UPSC 2019 Maths Optional Paper 2 Q3c — Step-by-Step Solution

10 marks · Section A

Cauchy's theorem (Cauchy-Goursat) · Complex Analysis · asked 2× in 14 yrs · Read the full method →

Question

Evaluate the integral $\displaystyle\int_C \operatorname{Re}(z^2)\,dz$ from $0$ to $2+4i$ along the curve $C$ where $C$ is a parabola $y=x^2$ .

Parametrize $z=x+ix^2$ ; substitute $\operatorname{Re}(z^2)=x^2-y^2$ with $y=x^2$ ; reduce to two real integrals.

On $C:\ y=x^2$ , write $z=x+iy=x+ix^2$ , $x:0\to2$ . Endpoints: $x=0\Rightarrow z=0$ ; $x=2\Rightarrow z=2+4i$ . ✓ Then

dz=(1+2ix)\,dx.

z^2=(x+iy)^2=(x^2-y^2)+2ixy\ \Rightarrow\ \operatorname{Re}(z^2)=x^2-y^2.

On $C$ , $y=x^2$ , so

\operatorname{Re}(z^2)=x^2-(x^2)^2=x^2-x^4.

\int_C \operatorname{Re}(z^2)\,dz=\int_0^2 (x^2-x^4)(1+2ix)\,dx.

Split into real and imaginary parts:

=\underbrace{\int_0^2(x^2-x^4)\,dx}_{\text{real}}+\;2i\underbrace{\int_0^2 x(x^2-x^4)\,dx}_{\text{imag}}=\int_0^2(x^2-x^4)\,dx+2i\int_0^2(x^3-x^5)\,dx.

Real part:

\int_0^2(x^2-x^4)\,dx=\Big[\tfrac{x^3}{3}-\tfrac{x^5}{5}\Big]_0^2=\tfrac{8}{3}-\tfrac{32}{5}=\frac{40-96}{15}=-\frac{56}{15}.

Imaginary part:

2\int_0^2(x^3-x^5)\,dx=2\Big[\tfrac{x^4}{4}-\tfrac{x^6}{6}\Big]_0^2=2\Big(4-\tfrac{64}{6}\Big)=2\Big(4-\tfrac{32}{3}\Big)=2\cdot\Big(-\tfrac{20}{3}\Big)=-\frac{40}{3}.

\boxed{\;\int_C \operatorname{Re}(z^2)\,dz=-\frac{56}{15}-\frac{40}{3}\,i.\;}