Complex Numbers — JEE Main PYQs

188 previous year questions on Complex Numbers (Mathematics), newest first. Open any question for its complete step-by-step solution.

Let $u = \frac{2 z + i}{z - ki}, z = x + iy$ and $k > 0$ . If the curve represented by $Re (u) + Im (u) = 1$ intersects the $y$ -axis at the points $P$ and $Q$ …
JEE Main 2020View solution →
If a and b are real numbers such that $(2 + α)^{4} = a + b α$ , where $α = \frac{- 1 + i 3}{2}$ , then $a + b$ is equal to:
JEE Main 2020View solution →
If $α$ and $β$ be the roots of the equation $x^{2} - 2 x + 2 = 0$ , then the least value of $n$ for which $(\frac{α}{β})^{n} = 1$ is:
JEE Main 2019View solution →
If $z = \frac{3}{2} + \frac{i}{2} (i = - 1)$ , then $(1 + i z + z^{5} + i z^{8})^{9}$ is equal to:
JEE Main 2019View solution →
All the points in the set $S = {\frac{α + i}{α - i} : α \in R} (i = - 1)$ lie on a :
JEE Main 2019View solution →
Let z be a complex number such that $∣ z ∣ + z = 3 + i$ (where $i = - 1$ ). Then $∣ z ∣$ is equal to:
JEE Main 2019View solution →
The equation $∣ z - i ∣ = ∣ z - 1∣, i = - 1$ , represents:
JEE Main 2019View solution →
If $z$ and $w$ are two complex numbers such that $∣ z w ∣ = 1$ and $ar g (z) - ar g (w) = \frac{π}{2}$ , then
JEE Main 2019View solution →
Let $(- 2 - \frac{1}{3} i)^{3} = \frac{x + iy}{27} (i = - 1)$ , where x and y are real numbers, then $y - x$ equals:
JEE Main 2019View solution →
If $\frac{Z - α}{Z + α} (α \in R)$ is a purely imaginary number and $∣ Z ∣ = 2$ , then a value of $α$ is:
JEE Main 2019View solution →
Let $z \in C$ with $Im (z) = 10$ and it satisfies $\frac{2 z - n}{2 z + n} = 2 i - 1$ for some natural number $n$ . Then :
JEE Main 2019View solution →
Let $z_{1}$ and $z_{2}$ be two complex numbers satisfying $∣ z_{1} ∣ = 9$ and $∣ z_{2} - 3 - 4 i ∣ = 4$ . Then the minimum value of $∣ z_{1} - z_{2} ∣$ is :
JEE Main 2019View solution →
Let $z \in C$ be such that $∣ z ∣ < 1$ . If $w = \frac{5 + 3 z}{5 ( 1 - z )}$ , then
JEE Main 2019View solution →
Let $z_{0}$ be a root of the quadratic equation, $x^{2} + x + 1 = 0$ . If $z = 3 + 6 iz_{0}^{81} - 3 iz_{0}^{93}$ , then $ar g z$ is equal to
JEE Main 2019View solution →
Let $A = {θ \in (- \frac{π}{2}, π) : \frac{3 + 2 i s i n θ}{1 - 2 i s i n θ}$ is purely imaginary $}$ . Then the sum of the elements in A is:
JEE Main 2019View solution →
Let $α$ and $β$ be two roots of the equation $x^{2} + 2 x + 2 = 0$ , then $α^{15} + β^{15}$ is equal to:
JEE Main 2019View solution →
If $a > 0$ and $z = \frac{( 1 + i ) ^{2}}{a - i}$ , has magnitude $\frac{2}{5}$ , then $\overset{z}{ˉ}$ is equal to:
JEE Main 2019View solution →
If $α, β \in C$ are the distinct roots, of the equation $x^{2} - x + 1 = 0$ , then $α^{101} + β^{107}$ is equal to :
JEE Main 2018View solution →
Let $ω$ be a complex number such that $2 ω + 1 = z$ where $z = - 3$ . If
$111 1 - ω^{2} - 1 ω^{2} 1 ω^{2} ω^{7} = 3 k$ ,
then k is equal to :
JEE Main 2017View solution →
A value of $θ$ for which $\frac{2 + 3 i s i n θ}{1 - 2 i s i n θ}$ is purely imaginary, is:
JEE Main 2016View solution →
A complex number z is said to be unimodular if $∣ z ∣ = 1$ . Suppose $z_{1}$ and $z_{2}$ are complex numbers such that $\frac{z _{1} - 2 z _{2}}{2 - z _{1} z ˉ _{2}}$ is unimodular and $z_{2}$ is not unimodular. Then the point $z_{1}$ lies on $a$ :
JEE Main 2015View solution →
If z is a complex number such that $∣ z ∣ \geq 2$ , then the minimum value of $z + \frac{1}{2}$
JEE Main 2014View solution →
If z is a complex number of unit modulus and argument $θ$ , then $ar g (\frac{1 + z}{1 + z})$ equals
JEE Main 2013View solution →
If $z \neq = 1$ and $\frac{z ^{2}}{z - 1}$ is real, then the point represented by the complex number $z$ lies
JEE Main 2012View solution →
Let $α, β$ be real and $z$ be a complex number. If $z^{2} + α z + β = 0$ has two distinct roots on the line $Re z = 1$ , then it is necessary that
JEE Main 2011View solution →
If $ω (\neq = 1)$ is a cube root of unity, and $(1 + ω)^{7} = A + B ω$ . Then $(A, B)$ equals
JEE Main 2011View solution →
The number of complex numbers $z$ such that $∣ z - 1∣ = ∣ z + 1∣ = ∣ z - i ∣$ equals
JEE Main 2010View solution →
If $Z - \frac{4}{Z} = 2$ , then the maximum value of $∣ Z ∣$ is equal to
JEE Main 2009View solution →
The conjugate of a complex number is $\frac{1}{i - 1}$ . Then the complex number is
JEE Main 2008View solution →
If $∣ z + 4∣ \leq 3$ , then the maximum value of $∣ z + 1∣$ is
JEE Main 2007View solution →
The value of $\sum_{k = 1}^{10} (sin \frac{2 k π}{11} + i cos \frac{2 k π}{11})$ is
JEE Main 2006View solution →
If $z^{2} + z + 1 = 0$ , where $z$ is a complex number, then the value of $(z + \frac{1}{z})^{2} + (z^{2} + \frac{1}{z ^{2}})^{2} + (z^{3} + \frac{1}{z ^{3}})^{2} + \dots + (z^{6} + \frac{1}{z ^{6}})^{2}$ is
JEE Main 2006View solution →
If the cube roots of unity are $1, ω, ω^{2}$ then the roots of the equation
$(x - 1)^{3} + 8 = 0$ , are
JEE Main 2005View solution →
If $z_{1}$ and $z_{2}$ are two non-zero complex numbers such that $∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$ then $ar g z_{1} - ar g z_{2}$ is equal to
JEE Main 2005View solution →
If $ω = \frac{z}{z - \frac{1}{3} i}$ and $∣ ω ∣ = 1$ , then $z$ lies on
JEE Main 2005View solution →
Let $z, w$ be complex numbers such that $\overline{z} + i \overline{w} = 0$ and arg $zw = π$ . Then arg z equals
JEE Main 2004View solution →
If $z = x - i y$ and $z^{\frac{1}{3}} = p + i q$ , then $\frac{( \frac{x}{p} + \frac{y}{q} )}{( p ^{2} + q ^{2} )}$ is equal to
JEE Main 2004View solution →
If $z^{2} - 1 = ∣ z ∣^{2} + 1$ , then $z$ lies on
JEE Main 2004View solution →