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

Q 1 :
If $z_{1}, z_{2}$ are two distinct complex number such that $| \frac{z_{1} - 2 z_{2}}{\frac{1}{2} - z_{1} {\bar{z}}_{2}} |$ $= 2$ , then [2024]

$z_{1}$ lies on a circle of radius $\frac{1}{2}$ and $z_{2}$ lies on a circle of radius 1.
either $z_{1}$ lies on a circle of radius 1 or $z_{2}$ lies on a circle of radius $\frac{1}{2} .$
both $z_{1}$ and $z_{2}$ lie on the same circle.
either $z_{1}$ lies on a circle of radius $\frac{1}{2}$ or $z_{2}$ lies on a circle of radius 1.

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(2)

We have, $| \frac{z_{1} - 2 z_{2}}{\frac{1}{2} - z_{1} {\bar{z}}_{2}} |$ $= 2$

$\Rightarrow | z_{1} - 2 z_{2} | = | 1 - 2 z_{1} {\bar{z}}_{2} |$

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

$\Rightarrow (z_{1} - 2 z_{2}) ({\bar{z}}_{1} - 2 {\bar{z}}_{2}) = (1 - 2 z_{1} {\bar{z}}_{2}) (1 - 2 {\bar{z}}_{1} z_{2})$

$\Rightarrow | z_{1} |^{2} + 4 | z_{2} |^{2} - 2 {\bar{z}}_{1} z_{2} - 2 {\bar{z}}_{2} z_{1}$

$= 1 + 4 | z_{1} |^{2} | z_{2} |^{2} - 2 z_{1} {\bar{z}}_{2} - 2 {\bar{z}}_{1} z_{2}$

$\Rightarrow | z_{1} |^{2} + 4 | z_{2} |^{2} - 4 | z_{1} |^{2} | z_{2} |^{2} - 1 = 0$

$\Rightarrow (| z_{1} |^{2} - 1) (1 - 4 | z_{2} |^{2}) = 0$

$\Rightarrow | z_{1} | = 1, | z_{2} | = \frac{1}{2}$

Q 2 :
Let $r$ and $θ$ respectively be the modulus and amplitude of the complex number $z = 2 - i (2 \tan \frac{5 π}{8}),$ then $(r, θ)$ is equal to [2024]

$(2 \sec \frac{3 π}{8}, \frac{5 π}{8})$
$(2 \sec \frac{3 π}{8}, \frac{3 π}{8})$
$(2 \sec \frac{11 π}{8}, \frac{11 π}{8})$
$(2 \sec \frac{5 π}{8}, \frac{3 π}{8})$

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(2)

We have, $z = 2 - i (2 \tan \frac{5 π}{8})$

$= 2 + 2 i \tan (π - \frac{5 π}{8}) \Rightarrow z = 2 + 2 i \tan \frac{3 π}{8}$

$\Rightarrow z = 2 (\cos \frac{3 π}{8} + i \sin \frac{3 π}{8}) \sec \frac{3 π}{8}$

So, $r = 2 \sec \frac{3 π}{8}, θ = \frac{3 π}{8}$

Q 3 :
Let the complex numbers $α$ and $\frac{1}{\bar{α}}$ lie on the circles $| z - z_{0} |^{2} = 4$ and $| z - z_{0} |^{2} = 16$ respectively, where $z_{0} = 1 + i .$ Then, the value of $100 | α |^{2}$ is ________. [2024]

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(20)

Given $| z - z_{0} |^{2} = 4$ ...(i)

$| z - z_{0} |^{2} = 16$ ...(ii)

$α$ lies on (i), $∴$ $| α - z_{0} |^{2} = 4$

$\Rightarrow (α - z_{0}) (\bar{α} - {\bar{z}}_{0}) = 4 \Rightarrow α \bar{α} - α {\bar{z}}_{0} - z_{0} \bar{α} + | z_{0} |^{2} = 4$

$\Rightarrow | α |^{2} - α {\bar{z}}_{0} - z_{0} \bar{α} + 2 = 4 (∵ | z_{0} | = \sqrt{2})$

$\Rightarrow | α |^{2} - a {\bar{z}}_{0} - z_{0} \bar{α} = 2$ ...(iii)

$\frac{1}{\bar{α}}$ lies on (ii)

$∴$ $| \frac{1}{\bar{α}} - z_{0} |^{2} = 16 \Rightarrow (\frac{1}{\bar{α}} - z_{0}) (\frac{1}{α} - {\bar{z}}_{0}) = 16$

$\Rightarrow (1 - \bar{α} z_{0}) (1 - α {\bar{z}}_{0}) = 16 α \bar{α}$

$\Rightarrow 1 - \bar{α} z_{0} - α {\bar{z}}_{0} + α \bar{α}$ $z_{0} {\bar{z}}_{0} = 16 | α |^{2}$

$\Rightarrow 1 - \bar{α} z_{0} - α {\bar{z}}_{0} = 14 | α |^{2}$ ...(iv)

From (iii) and (iv), we get

$\Rightarrow 15 | α |^{2} = 3 \Rightarrow | α |^{2} = \frac{3}{15} = \frac{1}{5}$

$∴$ $100 | α |^{2} = 100 \times \frac{1}{5} = 20$

Q 4 :
If $α$ denotes the number of solutions of $| 1 - i |^{x}$ $= 2^{x}$ and $β = (\frac{| z |}{\arg (z)}),$ where $z = \frac{π}{4} {(1 + i)}^{4}$ $[\frac{1 - \sqrt{π} i}{\sqrt{π} + i} + \frac{\sqrt{π} - i}{1 + \sqrt{π} i}],$ $i = \sqrt{- 1},$ then the distance of the point $(α, β)$ from the line $4 x - 3 y = 7$ is __________ [2024]

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(3)

We have, $| 1 - i |^{x}$ $= 2^{x}$

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

$2^{x / 2} = 2^{x} \Rightarrow \frac{x}{2} = x \Rightarrow 2 x - x = 0 \Rightarrow x = 0$

$\Rightarrow α = 1$

and $β = (\frac{| z |}{a r g (z)}),$ where $z = \frac{π}{4} {(1 + i)}^{4}$ $[\frac{1 - \sqrt{π} i}{\sqrt{π} + i} + \frac{\sqrt{π} - i}{1 + \sqrt{π} i}]$

$z = 2 π i \Rightarrow \arg (z) = \frac{π}{2}$

and $| z | = 2 π$

$β = \frac{| z |}{\arg (z)} = \frac{2 π}{π / 2} = 4 \Rightarrow β = 4$

Now, the distance of point $(α, β)$ from the line $4 x - 3 y = 7$ is given by

$d = \frac{| 4 \times 1 - 3 \times 4 - 7 |}{\sqrt{16 + 9}} = \frac{| 4 - 12 - 7 |}{5} = \frac{15}{5} = 3$

Q 5 :
The area (in sq. units) of the region

$S = {z \in C : | z - 1 | \leq 2; (z + \bar{z}) + i (z - \bar{z}) \leq 2, Im (z) \geq 0}$ is [2024]

$\frac{7 π}{4}$
$\frac{3 π}{2}$
$\frac{7 π}{3}$
$\frac{17 π}{8}$

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(2)

$| z - 1 |$ $\leq 2 \Rightarrow {(x - 1)}^{2} + y^{2} \leq 4$

$z + \bar{z} + i (z - \bar{z}) \leq 2$

$\Rightarrow x + i y + x - i y + i (x + i y - x + i y) \leq 2$

$\Rightarrow x - y \leq 1$

$Im (z) \geq 0$

$\Rightarrow y \geq 0$

$∴$ $Required area = (\frac{\frac{3 π}{4}}{2 π})$ $(π) {(2)}^{2} = \frac{3}{2} π sq. units$

Q 6 :
Let $S_{1} = {z \in C : | z | \leq 5}$ , $S_{2} = {z \in C : Im (\frac{z + 1 - \sqrt{3} i}{1 - \sqrt{3} i}) \geq 0}$ and $S_{3} = {z \in C : Re (z) \geq 0} .$ Then the area of the region $S_{1} \cap S_{2} \cap S_{3}$ is : [2024]

$\frac{125 π}{12}$
$\frac{125 π}{24}$
$\frac{125 π}{6}$
$\frac{125 π}{4}$

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(1)

$Let z = x + i y be any complex number$

$S_{1} = {z \in C : | z | \leq 5}$

$S_{1} : x^{2} + y^{2} \leq 25$ ...(i)

$S_{2} : Im [\frac{x + iy}{1 - \sqrt{3} i} + 1] \geq 0$

$i.e.,$ $S_{2} : Im [\frac{(x + iy) (1 + \sqrt{3} i)}{4} + 1] \geq 0$

$\Rightarrow \sqrt{3} x + y \geq 0$ ...(ii)

$S_{3} : {z \in C : Re (z) \geq 0}$

$\Rightarrow x \geq 0$ ...(iii)

$Now,$ $\sqrt{3} x + y = 0$

$\Rightarrow y = - \sqrt{3} x$

$\Rightarrow \tan θ = 120 °$

$∴$ $Required area = \frac{25 π}{2} - \frac{1}{12} {(5)}^{2} π$

$= \frac{25 π}{2} - \frac{25 π}{12}$

$= \frac{125 π}{12}$

Q 7 :
The sum of the square of the modulus of the elements in the set ${z = a + i b : a, b \in Z, z \in C, | z - 1 | \leq 1, | z - 5 | \leq | z - 5 i |}$ is ____________. [2024]

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(9)

$We have, | z - 1 | \leq 1$

$\Rightarrow {(a - 1)}^{2} + b^{2} \leq 1$

$and | z - 5 | \leq | z - 5 i |$

$\Rightarrow {(a - 5)}^{2} + b^{2} \leq {(a)}^{2} + {(b - 5)}^{2}$

$\Rightarrow a \geq b$

$z = 0 + 0 i, 1 + 0 i, 2 + 0 i, 1 + i, 1 - i$

$\Rightarrow | z | = 0, 1, 2, \sqrt{2}, \sqrt{2}$

$∴$ $Sum of | z |^{2} = 0 + 1 + 4 + 2 + 2 = 9$

Q 8 :
Let z be a complex number such that |z| = 1. If $\frac{2 + k^{2} z}{k + \bar{z}} = k z, k \in R$ , then the maximum distance of from the circle |z – (1 + 2i)| = 1 is : [2025]

$\sqrt{3} + 1$
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$\sqrt{5} + 1$
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(3)

We have, $\frac{2 + k^{2} z}{k + \bar{z}} = k z$

$\Rightarrow 2 + k^{2} z = k^{2} z + k z \bar{z}$

$\Rightarrow | z |^{2} k = 2$ [ $∵ z \bar{z} = | z |^{2}$ ]

$\Rightarrow k = 2$ [ $∵ | z | = 1$ ]

The centre of the given circle is (1, 2) and its radius is 1.

Now, $k + i k^{2} = 2 + 4 i$ .

Maximum distance = $O P + r = \sqrt{1 + 4} + 1 = \sqrt{5} + 1$ .

Q 9 :
If $z_{1}, z_{2}, z_{3} \in C$ are the vertices of an equilateral triangle, whose centroid is $z_{0}$ , then $\sum_{k = 1}^{3} {(z_{k} - z_{0})}^{2}$ is equal to [2025]

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i
– i
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(1)

Centroid of triangle is $z_{0}$ , then $z_{1} + z_{2} + z_{3} = 3 z_{0}$

$\Rightarrow {(z_{1} + z_{2} + z_{3})}^{2} = 9 z_{0}^{2}$

$\Rightarrow z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 3 (z_{1}^{2} + z_{2}^{2} + z_{3}^{2}) = 9 z_{0}^{2}$ [ $∵$ For equilateral $△, z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1} = z_{1}^{2} + z_{2}^{2} + z_{3}^{2}$ ]

$\Rightarrow z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3 z_{0}^{2}$

Now, $\sum_{k = 1}^{3} {(z_{k} - z_{0})}^{2} = {(z_{1} - z_{0})}^{2} + {(z_{2} - z_{0})}^{2} + {(z_{3} - z_{0})}^{2}$

$= z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 3 z_{0}^{2} - 2 (z_{1} + z_{2} + z_{3}) z_{0}$

$= 6 z_{0}^{2} - 6 z_{0}^{2} = 0$ .

Q 10 :
If the locus of $z \in C$ , such that $Re (\frac{z - 1}{2 z + i}) + R e (\frac{\bar{z} - 1}{2 \bar{z} + i}) = 2$ , is a circle of radius r and center (a, b), then $\frac{15 a b}{r^{2}}$ is equal to: [2025]

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(4)

We have, $Re (\frac{z - 1}{2 z + i}) + R e (\frac{\bar{z} - 1}{2 \bar{z} + i}) = 2$

$\Rightarrow Re (\frac{z - 1}{2 z + i}) + R e (\frac{\bar{z - 1}}{2 z + i}) = 2$ $[∵ (\frac{\bar{z - 1}}{2 z + i}) = \frac{\bar{z} - 1}{2 \bar{z} - i}]$

$\Rightarrow 2 Re (\frac{z - 1}{2 z + i}) = 2 \Rightarrow Re (\frac{z - 1}{2 z + i}) = 1$

Let z = x + iy

$Re (\frac{(x - 1) + i y}{2 x + i (2 y + 2)}) = 1$

$\Rightarrow R e [\frac{((x - 1) + i y) (2 x - i (2 y + 1))}{(2 x + i (2 y + 1)) (2 x - i (2 y + 1))}] = 1$

$\Rightarrow \frac{2 x (x - 1) + y (2 y + 1)}{4 x^{2} + {(2 y + 1)}^{2}} = 1$

$\Rightarrow 2 x^{2} - 2 x + 2 y^{2} + y = 4 x^{2} + 4 y^{2} + 1 + 4 y$

$\Rightarrow 2 x^{2} + 2 y^{2} + 3 y + 2 x + 1 = 0$

$\Rightarrow x^{2} + y^{2} + x + \frac{3}{2} y + \frac{1}{2} = 0$

$∴ Centre = (\frac{- 1}{2}, \frac{- 3}{4}) and r = \sqrt{\frac{1}{4} + \frac{9}{16} - \frac{1}{2}} = \frac{\sqrt{5}}{4}$

$\Rightarrow a = \frac{- 1}{2}, b = \frac{- 3}{4}, r^{2} = \frac{5}{16}$

$∴ \frac{15 a b}{r^{2}} = 15 \times (\frac{- 1}{2}) \times (\frac{- 3}{4}) \times \frac{16}{5} = 18$ .

Topic Question Set

Q 1 :
If $z_{1}, z_{2}$ are two distinct complex number such that $| \frac{z_{1} - 2 z_{2}}{\frac{1}{2} - z_{1} {\bar{z}}_{2}} |$ $= 2$ , then [2024]

Q 2 :
Let $r$ and $θ$ respectively be the modulus and amplitude of the complex number $z = 2 - i (2 \tan \frac{5 π}{8}),$ then $(r, θ)$ is equal to [2024]

Q 3 :
Let the complex numbers $α$ and $\frac{1}{\bar{α}}$ lie on the circles $| z - z_{0} |^{2} = 4$ and $| z - z_{0} |^{2} = 16$ respectively, where $z_{0} = 1 + i .$ Then, the value of $100 | α |^{2}$ is ________. [2024]

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Q 5 :
The area (in sq. units) of the region

$S = {z \in C : | z - 1 | \leq 2; (z + \bar{z}) + i (z - \bar{z}) \leq 2, Im (z) \geq 0}$ is [2024]

Q 6 :
Let $S_{1} = {z \in C : | z | \leq 5}$ , $S_{2} = {z \in C : Im (\frac{z + 1 - \sqrt{3} i}{1 - \sqrt{3} i}) \geq 0}$ and $S_{3} = {z \in C : Re (z) \geq 0} .$ Then the area of the region $S_{1} \cap S_{2} \cap S_{3}$ is : [2024]

Q 7 :
The sum of the square of the modulus of the elements in the set ${z = a + i b : a, b \in Z, z \in C, | z - 1 | \leq 1, | z - 5 | \leq | z - 5 i |}$ is ____________. [2024]

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Q 8 :
Let z be a complex number such that |z| = 1. If $\frac{2 + k^{2} z}{k + \bar{z}} = k z, k \in R$ , then the maximum distance of from the circle |z – (1 + 2i)| = 1 is : [2025]

Q 9 :
If $z_{1}, z_{2}, z_{3} \in C$ are the vertices of an equilateral triangle, whose centroid is $z_{0}$ , then $\sum_{k = 1}^{3} {(z_{k} - z_{0})}^{2}$ is equal to [2025]

Q 10 :
If the locus of $z \in C$ , such that $Re (\frac{z - 1}{2 z + i}) + R e (\frac{\bar{z} - 1}{2 \bar{z} + i}) = 2$ , is a circle of radius r and center (a, b), then $\frac{15 a b}{r^{2}}$ is equal to: [2025]

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Topic Question Set

Q 1 : If z1,z2 are two distinct complex number such that |z1-2z212-z1z¯2|=2, then [2024]

Q 2 : Let r and θ respectively be the modulus and amplitude of the complex number z=2-i(2tan5π8), then (r,θ) is equal to [2024]

Q 3 : Let the complex numbers α and 1α¯ lie on the circles |z-z0|2=4 and |z-z0|2=16 respectively, where z0=1+i. Then, the value of 100|α|2 is ________. [2024]

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Q 4 : If α denotes the number of solutions of |1-i|x=2x and β=(|z|arg(z)), where z=π4(1+i)4[1-πiπ+i+π-i1+πi],i=-1, then the distance of the point (α,β) from the line 4x-3y=7 is __________ [2024]

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Q 5 : The area (in sq. units) of the region S={z∈C:|z-1|≤2; (z+z¯)+i(z-z¯)≤2, Im(z)≥0} is [2024]

Q 6 : Let S1={z∈C:|z|≤5}, S2={z∈C:Im(z+1-3i1-3i)≥0} and S3={z∈C:Re(z)≥0}. Then the area of the region S1∩S2∩S3 is : [2024]

Q 7 : The sum of the square of the modulus of the elements in the set {z=a+ib:a,b∈Z,z∈C,|z-1|≤1,|z-5|≤|z-5i|} is ____________. [2024]

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Q 8 : Let z be a complex number such that |z| = 1. If 2+k2zk+z–=kz, k∈R, then the maximum distance of from the circle |z – (1 + 2i)| = 1 is : [2025]

Q 9 : If z1,z2,z3∈C are the vertices of an equilateral triangle, whose centroid is z0, then ∑k=13(zk–z0)2 is equal to [2025]

Q 10 : If the locus of z∈C, such that Re(z–12z+i)+Re(z––12z–+i)=2, is a circle of radius r and center (a, b), then 15abr2 is equal to: [2025]

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Q 1 :
If $z_{1}, z_{2}$ are two distinct complex number such that $| \frac{z_{1} - 2 z_{2}}{\frac{1}{2} - z_{1} {\bar{z}}_{2}} |$ $= 2$ , then [2024]

Q 2 :
Let $r$ and $θ$ respectively be the modulus and amplitude of the complex number $z = 2 - i (2 \tan \frac{5 π}{8}),$ then $(r, θ)$ is equal to [2024]

Q 3 :
Let the complex numbers $α$ and $\frac{1}{\bar{α}}$ lie on the circles $| z - z_{0} |^{2} = 4$ and $| z - z_{0} |^{2} = 16$ respectively, where $z_{0} = 1 + i .$ Then, the value of $100 | α |^{2}$ is ________. [2024]

Q 5 :
The area (in sq. units) of the region

$S = {z \in C : | z - 1 | \leq 2; (z + \bar{z}) + i (z - \bar{z}) \leq 2, Im (z) \geq 0}$ is [2024]

Q 6 :
Let $S_{1} = {z \in C : | z | \leq 5}$ , $S_{2} = {z \in C : Im (\frac{z + 1 - \sqrt{3} i}{1 - \sqrt{3} i}) \geq 0}$ and $S_{3} = {z \in C : Re (z) \geq 0} .$ Then the area of the region $S_{1} \cap S_{2} \cap S_{3}$ is : [2024]

Q 7 :
The sum of the square of the modulus of the elements in the set ${z = a + i b : a, b \in Z, z \in C, | z - 1 | \leq 1, | z - 5 | \leq | z - 5 i |}$ is ____________. [2024]

Q 8 :
Let z be a complex number such that |z| = 1. If $\frac{2 + k^{2} z}{k + \bar{z}} = k z, k \in R$ , then the maximum distance of from the circle |z – (1 + 2i)| = 1 is : [2025]

Q 9 :
If $z_{1}, z_{2}, z_{3} \in C$ are the vertices of an equilateral triangle, whose centroid is $z_{0}$ , then $\sum_{k = 1}^{3} {(z_{k} - z_{0})}^{2}$ is equal to [2025]

Q 10 :
If the locus of $z \in C$ , such that $Re (\frac{z - 1}{2 z + i}) + R e (\frac{\bar{z} - 1}{2 \bar{z} + i}) = 2$ , is a circle of radius r and center (a, b), then $\frac{15 a b}{r^{2}}$ is equal to: [2025]