JEE Mains JEE Advance BITSAT
JEE Mains | PYQ

Year

All years 2026 2025 2024 2023 2022 2021

Let \( x \) and \( y \) be real numbers such that \( 50\!\left(\dfrac{2x}{1+3i} - \dfrac{y}{1-2i}\right) = 31 + 17i \), \( i = \sqrt{-1} \). Then the value of \( 10(x-3y) \) is:

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

Let \( S=\{z \in \mathbb{C}: z^2+4z+16=0\} \). Then is equal to:

If \( x^2 + x + 1 = 0 \), then the value of \( \left(x + \frac{1}{x}\right)^4 + \left(x^2 + \frac{1}{x^2}\right)^4 + \left(x^3 + \frac{1}{x^3}\right)^4 + \cdots + \left(x^{25} + \frac{1}{x^{25}}\right)^4 \) is:

Let \( z \) be the complex number satisfying \( |z-5| \le 3 \) and having maximum positive principal argument. Then \( 34 \left|\frac{5z-12}{5iz+16}\right|^{2} \) is equal to:

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

Let \( z = (1+i)(1+2i)(1+3i)\ldots(1+ni) \), where \( i=\sqrt{-1} \). If \( |z|^{2}=44200 \), then \( n \) is equal to ________ .

Let \( z \) be a complex number such that \( \lvert z-6 \rvert = 5 \) and \( \lvert z + 2 - 6i \rvert = 5 \). Then the value of \( z^{3} + 3z^{2} - 15z + 141 \) is equal to:

Let $S = \left\{ z \in C : \left\{ \left| \frac{z - 6i}{z - 2i} \right| = 1 \text{ and } \left| \frac{z - 8 + 2i}{z + 2i} \right| = \frac{3}{5} \right\} \right\}$. Then $\sum_{z \in S} |z|^2$ is equal to:

Let \( S=\{z\in\mathbb{C}:\,4z^{2}+\overline{z}=0\} \). Then \( \displaystyle \sum_{z\in S}|z|^{2} \) is equal to:

Let \( S=\{z\in \mathbb{C}:z^2+\sqrt{6}iz-3=0\} \). Then \( \sum\limits_{z\in S} z^8 \) is equal to:

Let \( z \) be a complex number such that \( |z+2| = |z-2| \) and \( \arg\!\left(\frac{z+3}{z-i}\right) = \frac{\pi}{4} \). Then \( |z|^{2} \) is equal to:

Let \( \alpha = \frac{-1 + i\sqrt{3}}{2} \) and \( \beta = \frac{-1 - i\sqrt{3}}{2} \), \( i = \sqrt{-1} \). If \( (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10} \), then \( m \) is ______________.

The number of values of \( z \in \mathbb{C} \), satisfying the equations \( \lvert z - (4 + 8i) \rvert = \sqrt{10} \) and \( \lvert z - (3 + 5i) \rvert + \lvert z - (5 + 11i) \rvert = 4\sqrt{5} \), is:

Let the set of all values of \( k \in \mathbb{R} \) such that the equation \( {z}(\overline{z}+2+i)+k(2+3i)=0, \; z \in \mathbb{C} \), has at least one solution, be the interval \( [\alpha,\beta] \). Then \( 9(\alpha+\beta) \) is equal to:

Let $S = \{z : 3 \le |2z - 3(1 + i)| \le 7\}$ be a set of complex numbers. Then $\min_{z \in S} \left| (z + \frac{1}{2}(5 + 3i)) \right|$ is equal to:

Let the circles \( C_1: |z| = r \) and \( C_2: |z - 3 - 4i| = 5, \ z \in \mathbb{C}, \) be such that \( C_2 \) lies within \( C_1 \). If \( z_1 \) moves on \( C_1, z_2 \) moves on \( C_2 \) and \( \min |z_1 - z_2| = 2 \), then \( \max |z_1 - z_2| \) is equal to:

← PreviousQuestion 6 of 17Next →

JEE Main 2026 23rd Jan Evening Shift

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