Complex Number JEE Main previous year questions with solutions

5 solved JEE Main questions on Complex Number, free to read — no sign-in needed. The full chapter has 218 questions; sign in to attempt the remaining 213 in the exam simulator.

1. Q1JEE Main 2026 (06 Apr, Shift 1)Algebra of complex numbers
   Let the set of all values of $k \in \mathbb{R}$ such that the equation $z(\bar{z} + 2 + i) + k(2 + 3i) = 0$, $z \in \mathbb{C}$, has at least one solution, be the interval $[\alpha, \beta]$. Then $9(\alpha + \beta)$ is equal to:
   1. A.$-10$
   2. B.$-8$
   3. C.$10\sqrt{13}$
   4. D.$8\sqrt{13}$
   Show answer & solution

   Answer: (A)

   Let $z = x + iy$, then $\bar{z} = x - iy$. Substituting $z$ into the given equation: $(x + iy)(x - iy + 2 + i) + k(2 + 3i) = 0$ $x^2 + y^2 + 2x + ix + 2iy - y + 2k + 3ki = 0$ Separating the real and imaginary parts, we get: Real part: $x^2 + y^2 + 2x - y + 2k = 0$ Imaginary part: $x + 2y + 3k = 0 \Rightarrow x = -2y - 3k$ Substituting $x$ into the real part equation: $(-2y - 3k)^2 + y^2 + 2(-2y - 3k) - y + 2k = 0$ $4y^2 + 12ky + 9k^2 + y^2 - 4y - 6k - y + 2k = 0$ $5y^2 + (12k - 5)y + 9k^2 - 4k = 0$ For the equation to have at least one solution $z \in \mathbb{C}$, there must be at least one real value of $y$. Thus, the discriminant of this quadratic equation in $y$ must be non-negative ($\Delta \ge 0$): $\Delta = (12k - 5)^2 - 4(5)(9k^2 - 4k) \ge 0$ $144k^2 - 120k + 25 - 180k^2 + 80k \ge 0$ $-36k^2 - 40k + 25 \ge 0$ $36k^2 + 40k - 25 \le 0$ The values of $k$ lie in the interval $[\alpha, \beta]$, where $\alpha$ and $\beta$ are the roots of the equation $36k^2 + 40k - 25 = 0$. The sum of the roots is given by: $\alpha + \beta = -\dfrac{40}{36} = -\dfrac{10}{9}$ Therefore, $9(\alpha + \beta) = 9 \left( -\dfrac{10}{9} \right) = -10$. Answer: $-10$
2. Q2JEE Main 2025 (08 Apr, Shift 2)Conjugate, modulus and argument
   Let $\mathrm{A}=$ $\left\{\theta \in[0,2 \pi]: 1+10 \operatorname{Re}\left(\frac{2 \cos \theta+i \sin \theta}{\cos \theta-3 i \sin \theta}\right)=0\right\} .$ Then $\sum_{\theta \in A} \theta^2$ is equal to
   1. A.$\frac{21}{4} \pi^2$
   2. B.$8 \pi^2$
   3. C.$\frac{27}{4} \pi^2$
   4. D.$6 \pi^2$
   Show answer & solution

   Answer: (A)

   $\begin{aligned} & 1+10 \operatorname{Re}\left(\frac{2 \cos \theta+i \sin \theta}{\cos \theta-3 i \sin \theta}\right)=0 \\ & \therefore \mathrm{z}+\overline{\mathrm{z}}=2 \operatorname{Re}(\mathrm{z}) \\ & \frac{2 \cos \theta+\mathrm{i} \sin \theta}{\cos \theta-3 \mathrm{i} \sin \theta}+\frac{2 \cos \theta-\mathrm{i} \sin \theta}{\cos \theta+3 \mathrm{i} \sin \theta}=2 \times\left(\frac{-1}{10}\right) \\ & \frac{\left(2 \cos ^2 \theta-3 \sin ^2 \theta\right)+\left(2 \cos ^2 \theta\right)-\left(3 \sin ^2 \theta\right)}{\cos ^2 \theta+9 \sin ^2 \theta}=\frac{-2}{10} \\ & \Rightarrow \frac{2 \cos ^2 \theta-3 \sin ^2 \theta}{\cos ^2 \theta+9 \sin ^2 \theta}=\frac{-1}{10} \\ & \Rightarrow 20 \cos ^2 \theta-30 \sin ^2 \theta=-\cos ^2 \theta-9 \sin ^2 \theta \\ & 21 \cos ^2 \theta-21 \sin ^2 \theta=0 \\ & \Rightarrow \cos 2 \theta=0 \\ & 2 \theta=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \\ & \Rightarrow \sum \theta^2=\frac{\pi^2}{16}+\frac{9 \pi^2}{16}+\frac{25 \pi^2}{16}+\frac{49 \pi^2}{16}=\frac{84 \pi^2}{16}=\frac{21 \pi^2}{4}\end{aligned}$
3. Q3JEE Main 2024 (30 Jan, Shift 2)Cube Root of Unity
   If $z$ is a complex number, then the number of common roots of the equation ${z}^{1985}+{z}^{100}+1=0$ and ${z}^{3}+2{z}^{2}+2z+1=0$, is equal to :
   1. A.$1$
   2. B.$2$
   3. C.$0$
   4. D.$3$
   Show answer & solution

   Answer: (B)

   Given, ${z}^{1985}+{z}^{100}+1=0\&{z}^{3}+2{z}^{2}+2z+1=0$ Now solving, ${z}^{3}+2{z}^{2}+2z+1=0$ $\Rightarrow \left(z+1\right)\left({z}^{2}-z+1\right)+2z\left(z+1\right)=0$ $\Rightarrow \left(z+1\right)\left({z}^{2}+z+1\right)=0$ $\Rightarrow z=-1,z=\omega ,{\omega }^{2}$ Now putting $z=\omega$ in ${z}^{1985}+{z}^{100}+1$ we get, $\Rightarrow {\omega }^{1985}+{\omega }^{100}+1$ $\Rightarrow {\omega }^{2}+\omega +1=0$ $\left\{\text{as}{\omega }^{3n}=1\right\}$ Also, $z={\omega }^{2}$ $\Rightarrow {\omega }^{3970}+{\omega }^{200}+1$ $\Rightarrow \omega +{\omega }^{2}+1=0$ Also, $z=-1$ will not satisfy the equation ${z}^{1985}+{z}^{100}+1$ Hence, there are two common root
4. Q4JEE Main 2023 (31 Jan, Shift 2)Euler Form and De Moivres Theorem
   The complex number $z=\dfrac{i-1}{\cos \dfrac{\pi }{3}+i\sin \dfrac{\pi }{3}}$ is equal to:
   1. A.$\sqrt{2}i\left(\cos \dfrac{5\pi }{12}-i\sin \dfrac{5\pi }{12}\right)$
   2. B.$\cos \dfrac{\pi }{12}-i\sin \dfrac{\pi }{12}$
   3. C.$\sqrt{2}\left(\cos \dfrac{\pi }{12}+i\sin \dfrac{\pi }{12}\right)$
   4. D.$\sqrt{2}\left(\cos \dfrac{5\pi }{12}+i\sin \dfrac{5\pi }{12}\right)$
   Show answer & solution

   Answer: (D)

   Given, $z=\dfrac{i-1}{\cos \dfrac{\pi }{3}+i\sin \dfrac{\pi }{3}}$ $\Rightarrow z=\dfrac{\sqrt{2}\left(\dfrac{-1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}}\right)}{\cos \dfrac{\pi }{3}+i\sin \dfrac{\pi }{3}}$ Now argument of complex number $\dfrac{-1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}}$ is $\dfrac{3\pi }{4}$ So, by euler's formula ${e}^{i\theta }=\cos \theta +i\sin \theta$ we get, $\Rightarrow z=\dfrac{\sqrt{2}{e}^{i3\pi /4}}{{e}^{i\pi /3}}$ $\Rightarrow z=\sqrt{2}{e}^{i5\pi /12}$ $\Rightarrow z=\sqrt{2}\left(\cos \dfrac{5\pi }{12}+i\sin \dfrac{5\pi }{12}\right)$
5. Q5JEE Main 2022 (26 Jul, Shift 2)Geometry of Complex Number
   If $z=x+iy$ satisfies $\left|z\right|-2=0$ and $\left|z-i\right|-\left|z+5i\right|=0$, then
   1. A.$x+2y-4=0$
   2. B.${x}^{2}+y-4=0$
   3. C.$x+2y+4=0$
   4. D.${x}^{2}-y+3=0$
   Show answer & solution

   Answer: (C)

   Given, $z=x+iy$ Also, $\left|z\right|=2$ So, ${x}^{2}+{y}^{2}=4...\left(1\right)$ Now, $\left|z-i\right|=\left|z+5i\right|$ $\Rightarrow {\left|z-i\right|}^{2}={\left|z+5i\right|}^{2}$ $\Rightarrow {x}^{2}+{\left(y-1\right)}^{2}={x}^{2}+{\left(y+5\right)}^{2}$ $\Rightarrow y=-2...\left(2\right)$ So, $x=0$ by solving equation $\left(1\right)\&\left(2\right)$ Hence, only $x+2y+4=0$ is true, as point only satisfy this equation.