← 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\int_C (z^2+3z)\,dz counterclockwise from (2,0)(2,0) to (0,2)(0,2) along the curve CC, where CC is the circle z=2|z|=2.

Technique

Fundamental theorem for contour integrals: an entire integrand has a global antiderivative, so Cf=F(end)F(start)\int_C f=F(\text{end})-F(\text{start}).

Solution

Step 1 — The integrand is entire, so the integral is path-independent

f(z)=z2+3zf(z)=z^2+3z is a polynomial, hence entire (analytic on all of C\mathbb{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)=z33+3z22,  F(z)=z2+3z.\int_C f(z)\,dz=F(z_{\text{end}})-F(z_{\text{start}}),\qquad F(z)=\frac{z^3}{3}+\frac{3z^2}{2},\ \ F'(z)=z^2+3z.

The curve CC is the quarter-arc of z=2|z|=2 traversed counterclockwise from (2,0)(2,0) to (0,2)(0,2) — that is, from zstart=2z_{\text{start}}=2 to zend=2iz_{\text{end}}=2i. (Counterclockwise on z=2|z|=2 takes 22i2\to 2i through the first quadrant; the specific arc does not matter since ff is entire, but it is consistent here.)

Step 2 — Evaluate the antiderivative at the endpoints

F(2i)=(2i)33+3(2i)22=8i33+34i22=8i3+122=68i3.F(2i)=\frac{(2i)^3}{3}+\frac{3(2i)^2}{2} =\frac{8i^3}{3}+\frac{3\cdot 4i^2}{2} =\frac{-8i}{3}+\frac{-12}{2} =-6-\frac{8i}{3}. F(2)=233+3222=83+6=83+183=263.F(2)=\frac{2^3}{3}+\frac{3\cdot 2^2}{2}=\frac{8}{3}+6=\frac{8}{3}+\frac{18}{3}=\frac{26}{3}.

Step 3 — Subtract

Cfdz=F(2i)F(2)=(68i3)263=62638i3=18+2638i3=4438i3.\int_C f\,dz=F(2i)-F(2)=\Big(-6-\tfrac{8i}{3}\Big)-\tfrac{26}{3} =-6-\frac{26}{3}-\frac{8i}{3} =-\frac{18+26}{3}-\frac{8i}{3} =-\frac{44}{3}-\frac{8i}{3}.

Answer

  C(z2+3z)dz=44383i=44+8i3.  \boxed{\;\int_C (z^2+3z)\,dz=-\frac{44}{3}-\frac{8}{3}i=-\frac{44+8i}{3}.\;}
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