Complex numbers JEE Main PYQs (2020–2026)

Solved Complex numbers PYQs

3 recent questions with answers. The full set of 154 is free to practise.

1. JEE Main 2026EasyAlgebra of Complex Numbers & Modulus-Conjugate Properties
   Q1

   Let $S=\left\{z\in \mathbb{C}:z^2+\sqrt{6}iz-3=0\right\}$ . Then ${\sum }_{z\in S}^{}{z}^8$ is equal to :

   • A162
   • B184
   • C262
   • D324

   Show answer & solution

   Correct answer: (A)

   $$\begin{array}{rl}& z=\frac{-\sqrt{6}i\pm \sqrt{-6+12}}{2}\\ & =\frac{-\sqrt{6}i\pm \sqrt{6}}{2}\\ & =\frac{\sqrt{6}}{2}(\pm 1-i)\end{array}$$
   
   $$\begin{array}{rl}& \alpha =\frac{-\sqrt{6}}{2}(1+i),\beta =\frac{\sqrt{6}}{2}(1-i)\\ & \alpha =\frac{-\sqrt{6}}{2}\times \sqrt{2}{e}^{i\pi /4},\beta =+\frac{\sqrt{6}}{2}\times \sqrt{2}-i\pi /4\\ & \alpha =-\sqrt{3}{e}^{i\pi /4},\beta =+\sqrt{3}{e}^{-i\pi /4}\end{array}$$
   
   $$\begin{array}{rl}& {\alpha }^8+{\beta }^8=81\cdot e^{i2\pi }+81\cdot e^{i(-2\pi )}\\ & =81(1+1)=162\end{array}$$
2. JEE Main 2026EasySquare Root and Quadratic Equations with Complex Coefficients
   Q2

   Let $z_1,z_2\in \mathbb{C}$ be the distinct solutions of the equation $z^2+4z-(1+12i)=0$ . Then ${\left|z_1\right|}^2+{\left|z_2\right|}^2$ is equal to :

   • A18
   • B22
   • C29
   • D34

   Show answer & solution

   Correct answer: (D)

   Given the quadratic equation
   
   $$z^2+4z-(1+12i)=0$$
   
   let its roots be $z_1$ and $z_2$ .
   
   We need to find
   
   $$|z_1{|}^2+|z_2{|}^2$$
   
   Using Vieta’s formulas for
   
   $$a{z}^2+bz+c=0,$$
   
   we have
   
   $$z_1+z_2=-\frac{b}{a}=-4$$
   
   and
   
   $$z_1{z}_2=\frac{c}{a}=-(1+12i)=-1-12i$$
   
   Now solve the equation to identify the roots.
   
   $$z^2+4z-(1+12i)=0$$
   
   Using the quadratic formula,
   
   $$z=\frac{-4\pm \sqrt{16+4(1+12i)}}{2}$$
   
   $$z=\frac{-4\pm \sqrt{20+48i}}{2}$$
   
   Now observe that
   
   $$(6+4i{)}^2=36+48i-16=20+48i$$
   
   So,
   
   $$\sqrt{20+48i}=6+4i$$
   
   Hence the roots are
   
   $$z_1=\frac{-4+(6+4i)}{2}=1+2i$$
   
   and
   
   $$z_2=\frac{-4-(6+4i)}{2}=-5-2i$$
   
   Now compute their moduli squared:
   
   $$|z_1{|}^2=|1+2i{|}^2=1^2+2^2=5$$
   
   $$|z_2{|}^2=|-5-2i{|}^2$$
   
   $$=(-5{)}^2+(-2{)}^2=25+4=29$$
   
   Therefore,
   
   $$|z_1{|}^2+|z_2{|}^2=5+29=34$$
   
   So the correct answer is
   
   $$\boxed{34}$$
   
   Hence, Option D is correct.
3. JEE Main 2026EasyDe Moivre's Theorem & Roots of Unity
   Q3

   If $z=\frac{\sqrt{3}}{2}+\frac{i}{2},i=\sqrt{-1}$ , then ${\left(z^{201}-i\right)}^8$ is equal to

   • A1
   • B0
   • C-1
   • D256

   Show answer & solution

   Correct answer: (D)

   $$z=\frac{\sqrt{3}}{2}+\frac{i}{2}$$
   
   Convert into Euler form $z=r{e}^{i\theta }$ where $r=|z|$ and $\theta =arg(z)={\tan }^{-1}\left(\frac{y}{x}\right)$ if $z=x+iy$ .
   
   $$r=|z|=\sqrt{{\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2}=\sqrt{\frac{3}{4}+\frac{1}{4}}=1$$
   
   Since $z$ lies in the first quadrant in the complex plane,
   
   $$\begin{array}{rl}& \text{ So, }\theta ={\tan }^{-1}\left(\frac{1/2}{\sqrt{3}/2}\right)=\frac{\pi }{6}\\ & z=e^{i\frac{\pi }{6}}\end{array}$$
   
   To calculate ${\left(z^{201}-i\right)}^8={\left({\left(e^{i\frac{\pi }{6}}\right)}^{201}-i\right)}^8$
   
   $$={\left(e^{i\frac{201\pi }{6}}-i\right)}^8$$
   
   Using De Moivre’s theorem, $e^{i\theta }=cos\theta +i\sin \theta$
   
   $$\begin{array}{rl}& {\left(z^{201}-i\right)}^8={\left(\cos \left(\frac{201\pi }{6}\right)+i\sin \left(\frac{201\pi }{6}\right)-i\right)}^8\\ & ={\left(\cos \left(\frac{67\pi }{2}\right)+i\sin \left(\frac{67\pi }{2}\right)-i\right)}^8\\ & =(0+i(-1)-i{)}^8\\ & =(-2i{)}^8=2^8(i{)}^8\end{array}$$
   
   Since $i=\sqrt{-1}\Rightarrow i^2=-1\Rightarrow i^4=1\Rightarrow i^8=1$
   
   $\therefore {\left(z^{201}-i\right)}^8=2^8=256$ .

Topic-wise question count

Where JEE Main has asked from Complex numbers, most frequent first.

• Algebra of Complex Numbers & Modulus-Conjugate Properties44
• Geometry of Complex Numbers & Locus Problems38
• Equations of Lines and Circles in Complex Plane17
• Triangle Inequality & Distance Relations16
• De Moivre's Theorem & Roots of Unity16
• Square Root and Quadratic Equations with Complex Coefficients14
• Polar and Euler Form & Argument9

Year-wise Complex numbers questions in JEE Main

Number of questions from this chapter in each year's paper(s).