + 2πi * [(8z^2 - 2)/(z(z+1)) {at z = 1}

= 2πi * 2 + 2πi * 3

= 10πi.

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I hope this helps!



Note: If you know how to use the residue theorem, you multiply the sum of the residues at the simple poles 0 and 1 by 2πi.

Answer 2

We have 3 poles : z=0, z=1 and z=-1

2 of them are situated inside the given triangle (z=0 and z=1)

So we have to calculate the 2 corresponding residues



z=0 : 2

z=1 : 3

the sum of the two is 5



we have to mulitiply by 2 pi i in the residu theorem to find the final result :



10 pi i = 31.4... i

Answer 3

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