Complex Number JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Complex Number, free to read — no sign-in needed. The full chapter has 39 questions; sign in to attempt the remaining 34 in the exam simulator.

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1. Q1JEE Advanced Adv 2025 (Paper 2)
   For a non-zero complex number $z$, let $\arg (z)$ denote the principal argument of $z$, with $-\pi \lt \arg (z) \leq \pi$. Let $\omega$ be the cube root of unity for which $0 \lt \arg (\omega) \lt \pi$. Let $\alpha=\arg \left(\sum_{n=1}^{2025}(-\omega)^n\right)$. Then the value of $\frac{3 \alpha}{\pi}$ is ___________
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   Answer: -2

   $\begin{aligned} & \alpha=\arg \left(-\omega+\omega^2-\omega^3+\ldots \ldots \ldots+(-\omega)^{2025}\right) \\ & \alpha=\arg \left(\frac{-\omega\left((-\omega)^{2025}-1\right)}{-\omega-1}\right) \\ & \alpha=\arg \left(\frac{-\omega}{-\omega-1}(-2)\right) \\ & \alpha=\arg \left(\frac{-2 \omega}{\omega+1}\right) \\ & \alpha=\arg \left(\frac{-2 \omega}{-\omega^2}\right) \\ & \alpha=\arg \left(\frac{2}{\omega}\right) \\ & \alpha=\arg \left(2 \omega^2\right) \\ & \alpha=\frac{-2 \pi}{3} \\ & \frac{3 \alpha}{\pi}=-2\end{aligned}$
2. Q2JEE Advanced Adv 2025 (Paper 1)
   Let $\mathbb{R}$ denote the set of all real numbers. Let $z_1=1+2 i$ and $z_2=3 i$ be two complex numbers, where $i=\sqrt{-1}$. Let $S=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}:\left|x+i y-z_1\right|=2\left|x+i y-z_2\right|\right\}$. Then which of the following statements is (are) TRUE?
   1. A.$S$ is a circle with centre $\left(-\frac{1}{3}, \frac{10}{3}\right)$
   2. B.$S$ is a circle with centre $\left(\frac{1}{3}, \frac{8}{3}\right)$
   3. C.$S \text { is a circle with radius } \frac{\sqrt{2}}{3}$
   4. D.$S \text { is a circle with radius } \frac{2 \sqrt{2}}{3}$
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   Answer: A,D

   $\begin{aligned} &\begin{aligned} & |x+i y-1-2 i|=2|x+i y-3 i| \\ & \Rightarrow \quad(x-1)^2+(y-2)^2=4\left(x^2+(y-3)^2\right) \\ & \Rightarrow \quad 3 x^2+3 y^2+2 x-20 y+31=0 \\ & \Rightarrow \quad x^2+y^2+\frac{2 x}{3}-\frac{20 y}{3}+\frac{31}{3}=0 \end{aligned}\\ &\therefore \mathrm{S} \text { is a circle with centre }\left(-\frac{1}{3}, \frac{10}{3}\right) \text { and radius }=\sqrt{\frac{1}{9}+\frac{100}{9}-\frac{31}{3}}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3} \end{aligned}$
3. Q3JEE Advanced Adv 2024 (Paper 1)
   Let $S=\{a+b \sqrt{2}: a, b \in \mathbb{Z}\}, T_1=\left\{(-1+\sqrt{2})^n: n \in \mathbb{N}\right\}$, and $T_2=\left\{(1+\sqrt{2})^n: n \in \mathbb{N}\right\}$. Then which of the following statements is (are) TRUE?
   1. A.$\mathbb{Z} \cup T_1 \cup T_2 \subset S$
   2. B.$T_1 \cap\left(0, \frac{1}{2024}\right)=\phi$, where $\phi$ denotes the empty set.
   3. C.$T_2 \cap(2024, \infty) \neq \phi$
   4. D.For any given $a, b \in \mathbb{Z}, \cos (\pi(a+b \sqrt{2}))+i \sin (\pi(a+b \sqrt{2})) \in \mathbb{Z}$ if and only if $b=0$, where $i=\sqrt{-1}$.
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   Answer: A,C,D

   (1) $\begin{aligned}& (-1+\sqrt{2})^{\mathrm{n}}=\mathrm{m}+\sqrt{2} \mathrm{n}, \mathrm{m}, \text{ where, } \mathrm{n} \in \mathbb{Z} \\ & (1+\sqrt{2})^{\mathrm{n}}=\mathrm{m}_1+\sqrt{2} \mathrm{n}_1, \text{ where, } \mathrm{~m}_1, \mathrm{n}_1 \in \mathbb{Z} \\& \Rightarrow \mathbb{Z} \cup \mathrm{T}_1 \cup \mathrm{T}_2 \subseteq \mathrm{S}\end{aligned}$ but $b \sqrt{2} \in S$ for negative $b \in \mathbb{Z}$. So $\mathbb{Z} \cup \mathrm{T}_1 \cup \mathrm{T}_2 \subset \mathrm{S}$ (2) $\begin{aligned} & (\sqrt{2}-1)^{\mathrm{n}}=\frac{1}{(\sqrt{2}+1)^{\mathrm{n}}} \lt \frac{1}{2024} \\ & \Rightarrow 2024 \lt (\sqrt{2}+1)^{\mathrm{n}}, \text{ where, } \exists \mathrm{n} \in \mathbb{N} \\ & \Rightarrow \mathrm{T}_1 \cap\left(0, \frac{1}{2024}\right) \neq \phi\end{aligned}$ (3) $\begin{aligned} & (1+\sqrt{2})^{\mathrm{n}}\gt 2024, \text{ where, } \exists \mathrm{n} \in \mathbb{N} \\ & \Rightarrow \mathrm{T}_2 \cap(2024, \infty) \neq \phi\end{aligned}$ (4) $\sin (\pi(a+b \sqrt{2})=0) \Rightarrow b=0, \text{ where, } a \in \mathbb{Z}$ $\Rightarrow$ Options (1), (3), (4) are Correct.
4. Q4JEE Advanced Adv 2023 (Paper 2)
   Let ${A}_{1},{A}_{2},{A}_{3},\ldots ,{A}_{8}$ be the vertices of a regular octagon that lie on a circle of radius $2$. Let $P$ be a point on the circle and let $P{A}_{i}$ denote the distance between the points $P$ and ${A}_{i}$ for $i=1,2,\ldots ,8$. If $P$ varies over the circle, then the maximum value of the product $P{A}_{1}\cdot P{A}_{2}\ldots \cdot P{A}_{8}$, is
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   Answer: 512

   Given, ${A}_{1},{A}_{2},{A}_{3},\ldots ,{A}_{8}$ vertices of a regular octagon lying on a circle of radius $2$. Now using the concept of ${n}^{th}$ root of unity, Let any point $P$ be, $Z=(2)(1{)}^{1/8}$ $\Rightarrow {Z}^{8}={2}^{8}\times 1$ $\Rightarrow {Z}^{8}-{2}^{8}=0$ $\Rightarrow Z=2,2\alpha ,2{\alpha }^{2},2{\alpha }^{3},\ldots ,2{\alpha }^{7};\left\{\text{here}\alpha ={e}^{i\dfrac{2\pi }{8}}\right\}$ $\Rightarrow {Z}^{8}-{2}^{8}=(Z-2)(Z-2\alpha )\left(Z-2{\alpha }^{2}\right)\left(Z-2{\alpha }^{3}\right)\ldots \left(Z-2{\alpha }^{7}\right)$ $\Rightarrow \left|{Z}^{8}-{2}^{8}\right|=|Z-2||Z-2\alpha |\ldots .\left|Z-2{\alpha }^{7}\right|$ $\text{But}\left|{Z}^{8}+\left(-{2}^{8}\right)\right|\leq |Z{|}^{8}+{2}^{8}$ $\begin{matrix}\Rightarrow |Z-2||Z-2\alpha |\ldots \left|Z-2{\alpha }^{7}\right|\leq |Z{|}^{8}+{2}^{8} \\ \leq {2}^{8}+{2}^{8} \\ \leq {2}^{9}\end{matrix}$ $\Rightarrow Max⁡\left(P{A}_{1}\cdot P{A}_{2}\ldots .P{A}_{8}\right)={2}^{9}=512$
5. Q5JEE Advanced Adv 2023 (Paper 1)
   Let $z$ be a complex number satisfying ${\left|z\right|}^{3}+2{z}^{2}+4\bar{z}-8=0$, where $\bar{z}$ denotes the complex conjugate of $z$. Let the imaginary part of $z$ be nonzero. Match each entry in List-I to the correct entries in List-II. List-I List-II $\left(P\right)$ ${\left|z\right|}^{2}$ is equal to $\left(1\right)$ $12$ $\left(Q\right)$ ${\left|z-\bar{z}\right|}^{2}$ is equal to $\left(2\right)$ $4$ $\left(R\right)$ ${\left|z\right|}^{2}+{\left|z+\bar{z}\right|}^{2}$ is equal to $\left(3\right)$ $8$ $\left(S\right)$ ${\left|z+1\right|}^{2}$ is equal to $\left(4\right)$ $10$ $\left(5\right)$ $7$ The correct option is
   1. A.$\left(P\right)\rightarrow \left(1\right)\left(Q\right)\rightarrow \left(3\right)\left(R\right)\rightarrow \left(5\right)\left(S\right)\rightarrow \left(4\right)$
   2. B.$\left(P\right)\rightarrow \left(2\right)\left(Q\right)\rightarrow \left(1\right)\left(R\right)\rightarrow \left(3\right)\left(S\right)\rightarrow \left(5\right)$
   3. C.$\left(P\right)\rightarrow \left(2\right)\left(Q\right)\rightarrow \left(4\right)\left(R\right)\rightarrow \left(5\right)\left(S\right)\rightarrow \left(1\right)$
   4. D.$\left(P\right)\rightarrow \left(2\right)\left(Q\right)\rightarrow \left(3\right)\left(R\right)\rightarrow \left(5\right)\left(S\right)\rightarrow \left(4\right)$
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   Answer: (B)

   Given, ${\left|z\right|}^{3}+2{z}^{2}+4\bar{z}-8=0.......\left(1\right)$ Now on taking conjugate both side we get, ${\left|z\right|}^{3}+2{\bar{z}}^{2}+4z-8=0........\left(2\right)$ $\left\{\text{Note}\left|\bar{z}\right|=\left|z\right|\right\}$ Now subtracting both equations we get, $2\left({z}^{2}-{\bar{z}}^{2}\right)+4\left(\bar{z}-z\right)=0$ $\Rightarrow \left(z-\bar{z}\right)\left[2\left(z+\bar{z}\right)-4\right]=0$ $∵z=\bar{z}$ (not possible) or $4x=4\Rightarrow x=1$ $\left\{\text{as}z+\bar{z}=2x\right\}$ So, $z=1+\lambda i$ $\Rightarrow \left|z\right|=\sqrt{1+{\lambda }^{2}}\&\bar{z}=1-\lambda i$ Now putting the value of $z,\left|z\right|\&\bar{z}$ in given equation we get, ${\left(1+{\lambda }^{2}\right)}^{3/2}+2\left(1-{\lambda }^{2}+2\lambda i\right)+4\left(1-\lambda i\right)-8=0$ $\Rightarrow {\left(1+{\lambda }^{2}\right)}^{3/2}+2\left(1-{\lambda }^{2}\right)=4$ $\Rightarrow {\left(1+{\lambda }^{2}\right)}^{3/2}=2\left(1+{\lambda }^{2}\right)$ $\Rightarrow \left(1+{\lambda }^{2}\right)\left[\sqrt{1+{\lambda }^{2}}-2\right]=0$ $\Rightarrow {\lambda }^{2}=3$ Now solving, $\left(P\right)$ ${\left|z\right|}^{2}=1+{\lambda }^{2}=1+3=4$ $\left(Q\right)$ ${\left|z-\bar{z}\right|}^{2}={\left|1+\lambda i-\left(1-\lambda i\right)\right|}^{2}={\left|2\lambda i\right|}^{2}=4{\lambda }^{2}=12$ $\left(R\right)$ ${\left|z\right|}^{2}+{\left|z+\bar{z}\right|}^{2}=4+{\left|\left(1+\lambda i\right)+\left(1-\lambda i\right)\right|}^{2}=4+4=8$ $\left(S\right)$ ${\left|z+1\right|}^{2}={\left|1+\lambda i+1\right|}^{2}=4+{\lambda }^{2}=4+3=7$ $∴P\rightarrow \left(2\right),Q\rightarrow \left(1\right),R\rightarrow \left(3\right),S\rightarrow \left(5\right)$