← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q1d — Step-by-Step Solution

10 marks · Section A

Cauchy-Riemann equations (necessary and sufficient) · Complex Analysis · asked 6× in 14 yrs · Read the full method →

Question

Suppose f(z)f(z) is an analytic function on a domain DD in C\mathbb C and satisfies the equation Im⁡f(z)=(Re⁡f(z))2, z∈D\operatorname{Im} f(z)=(\operatorname{Re} f(z))^2,\ z\in D. Show that f(z)f(z) is constant in DD.

Technique

Cauchy–Riemann equations applied to the constraint v=u2v=u^2; the algebraic factor 1+4u2>01+4u^2>0 forces all first partials to vanish on the connected domain.

Solution

Write f=u+ivf=u+iv with u=Re⁡fu=\operatorname{Re}f, v=Im⁡fv=\operatorname{Im}f real-valued and C∞C^\infty (analytic ⇒\Rightarrow harmonic, hence smooth). The hypothesis is

v=u2on D.v=u^2\qquad\text{on }D.

Step 1 — Cauchy–Riemann equations

Analyticity gives

ux=vy,uy=−vx.(CR)u_x=v_y,\qquad u_y=-v_x.\tag{CR}

Step 2 — Differentiate the relation v=u2v=u^2

vx=2u ux,vy=2u uy.v_x=2u\,u_x,\qquad v_y=2u\,u_y.

Substitute into (CR):

ux=vy=2u uy,uy=−vx=−2u ux.u_x=v_y=2u\,u_y,\qquad u_y=-v_x=-2u\,u_x.

Step 3 — Solve the linear system for (ux,uy)(u_x,u_y)

From the two equations,

ux=2u uyanduy=−2u ux.u_x=2u\,u_y\quad\text{and}\quad u_y=-2u\,u_x.

Substitute the second into the first:

ux=2u (−2u ux)=−4u2 ux ⟹ ux(1+4u2)=0.u_x=2u\,(-2u\,u_x)=-4u^2\,u_x\ \Longrightarrow\ u_x(1+4u^2)=0.

Since 1+4u2≥1>01+4u^2\ge 1>0 everywhere, we must have ux=0u_x=0. Then uy=−2u ux=0u_y=-2u\,u_x=0 as well.

Step 4 — Conclude

ux=uy=0u_x=u_y=0 on the domain (connected open set) DD, so uu is constant on DD. By (CR), vx=−uy=0v_x=-u_y=0 and vy=ux=0v_y=u_x=0, so vv is constant too (consistently, v=u2=v=u^2=const). Hence

Answer

 f=u+iv is constant on D. \boxed{\,f=u+iv\ \text{is constant on }D.\,}
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