← 2020 Paper 2

UPSC 2020 Maths Optional Paper 2 Q4a — Step-by-Step Solution

15 marks · Section A

Cauchy-Riemann equations (necessary and sufficient) · Complex Analysis · asked 5× in 13 yrs · Read the full method →

Question

If v(r,θ)=(r−1r)sin⁡θv(r,\theta)=\left(r-\frac1r\right)\sin\theta, r≠0r\ne0, then find an analytic function f(z)=u(r,θ)+iv(r,θ)f(z)=u(r,\theta)+iv(r,\theta).

Technique

Polar Cauchy–Riemann equations ur=1rvθ, uθ=−rvru_r=\tfrac1r v_\theta,\ u_\theta=-r v_r; integrate to get uu; recombine via e±iθe^{\pm i\theta} to identify f(z)f(z) in closed form.

Solution

We are given the imaginary part vv in polar coordinates and must reconstruct uu (its harmonic conjugate) and hence ff. Use the polar Cauchy–Riemann equations: for f=u+ivf=u+iv analytic,

ur=1r vθ,1r uθ=−vr  (i.e. uθ=−r vr).u_r=\frac1r\,v_\theta,\qquad \frac1r\,u_\theta=-v_r\ \ \Big(\text{i.e. } u_\theta=-r\,v_r\Big).

Step 1 — Partial derivatives of vv

v=(r−1r)sin⁡θ.v=\Big(r-\tfrac1r\Big)\sin\theta. vr=(1+1r2)sin⁡θ,vθ=(r−1r)cos⁡θ.v_r=\Big(1+\tfrac1{r^2}\Big)\sin\theta,\qquad v_\theta=\Big(r-\tfrac1r\Big)\cos\theta.

Step 2 — Recover uru_r and uθu_\theta from CR

ur=1rvθ=1r(r−1r)cos⁡θ=(1−1r2)cos⁡θ.u_r=\frac1r v_\theta=\frac1r\Big(r-\tfrac1r\Big)\cos\theta=\Big(1-\tfrac1{r^2}\Big)\cos\theta. uθ=−r vr=−r(1+1r2)sin⁡θ=−(r+1r)sin⁡θ.u_\theta=-r\,v_r=-r\Big(1+\tfrac1{r^2}\Big)\sin\theta=-\Big(r+\tfrac1r\Big)\sin\theta.

Step 3 — Integrate to find uu

Integrate uru_r with respect to rr (holding θ\theta fixed):

u=∫(1−1r2)cos⁡θ dr=(r+1r)cos⁡θ+φ(θ),u=\int\Big(1-\tfrac1{r^2}\Big)\cos\theta\,dr=\Big(r+\tfrac1r\Big)\cos\theta+\varphi(\theta),

where φ(θ)\varphi(\theta) is an arbitrary function of θ\theta. Differentiate w.r.t. θ\theta and match the known uθu_\theta:

uθ=−(r+1r)sin⁡θ+φ′(θ)=!−(r+1r)sin⁡θ.u_\theta=-\Big(r+\tfrac1r\Big)\sin\theta+\varphi'(\theta)\stackrel{!}{=}-\Big(r+\tfrac1r\Big)\sin\theta.

Hence φ′(θ)=0\varphi'(\theta)=0, so φ=C\varphi=C (a real constant). Therefore

u(r,θ)=(r+1r)cos⁡θ+C.u(r,\theta)=\Big(r+\tfrac1r\Big)\cos\theta+C.

Step 4 — Assemble f=u+ivf=u+iv and express in zz

f=(r+1r)cos⁡θ+i(r−1r)sin⁡θ+C.f=\Big(r+\tfrac1r\Big)\cos\theta + i\Big(r-\tfrac1r\Big)\sin\theta + C.

Group terms using cos⁡θ+isin⁡θ=eiθ\cos\theta+i\sin\theta=e^{i\theta} and cos⁡θ−isin⁡θ=e−iθ\cos\theta-i\sin\theta=e^{-i\theta}:

f=r(cos⁡θ+isin⁡θ)+1r(cos⁡θ−isin⁡θ)+C=reiθ+1re−iθ+C.f=r(\cos\theta+i\sin\theta)+\frac1r(\cos\theta-i\sin\theta)+C =r e^{i\theta}+\frac1r e^{-i\theta}+C.

Since z=reiθz=re^{i\theta} and 1z=1re−iθ\dfrac1z=\dfrac1r e^{-i\theta},

  f(z)=z+1z+C(C∈R arbitrary).  \boxed{\;f(z)=z+\frac1z+C\quad(C\in\mathbb{R}\text{ arbitrary}).\;}

Verification

Write z=reiθz=re^{i\theta}. Then

z+1z=reiθ+1re−iθ=(r+1r)cos⁡θ+i(r−1r)sin⁡θ.z+\frac1z=re^{i\theta}+\frac1r e^{-i\theta} =\Big(r+\tfrac1r\Big)\cos\theta+i\Big(r-\tfrac1r\Big)\sin\theta.

The imaginary part is exactly (r−1r)sin⁡θ=v\big(r-\tfrac1r\big)\sin\theta=v ✓. The function z+1/zz+1/z is analytic on C∖{0}\mathbb{C}\setminus\{0\} (i.e. r≠0r\ne0, as required). u=(r+1r)cos⁡θu=\big(r+\tfrac1r\big)\cos\theta is harmonic and is the conjugate of vv (SymPy confirms uθ=−rvru_\theta=-rv_r and ur=1rvθu_r=\tfrac1r v_\theta). ■\blacksquare

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