Skip to main content

Physics Library

Free Physics Study Material for CSIR NET, IIT JAM, GATE, JEST & TIFR

Physics Library
by Pravegaa Education
Visit Pravegaa.com
Login / Register
Menu
Menu

Chapter 4 Complex Variables Previous Year Question with Solution

Pravegaa PYQ Series · Volume 1: Mathematical Physics

MP04: Complex Variables

33 previous-year questions with complete solutions, from IIT JAM, GATE, CSIR NET-JRF, JEST and TIFR, arranged from IIT JAM and GATE through to TIFR and from the most recent paper backwards. Click Show solution under any question.

  • IIT JAM: 8
  • GATE: 6
  • CSIR NET: 10
  • JEST: 5
  • TIFR GS: 4
MP04.1IIT JAM 2026, Q46NAT | 1 mark

The value of \((1-i\sqrt3)^3\) is . (Answer in integer)

Show solution

\(|1-i\sqrt3|=2\) and \(\arg(1-i\sqrt3)=-\pi/3\), so \(1-i\sqrt3=2e^{-i\pi/3}\). By de Moivre, \[ (1-i\sqrt3)^3=2^3e^{-i\pi}=8(\cos\pi-i\sin\pi)=-8 \] \[ \boxed{-8} \]

Answer: \(-8\)

MP04.2IIT JAM 2025, Q14MCQ | 2 marks

If \(\left(\dfrac{1-i}{1+i}\right)^{\frac{n}{2}}=-1\), where \(i=\sqrt{-1}\), one possible value of \(n\) is

  1. 2

  2. 4

  3. 6

  4. 8

Show solution

\[\frac{1-i}{1+i}=\frac{(1-i)^2}{(1+i)(1-i)}=\frac{-2i}{2}=-i=e^{-i\pi/2}.\] Then \[\left(\frac{1-i}{1+i}\right)^{n/2}=e^{-i\pi n/4}=-1\;\Rightarrow\;\frac{n\pi}{4}=\pi,3\pi,5\pi,\dots\;\Rightarrow\;n=4,12,20,\dots\] Check: \(n=2\) gives \(-i\), \(n=6\) gives \((-i)^3=i\), \(n=8\) gives \((-i)^4=1\), while \(n=4\) gives \((-i)^2=-1\). Hence \(\boxed{n=4}\).

Answer: (B) 4

MP04.3IIT JAM 2024, Q16MCQ | 2 marks

The roots of the polynomial, \(f(z)=z^4-8z^3+27z^2-38z+26\), are \(z_1,z_2,z_3,\,\&\,z_4\), where \(z\) is a complex variable. Which of the following statements is correct?

  1. \(\dfrac{z_1+z_2+z_3+z_4}{z_1z_2z_3z_4}=-\dfrac{4}{19}\)

  2. \(\dfrac{z_1+z_2+z_3+z_4}{z_1z_2z_3z_4}=\dfrac{4}{13}\)

  3. \(\dfrac{z_1z_2z_3z_4}{z_1+z_2+z_3+z_4}=-\dfrac{26}{27}\)

  4. \(\dfrac{z_1z_2z_3z_4}{z_1+z_2+z_3+z_4}=\dfrac{13}{19}\)

Show solution

For \(az^4+bz^3+cz^2+dz+e=0\), Vieta's relations give \[ z_1+z_2+z_3+z_4=-\frac ba=8,\qquad z_1z_2z_3z_4=\frac ea=26 . \] \[ \frac{z_1+z_2+z_3+z_4}{z_1z_2z_3z_4}=\frac{8}{26}=\boxed{\frac{4}{13}} \]

Answer: (B) \(4/13\)

MP04.4IIT JAM 2022, Q1MCQ | 1 mark

The equation \(z^2+\bar z^{2}=4\) in the complex plane (where \(\bar z\) is the complex conjugate of \(z\)) represents

  1. Ellipse

  2. Hyperbola

  3. Circle of radius \(2\)

  4. Circle of radius \(4\)

Show solution

Write \(z=x+iy\), so that \(\bar z=x-iy\). Then \begin{align*} z^2+\bar z^{2}&=(x+iy)^2+(x-iy)^2=(x^2-y^2+2ixy)+(x^2-y^2-2ixy)=2(x^2-y^2). \end{align*} Hence \(2(x^2-y^2)=4\Rightarrow x^2-y^2=2\), which is a rectangular hyperbola. \[\boxed{\text{Hyperbola}}\]

Answer: (B) Hyperbola

MP04.5IIT JAM 2021, Q41NAT | 1 mark

One of the roots of the equation, \(z^6-3z^4-16=0\) is given by \(z_1=2\). The value of the product of the other five roots is

Show solution

For \(a_6z^6+\dots+a_0=0\) the product of all six roots is \((-1)^6\,a_0/a_6=-16\). Hence \[ z_2z_3z_4z_5z_6=\frac{-16}{z_1}=\frac{-16}{2}=-8 . \] Check by factorising with \(u=z^2\): \(u^3-3u^2-16=(u-4)(u^2+u+4)\). The roots are \(z=\pm2\) and \(z=\pm\sqrt{u_{1,2}}\) with \(u_1u_2=4\). The other five roots are \(-2\) and the four roots \(\pm\sqrt{u_1},\pm\sqrt{u_2}\), whose product is \((-u_1)(-u_2)=4\), giving \(-2\times4=-8\). \[ \boxed{-8} \]

Answer: \(-8\)

MP04.6IIT JAM 2020, Q11MCQ | 2 marks

Which of the following statements is correct?

Given \(\dbinom{n}{m}=\dfrac{n!}{m!\,(n-m)!}\) is the binomial coefficient

  1. \(\cos n\theta=\cos^n\theta-\dbinom{n}{2}\cos^{n-2}\theta\sin^2\theta+\dbinom{n}{4}\cos^{n-4}\theta\sin^4\theta-\dots\)

Continue reading — it's free

Register with your mobile number to unlock the full chapter and all 100+ chapters of the Physics Library.

Register / Login with OTP

Already registered? Use the same button — just verify your number.