← 2020 Paper 2
UPSC 2020 Maths Optional Paper 2 Q1d — Step-by-Step Solution
10 marks · Section A
Cauchy's theorem (Cauchy-Goursat) · Complex Analysis · asked 2× in 13 yrs · Read the full method →
Question
Evaluate the integral ∫C(z2+3z)dz counterclockwise from (2,0) to (0,2) along the curve C, where C is the circle ∣z∣=2.
Technique
Fundamental theorem for contour integrals: an entire integrand has a global antiderivative, so ∫Cf=F(end)−F(start).
Solution
Step 1 — The integrand is entire, so the integral is path-independent
f(z)=z2+3z is a polynomial, hence entire (analytic on all of C). For an analytic function with an antiderivative, the contour integral depends only on the endpoints:
∫Cf(z)dz=F(zend)−F(zstart),F(z)=3z3+23z2, F′(z)=z2+3z.
The curve C is the quarter-arc of ∣z∣=2 traversed counterclockwise from (2,0) to (0,2) — that is, from zstart=2 to zend=2i. (Counterclockwise on ∣z∣=2 takes 2→2i through the first quadrant; the specific arc does not matter since f is entire, but it is consistent here.)
Step 2 — Evaluate the antiderivative at the endpoints
F(2i)=3(2i)3+23(2i)2=38i3+23⋅4i2=3−8i+2−12=−6−38i.
F(2)=323+23⋅22=38+6=38+318=326.
Step 3 — Subtract
∫Cfdz=F(2i)−F(2)=(−6−38i)−326=−6−326−38i=−318+26−38i=−344−38i.
Answer
∫C(z2+3z)dz=−344−38i=−344+8i.