Complex Numbers jee mains questions | JEE Main Maths Complex Numbers Previous Year Question

Q 1 :
Let $α$ and $β$ be the sum and the product of all the non-zero solutions of the equation ${(\bar{z})}^{2} + | z | = 0, z \in C .$ Then $4 (α^{2} + β^{2})$ is equal to [2024]

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(C)

We have, ${(\bar{z})}^{2} + | z | = 0$ ...(i)

Let $z = x + i y \Rightarrow \bar{z} = x - i y$

From (i), ${(x - i y)}^{2} + | x + i y | = 0$

$\Rightarrow x^{2} - y^{2} - 2 i x y + \sqrt{x^{2} + y^{2}} = 0$

$\Rightarrow (x^{2} - y^{2} + \sqrt{x^{2} + y^{2}}) - i (2 x y) = 0$

$\Rightarrow x^{2} - y^{2} + \sqrt{x^{2} + y^{2}} = 0$ and $2 x y = 0$

Case I: $x = 0$ and $y \neq 0$

$\Rightarrow - y^{2} + | y | = 0 \Rightarrow | y | = y^{2} \Rightarrow y = \pm 1$

Case II: $x \neq 0$ and $y = 0$

$\Rightarrow x^{2} + | x | = 0$

$\Rightarrow x = 0$ only $(∵ x \in R)$

$\Rightarrow x = 0, y = 0$ (Rejected) $\Rightarrow z = i, - i$ are solutions.

Hence, $α = i + (- i) = 0$ and $β = i (- i) = - i^{2} = 1$

Hence, $4 (α^{2} + β^{2}) = 4 (0 + 1) = 4$

Q 2 :
Consider the following two statements

Statement I : For any two non-zero complex numbers $z_{1}, z_{2},$ $(| z_{1} | + | z_{2} |)$ $| \frac{z_{1}}{| z_{1} |} + \frac{z_{2}}{| z_{2} |} |$ $\leq 2 (| z_{1} | + | z_{2} |),$ and

Statement II : If $x, y, z$ are three distinct complex numbers and a, b, c are three positive real numbers such that $\frac{a}{| y - z |} = \frac{b}{| z - x |} = \frac{c}{| x - y |},$ then $\frac{a^{2}}{y - z} + \frac{b^{2}}{z - x} + \frac{c^{2}}{x - y} = 1 .$

Between the above two statements, [2024]

Statement I is correct but Statement II is incorrect.
Both Statement I and Statement II are incorrect.
Both Statement I and Statement II are correct.
Statement I is incorrect but Statement II is correct.

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(A)

We have, $(| z_{1} | + | z_{2} |)$ $| \frac{z_{1}}{| z_{1} |} + \frac{z_{2}}{| z_{2} |} |$ $\leq (| z_{1} | + | z_{2} |) (\frac{| z_{1} |}{| z_{1} |} + \frac{| z_{2} |}{| z_{2} |})$

$\Rightarrow (| z_{1} | + | z_{2} |)$ $| \frac{z_{1}}{| z_{1} |} + \frac{z_{2}}{| z_{2} |} |$ $\leq 2 (| z_{1} | + | z_{2} |)$

Hence, statement-I is correct.

Now, let $\frac{a}{| y - z |} = \frac{b}{| z - x |} = \frac{c}{| x - y |} = λ$

$\Rightarrow a = λ | y - z |, b = λ | z - x |, c = λ | x - y |$

$\Rightarrow a^{2} = λ^{2} (y - z) (\bar{y} - \bar{z})$ [ $∵$ $| x - y |^{2} = (x - y) (\bar{x - y})$ for any complex number $x$ and $y$ ]

$b^{2} = λ^{2} (z - x) (\bar{z} - \bar{x})$

and $c^{2} = λ^{2} (x - y) (\bar{x} - \bar{y})$

Now, $\frac{a^{2}}{y - z} + \frac{b^{2}}{z - x} + \frac{c^{2}}{x - y} = λ^{2} (\bar{y} - \bar{z} + \bar{z} - \bar{x} + \bar{x} - \bar{y}) = 0$

Hence, statement-II is incorrect.

Q 3 :
Let $z$ be a complex number such that $| z + 2 | = 1$ and $Im$ $(\frac{z + 1}{z + 2})$ $= \frac{1}{5} .$ Then the value of $| R e (\bar{z + 2}) |$ is [2024]

$\frac{2 \sqrt{6}}{5}$
$\frac{\sqrt{6}}{5}$
$\frac{1 + \sqrt{6}}{5}$
$\frac{24}{5}$

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(A)

Let $z = x + i y$

$\Rightarrow | x + i y + 2 | = 1 \Rightarrow {(x + 2)}^{2} + y^{2} = 1$ ...(i)

Also, $Im$ $(\frac{(x + 1) + i y}{(x + 2) + i y})$ $= \frac{1}{5}$

$\Rightarrow Im$ $(\frac{[(x + 1) + i y] [(x + 2) - i y]}{{(x + 2)}^{2} + y^{2}})$ $= \frac{1}{5}$

$\Rightarrow \frac{(x + 2) y - (x + 1) y}{{(x + 2)}^{2} + y^{2}} = \frac{1}{5}$

$\Rightarrow \frac{y}{1} = \frac{1}{5}$ [Using (i)]

$\Rightarrow {(x + 2)}^{2} + \frac{1}{25} = 1 \Rightarrow {(x + 2)}^{2} = 1 - \frac{1}{25}$

$\Rightarrow (x + 2) = \pm \frac{2 \sqrt{6}}{5}$

$∴$ $| Re (\bar{z + 2}) |$ $=$ $| Re[(x + 2) - i y)] |$ $=$ $| (x + 2) |$ $= \frac{2 \sqrt{6}}{5} .$

Q 4 :
The sum of all possible values of $θ \in [- π, 2 π]$ , for which $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is purely imaginary, is equal to: [2024]

$2 π$
$4 π$
$3 π$
$5 π$

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(C)

We have, $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is purely imaginary.

$∴ \frac{1 + i \cos θ}{1 - 2 i \cos θ} + \frac{1 - i \cos θ}{1 + 2 i \cos θ} = 0$ $(∵ z + \bar{z} = 0)$

$\Rightarrow \frac{1 + i \cos θ + 2 i \cos θ - 2 \cos^{2} θ + 1 - i \cos θ - 2 i \cos θ - 2 \cos^{2} θ}{1 + 4 \cos^{2} θ} = 0$

$\Rightarrow 2 - 4 \cos^{2} θ = 0 \Rightarrow \cos^{2} θ = \frac{1}{2}$

$θ = n π \pm \frac{π}{4}$

$θ$ can be $\frac{π}{4}, \frac{- π}{4}, \frac{3 π}{4}, \frac{- 3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ $(∵ θ \in [- π, 2 π])$

$∴$ Required $= \frac{π}{4} - \frac{π}{4} + \frac{3 π}{4} - \frac{3 π}{4} + \frac{5 π}{4} + \frac{7 π}{4} = 3 π$

Q 5 :
Let $S = {z \in C : | z - 1 | = 1$ and $(\sqrt{2} - 1) (z + \bar{z}) - i (z - \bar{z}) = 2 \sqrt{2}} .$ Let $z_{1}, z_{2} \in S$ be such that $| z_{1} |$ $=$ $\max_{z \in S}$ $| z |$ and $| z_{2} |$ $=$ $\min_{z \in S}$ $| z | .$ Then $| \sqrt{2} z_{1} - z_{2} |^{2}$ equals: [2024]

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(B)

We have, $S = {z \in C : | z - 1 | = 1$ and $(\sqrt{2} - 1) (z + \bar{z}) - i (z - \bar{z}) = 2 \sqrt{2}}$

Let $z = x + i y$ and $\bar{z} = x - i y$

${(x - 1)}^{2} + y^{2} = 1$ ...(i)

$(\sqrt{2} - 1) x + y = \sqrt{2}$ ...(ii)

Equation (i) and (ii) intersect at (1, 1) and $(1 + \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$

$∴$ $S = {(1, 1), (1 + \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})}$

$| z_{1} |$ $= \max_{z \in S}$ $| z |$ at $(1 + \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$

$∴$ $z_{1} = 1 + \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$

$z_{2} = 1 + i$

$∴$ $\sqrt{2} z + 1 + i - 1 - i = \sqrt{2}$

$∴$ $| \sqrt{2} z_{1} - z_{2} |^{2} = 2$

Q 6 :
If $S = {z \in C :$ $| z - i |$ $=$ $| z + i |$ $=$ $| z - 1 |$ $}$ then $n (S)$ is: [2024]

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(A)

Given, $| z - i |$ $=$ $| z + i |$ $=$ $| z - 1 |$ ...(i)

Putting $z = x + i y$ in (i), we get $| x + i y - i |$ $=$ $| x + i y + i |$ $=$ $| x + i y - 1 |$

Now, $| x + i (y - 1) |$ $=$ $| x - 1 + i y |$

$\Rightarrow x^{2} + {(y - 1)}^{2} = {(x - 1)}^{2} + y^{2}$

$\Rightarrow x^{2} + y^{2} + 1 - 2 y = x^{2} + 1 - 2 x + y^{2} \Rightarrow 2 x - 2 y = 0$

$\Rightarrow x - y = 0$ ...(ii)

Again, $| x + i (y + 1) |$ $=$ $| x - 1 + i y |$

$\Rightarrow x^{2} + {(y + 1)}^{2}$ $=$ ${(x - 1)}^{2}$ $+ y^{2}$

$\Rightarrow x^{2} + y^{2} + 1 + 2 y = x^{2} + 1 - 2 x + y^{2} \Rightarrow 2 x + 2 y = 0$

$\Rightarrow x + y = 0$ ...(iii)

From (ii) and (iii), we get $x = 0$ and $y = 0$

Thus, $z = 0 + i 0 ∴ n (S) = 1$

Q 7 :
If $z = \frac{1}{2} - 2 i$ is such that $| z + 1 |$ $= α z + β (1 + i),$ $i = \sqrt{- 1}$ and $α, β \in R,$ then $α + β$ is equal to [2024]

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(C)

Given, $z = \frac{1}{2} - 2 i$ ...(i)

and $| z + 1 | = α z + β (1 + i)$ ...(ii)

From (i) and (ii), we get:

$| \frac{1}{2} - 2 i + 1 |$ $= α (\frac{1}{2} - 2 i) + β (1 + i)$

$\Rightarrow$ $| \frac{3}{2} - 2 i |$ $= α (\frac{1}{2} - 2 i) + β (1 + i)$

$\Rightarrow \sqrt{\frac{9}{4} + 4} = \frac{α}{2} - 2 α i + β + β i$ $[∵$ $| z |$ $= \sqrt{x^{2} + y^{2}}]$

$\Rightarrow \sqrt{\frac{25}{4}} = \frac{α}{2} + β + i (- 2 α + β) \Rightarrow \frac{5}{2} = \frac{α}{2} + β + i (- 2 α + β)$

$\Rightarrow \frac{5}{2} = \frac{α}{2} + β$ ...(iii) and $- 2 α + β = 0 \Rightarrow β = 2 α$ ...(iv)

Solving (iii) and (iv), we get $α = 1$ and $β = 2$

$∴$ $α + β = 1 + 2 = 3$

Q 8 :
If $z = x + i y, x y \neq 0,$ satisfies the equation $z^{2} + i \bar{z} = 0,$ then $| z^{2} |$ is equal to [2024]

$9$
$\frac{1}{4}$
$1$
$4$

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(C)

We have, $z^{2} + i \bar{z} = 0$

$\Rightarrow z^{2} = - i \bar{z} \Rightarrow$ $| z^{2} |$ $=$ $| - i \bar{z} |$

$\Rightarrow | z^{2} | = | - i | | \bar{z} | \Rightarrow | z^{2} | = 1 | z |$

$\Rightarrow | z |^{2} - | z | = 0$ $[∴ | \bar{z} | = | z |]$

$\Rightarrow$ $| z |$ $(| z | - 1) = 0$

$\Rightarrow | z | = 0 (not acceptable) or | z | = 1 \Rightarrow | z |^{2} = 1$

Q 9 :
Let $z_{1}$ and $z_{2}$ be two complex numbers such that $z_{1} + z_{2} = 5$ and $z_{1}^{3} + z_{2}^{3} = 20 + 15 i .$ Then $| z_{1}^{4} + z_{2}^{4} |$ equals [2024]

$30 \sqrt{3}$
$15 \sqrt{15}$
$75$
$25 \sqrt{3}$

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(C)

Given, $z_{1} + z_{2} = 5 \Rightarrow {(z_{1} + z_{2})}^{3} = 5^{3}$

$\Rightarrow 20 + 15 i + 3 z_{1} z_{2} (5) = 125$

$\Rightarrow z_{1} z_{2} = 7 - i$

Now $z_{1}^{4} + z_{2}^{4} = {[{(z_{1} + z_{2})}^{2} - 2 z_{1} z_{2}]}^{2} - 2 z_{1}^{2} z_{2}^{2}$

$= {[25 - 2 (7 - i)]}^{2} - 2 {(7 - i)}^{2}$

$= 625 + 4 {(7 - i)}^{2} - 100 (7 - i) - 2 {(7 - i)}^{2}$

$= 625 + (7 - i) [2 (7 - i) - 100]$

$= 625 + (7 - i) [14 - 2 i - 100]$

$= 625 + (7 - i) [- 86 - 2 i]$

$= 625 - 602 - 14 i + 86 i - 2 = 21 + 72 i$

$∴$ $| z_{1}^{4} + z_{2}^{4} | = \sqrt{441 + 5184} = \sqrt{5625} = 75$

Q 10 :
Let $P = {z \in C : | z + 2 - 3 i | \leq 1}$ and

   $Q = {z \in C : z (1 + i) + \bar{z} (1 - i) \leq - 8} .$

Let in $P \cap Q, | z - 3 + 2 i |$ be maximum and minimum at $z_{1}$ and $z_{2}$ respectively. If $| z_{1} |^{2} + 2 | z_{2} |^{2} = α + β \sqrt{2}$ , where $α, β$ are integers, then $α + β$ equals________ .   [2024]

Topic Question Set

Q 1 :
Let $α$ and $β$ be the sum and the product of all the non-zero solutions of the equation ${(\bar{z})}^{2} + | z | = 0, z \in C .$ Then $4 (α^{2} + β^{2})$ is equal to [2024]

Q 3 :
Let $z$ be a complex number such that $| z + 2 | = 1$ and $Im$ $(\frac{z + 1}{z + 2})$ $= \frac{1}{5} .$ Then the value of $| R e (\bar{z + 2}) |$ is [2024]

Q 4 :
The sum of all possible values of $θ \in [- π, 2 π]$ , for which $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is purely imaginary, is equal to: [2024]

Q 5 :
Let $S = {z \in C : | z - 1 | = 1$ and $(\sqrt{2} - 1) (z + \bar{z}) - i (z - \bar{z}) = 2 \sqrt{2}} .$ Let $z_{1}, z_{2} \in S$ be such that $| z_{1} |$ $=$ $\max_{z \in S}$ $| z |$ and $| z_{2} |$ $=$ $\min_{z \in S}$ $| z | .$ Then $| \sqrt{2} z_{1} - z_{2} |^{2}$ equals: [2024]

Q 6 :
If $S = {z \in C :$ $| z - i |$ $=$ $| z + i |$ $=$ $| z - 1 |$ $}$ then $n (S)$ is: [2024]

Q 7 :
If $z = \frac{1}{2} - 2 i$ is such that $| z + 1 |$ $= α z + β (1 + i),$ $i = \sqrt{- 1}$ and $α, β \in R,$ then $α + β$ is equal to [2024]

Q 8 :
If $z = x + i y, x y \neq 0,$ satisfies the equation $z^{2} + i \bar{z} = 0,$ then $| z^{2} |$ is equal to [2024]

Q 9 :
Let $z_{1}$ and $z_{2}$ be two complex numbers such that $z_{1} + z_{2} = 5$ and $z_{1}^{3} + z_{2}^{3} = 20 + 15 i .$ Then $| z_{1}^{4} + z_{2}^{4} |$ equals [2024]

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(36)

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Topic Question Set

Q 1 : Let α and β be the sum and the product of all the non-zero solutions of the equation (z¯)2+|z|=0, z∈C. Then 4(α2+β2) is equal to [2024]

Q 3 : Let z be a complex number such that |z+2|=1 and Im(z+1z+2)=15. Then the value of |Re(z+2¯)| is [2024]

Q 4 : The sum of all possible values of θ∈[-π,2π], for which 1+icosθ1-2icosθ is purely imaginary, is equal to: [2024]

Q 5 : Let S={z∈C:|z-1|=1 and (2-1)(z+z¯)-i(z-z¯)=22}. Let z1,z2∈S be such that |z1|=maxz∈S|z| and |z2|=minz∈S|z|. Then |2z1-z2|2 equals: [2024]

Q 6 : If S={z∈C:|z-i|=|z+i|=|z-1|} then n(S) is: [2024]

Q 7 : If z=12-2i is such that |z+1|=αz+β(1+i),i=-1 and α,β∈R, then α+β is equal to [2024]

Q 8 : If z=x+iy,xy≠0, satisfies the equation z2+iz¯=0, then |z2| is equal to [2024]

Q 9 : Let z1 and z2 be two complex numbers such that z1+z2=5 and z13+z23=20+15i. Then |z14+z24| equals [2024]

Q 10 : Let P={z∈C:|z+2-3i|≤1} and Q={z∈C:z(1+i)+z¯(1-i)≤-8}. Let in P∩Q,|z-3+2i| be maximum and minimum at z1 and z2 respectively. If |z1|2+2|z2|2=α+β2, where α,β are integers, then α+β equals________ . [2024]

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(36)

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Q 1 :
Let $α$ and $β$ be the sum and the product of all the non-zero solutions of the equation ${(\bar{z})}^{2} + | z | = 0, z \in C .$ Then $4 (α^{2} + β^{2})$ is equal to [2024]

Q 3 :
Let $z$ be a complex number such that $| z + 2 | = 1$ and $Im$ $(\frac{z + 1}{z + 2})$ $= \frac{1}{5} .$ Then the value of $| R e (\bar{z + 2}) |$ is [2024]

Q 4 :
The sum of all possible values of $θ \in [- π, 2 π]$ , for which $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is purely imaginary, is equal to: [2024]

Q 5 :
Let $S = {z \in C : | z - 1 | = 1$ and $(\sqrt{2} - 1) (z + \bar{z}) - i (z - \bar{z}) = 2 \sqrt{2}} .$ Let $z_{1}, z_{2} \in S$ be such that $| z_{1} |$ $=$ $\max_{z \in S}$ $| z |$ and $| z_{2} |$ $=$ $\min_{z \in S}$ $| z | .$ Then $| \sqrt{2} z_{1} - z_{2} |^{2}$ equals: [2024]

Q 6 :
If $S = {z \in C :$ $| z - i |$ $=$ $| z + i |$ $=$ $| z - 1 |$ $}$ then $n (S)$ is: [2024]

Q 7 :
If $z = \frac{1}{2} - 2 i$ is such that $| z + 1 |$ $= α z + β (1 + i),$ $i = \sqrt{- 1}$ and $α, β \in R,$ then $α + β$ is equal to [2024]

Q 8 :
If $z = x + i y, x y \neq 0,$ satisfies the equation $z^{2} + i \bar{z} = 0,$ then $| z^{2} |$ is equal to [2024]

Q 9 :
Let $z_{1}$ and $z_{2}$ be two complex numbers such that $z_{1} + z_{2} = 5$ and $z_{1}^{3} + z_{2}^{3} = 20 + 15 i .$ Then $| z_{1}^{4} + z_{2}^{4} |$ equals [2024]