datavorous/entrance-exam-dataset huggingface dataset:
Let \( y=y(x) \) be the solution of the differential equation \( \sin x \frac{d y}{d x}+y \cos x=4 x, x \in(0, \pi) . \) If \( y\left(\frac{\pi}{2}\right)=0 \), then \( y\left(\frac{\pi}{6}\right) \) is equal to

Options:
A \( -\frac{4}{9} \pi^{2} \)
B \( \frac{4}{9 \sqrt{3}} \pi^{2} \)
C \( \frac{-8}{9 \sqrt{3}} \pi^{2} \)
D \( -\frac{8}{9} \pi^{2} \)

The correct answer is:
\( -\frac{8}{9} \pi^{2} \)

Solution:
dydx+ycot x=4xsinx, x∈0, π
I.F. =e∫cotx=eln⁡sinx=sinx
∴y sinx=∫4xsinx.sinx dx
ysinx=2x2+c
∵yπ2=0
∴0×1=2×π22+c
⇒c=-π22
yπ6=2×π62-π2212=π218-π22×2
=π29-π2
=-8π29

Topic: Mathematics, Differential Equations, JEE Main
Let z0 be a root of quadratic equation, x2+x+1=0. If z=3+6iz081-3iz093 , then arg z is equal to:

Options:
A 0
B π4
C π6
D π3

The correct answer is:
π4

Solution:
z0=w or w2
When z0=w
z=3+6i×w81-3i×w93
=3+6i-3i=3+3i
When z0=w2
z=3+6i×w281-3i×w293
=3+6i-3i=3+3i
∴arg⁡z=tan-1ImzRez=tan-1⁡33=π4

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Let Z=113512310 and P=102210301. If Z=PQ-1, where Q is a square matrix of order 3, then the value of TradjQP is equal to (where Tr(A) represents the trace of a matrix A i.e. the sum of all the diagonal elements of the matrix A and adjB represents the adjoint matrix of matrix B)

Options:
A 3
B -1
C 4
D 65

The correct answer is:
-1

Solution:
Given, Z=PQ-1

P=-5, Z=10

Z=PQ⇒Q=-510=-12
ZQ=P.adjQ ⇒P.adjQ=-12Z
TradjQ.P=TrP.adjQ
=-12Tr(Z)
=-1

Topic: Mathematics, Determinants, JEE Main
Let z1=1, z2=2, z3=3 and z1+z2+z3=3+5i, then the value of Rez1z2-+z2z3-+z3z1- is equal to (Where z1, z2 and z3 are complex numbers)

Options:
A 1
B -1
C -12
D 0

The correct answer is:
0

Solution:
z1+z2+z32=32+5=14
⇒z1+z2+z3z1-+z2-+z3-=14
⇒z12+z22+z32+z1z2-+z2z3-+z3z1-+z1-z2+z2-z3+z3-z1=14
⇒1+4+9+2 Rez1z2-+z2z3-+z3z1-=14
⇒2 Rez1z2-+z2z3-+z3z1-=0
⇒ Rez1z2-+z2z3-+z3z1-=0

Topic: Mathematics, Complex Number, JEE Main
Let z1=3,z2=2 and z1+z2+z3=3+4i. If the real part of z1z2-+z2z3-+z3z1- is equal to 4, then z3 is equal to (where, i2=-1)

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
|z1+z2+z3|2=32+42=25
⇒z1+z2+z3z1-+z2-+z3-=25
⇒|z1|2+|z2|2+|z3|2+z1z2-+z2z3-+z3z1-+z1-z2+z2-z3+z3-z1=25
⇒9+4+z32+z1z2-+z2z3-+z3z1-+z1-z2+z2-z3+z3-z1=25
⇒z32+2real(z1z2-+z2z3-+z3z1-)=12
⇒|z3|2=4⇒z3=2

Topic: Mathematics, Complex Number, JEE Main
Let z=1+ai, be a complex number, a>0, such that z3 is a real number. Then, the sum 1+z+z2+….+z11 is equal to :

Options:
A 1365Â 3i
B -1365 3i
C -1250Â 3i
D 1250Â 3i

The correct answer is:
-1365 3i

Solution:
z=1+ai
Cubing both sides,
z3=1-a2+2ai+1-a2ai-2a2
z3=1-3a2+(3a-a3)i
∵ z3 is a real number.
a3-a2=0 ⇒ a= 3 a>0
1+z+z2…..z11=z12-1z-1=1+ 3i12-11+ 3i-1=1+ 3i12-13i
1+ 3i12=21212+32i12=212cos⁡π3+isin⁡π312=212 cos⁡4π+isin⁡4π=212
=212-13i=-13653i

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2016 (10 Apr Online)
Let $Z_1$ and $Z_2$ be any two complex number.
Statement 1: $\left|Z_1-Z_2\right| \geq\left|Z_1\right|-\left|Z_2\right|$
Statement 2: $\left|Z_1+Z_2\right| \leq\left|Z_1\right|+\left|Z_2\right|$

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true.

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.

Solution:
Statement $-1$ and 2 both are true.
It is fundamental property.
But Statement - 2 is not correct explanation for Statement $-1$.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2012 (07 May Online)
Let z1 and z2 be any two non-zero complex numbers such that 3z1=4z2. If z=3z12z2+2z23z1 then maximum value of z is
Note: In actual paper value of z was asked. Hence, none of the options given were correct. So we have modified the question as well as options.

Options:
A 72
B 92
C 52
D 12172

The correct answer is:
52

Solution:
Let arg3z12z2=θ, and we know that, if the argz=α, then arg1z=-α.
∴ arg2z23z1=-θ

Also, we know that a complex number w can be expressed as w=wcosα+isinα, where α is the argument of the complex number.

∴ z=32z1z2cosθ+isinθ+23z2z1cosθ-isinθ

Given, 3z1=4z2, ⇒z1z2=43,

⇒z=32×43cosθ+isinθ+23×34cosθ-isinθ

⇒z=2cosθ+isinθ+12cosθ-isinθ

⇒z=52cosθ+32sinθi

∴ z=254cos2θ+94sin2θ

Now, using sin2θ+cos2θ=1, we get

z=254cos2θ+941-cos2θ

⇒z=4cos2θ+94

And, we know that the maximum value of cosθ is 1.
∴ zmax=4+94=52.

Topic: Mathematics, Complex Number, JEE Main
Let z1 and z2 be any two non-zero complex numbers such that 3z1=4z2. If z=3z12z2+2z23z1 then maximum value of z is
Note: In actual paper value of z was asked. Hence, none of the options given were correct. So we have modified the question as well as options.

Options:
A 72
B 92
C 52
D 12172

The correct answer is:
52

Solution:
Let arg3z12z2=θ, and we know that, if the argz=α, then arg1z=-α.
∴ arg2z23z1=-θ

Also, we know that a complex number w can be expressed as w=wcosα+isinα, where α is the argument of the complex number.

∴ z=32z1z2cosθ+isinθ+23z2z1cosθ-isinθ

Given, 3z1=4z2, ⇒z1z2=43,

⇒z=32×43cosθ+isinθ+23×34cosθ-isinθ

⇒z=2cosθ+isinθ+12cosθ-isinθ

⇒z=52cosθ+32sinθi

∴ z=254cos2θ+94sin2θ

Now, using sin2θ+cos2θ=1, we get

z=254cos2θ+941-cos2θ

⇒z=4cos2θ+94

And, we know that the maximum value of cosθ is 1.
∴ zmax=4+94=52.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Let z1 and z2 be the nth roots of unity which are ends of a line segment that subtend a right angle at the origin. Then n must be of the form

Options:
A 4k+1
B 4k+2
C 4k+3
D 4k

The correct answer is:
4k

Solution:
11n=cosâ¡2rÏn+isinâ¡2rÏn, r=0, 1, â¦â¦, n-1
Let, z1=cosâ¡2r1Ïn+ isinâ¡2r1Ïn
And, z2=cosâ¡2r2Ïn+isinâ¡2r2Ïn .
Then, â z 1 O z 2 =amp z 1 z 2 =amp z 1 âamp z 2
=2r1-r2Ïn=Ï2 (Given)
â´ n=4 r1-r2=4Ã integer, so n is of the form 4k.

Topic: Mathematics, Complex Number, JEE Main
Let z1 and z2 be two complex number such that z1+z2=5 and z13+z23=20+15i. Then z14+z24 equals-

Options:
A 303
B 75
C 1515
D 253

The correct answer is:
75

Solution:
Given: z1+z2=5 & z13+z23=20+15i

Now, solving z13+z23=20+15i

⇒z1+z23-3z1z2z1+z2=20+15i

⇒53-3z1z25=20+15i

⇒125-15z1z2=20+15i

⇒z1z2=7-i

Now, finding z14+z24=z12+z222-2z1z22

⇒z14+z24=z1+z22-2z1z22-2z1z22

⇒z14+z24=25-27-i2-27-i2

⇒z14+z24=11+2i2-248-14i

⇒z14+z24=121-4+44i-96+28i

⇒z14+z24=21+72i

⇒z14+z24=212+722

⇒z14+z24=5625

⇒z14+z24=75

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Let z1 and z2 be two complex numbers satisfying |z1|=9 and |z2-3-4i|=4. Then the minimum value of z1-z2 is :

Options:
A 2
B 2
C 0
D 1

The correct answer is:
0

Solution:
We have,

C1:z1=9

C2:z2-3-4i=4

Centres of the circles are c1≡0,0 and c2≡3,4.

Radius of the circles are r1=9 and r2=4.

⇒c1c2=32+42=5 and r1-r2=5

Circles touch each other internally.
⇒z1-z2min=0 at the point of contact.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (12 Jan Shift 2)
Let z1 and z2 be two complex numbers such that argz1-z2=π4 and z1,z2 satisfy the equation z-3=Re(z). Then the imaginary part z1+z2 is equal to

Options:

The correct answer is:
6

Solution:
Let z1=x1+iy1,z2=x2+iy2
z1-z2=x1-x2+iy1-y2
∵argz1-z2=π4⇒tan-1y1-y2x1-x2=π4
⇒y1-y2=x1-x2----(1)
z1-3=Rez1⇒x1-32+y12=x12
z2-3=Rez2⇒x2-32+y22=x22
⇒x1-32-x2-32+y12-y22=x12-x22
⇒x1-x2x1+x2-6+y1-y2y1+y2=x1+x2x1-x2
⇒x1+x2-6+y1+y2=x1+x2
⇒y1+y2=6

Topic: Mathematics, Complex Number, JEE Main
Let z1 and z2 be two complex numbers such that argz1-z2=π4 and z1,z2 satisfy the equation z-3=Re(z). Then the imaginary part z1+z2 is equal to

Options:

The correct answer is:
6

Solution:
Let z1=x1+iy1,z2=x2+iy2
z1-z2=x1-x2+iy1-y2
∵argz1-z2=π4⇒tan-1y1-y2x1-x2=π4
⇒y1-y2=x1-x2----(1)
z1-3=Rez1⇒x1-32+y12=x12
z2-3=Rez2⇒x2-32+y22=x22
⇒x1-32-x2-32+y12-y22=x12-x22
⇒x1-x2x1+x2-6+y1-y2y1+y2=x1+x2x1-x2
⇒x1+x2-6+y1+y2=x1+x2
⇒y1+y2=6

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (27 Aug Shift 2)
Let z1 and z2 be two complex numbers such that z¯1=iz¯2 and argz1z¯2=π, then the argument of z1 is

Options:
A arg z2=π4
B arg z2=-3π4
C arg z1=π4
D arg z1=-3π4

The correct answer is:
arg z1=π4

Solution:
z¯1=iz¯2 ⇒z1=-iz2 ⇒iz1=z2⇒eiπ2z1=z2

Given, argz1z¯2=π

Let z1=reiθ ⇒ z2¯=re-iπ2+θ

So argei2θ+π2=π ⇒ 2θ+π2=π
⇒ θ=π4

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2022 (25 Jun Shift 2)
Let $Z_1$ and $Z_2$ be two roots of the equation $x^2+a Z+b=0$ being complex. Further, assume that the origin, $Z_1$ and $\mathrm{Z}_2$ form an equilateral triangle. Then

Options:
A
$a^2=4 b$

B
$a^2=b$

C
$a^2=2 b$

D
$a^2=3 b$

The correct answer is:
$a^2=3 b$

Solution:
$\mathrm{z}^2+\mathrm{az}+\mathrm{b}=0 ; \mathrm{z}_1+\mathrm{z}_2=-\mathrm{a} \& \mathrm{z}_1 \mathrm{z}_2=\mathrm{b}$
$0, \mathrm{z}, \mathrm{z}_2$ form an equilateral $\Delta$
$\therefore 0^2+\mathrm{z}_1{ }^2+\mathrm{z}_2{ }^2=0 . \mathrm{z}_1+\mathrm{z}_1 \cdot \mathrm{z}_2+\mathrm{z}_2 \cdot 0$
(for equation $\Delta, \mathrm{z}_1{ }^2+\mathrm{z}_2{ }^2+\mathrm{z}_3{ }^2=\mathrm{z}_1 \mathrm{z}_2+\mathrm{z}_2 \mathrm{z}_3+\mathrm{z}_3 \mathrm{z}_1$ ) $z_1^2+z_2^2=z_1 z_2$ or $\left(z_1+z_2\right)^2=3 z_1 z_2$
$\therefore \mathrm{a}^2=3 \mathrm{~b} \text {. }$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2003
Let z=1-i32,i=-1. Then the value of 21+z+1z3+z2+1z23+z3+1z33+…+z21+1z213 is______.

Options:

The correct answer is:
13

Solution:
Given:

z=1-i32

⇒z=--1+i32

⇒z=−ω

where, ω is the cube root of unity. So,

1+ω+ω2=0

ω3=1

Now, let

A=z+1z3+z2+1z23+z3+1z33+…+z21+1z213

⇒A=−ω−1ω3+ω2+1ω23+−ω3−1ω33+….+−ω21−1ω213

⇒A=-ω2+1ω3+ω4+1ω23+−1−113+….+−ω21−1ω213

⇒A=--ωω3+-ω2ω23+−1−13+….+−ω21−1ω213

⇒A=1+-1+−1−13+1+-1+−1−13….+−1−13

⇒A=31+-1+−1−13

⇒A=−8

Then,

21+A=21−8

=13

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (26 Aug Shift 1)
Let z=1+i and z1=1+iz¯z¯(1-z)+1z· Then 12π argz1 is equal to

Options:

The correct answer is:
9

Solution:
Given:

z=1+i

z¯=1-i

And, 1z=11+i×1-i1-i=1-i2

So,

z1=1+iz¯z¯(1-z)+1z

=1+i-i2(1-i)(-i)+1-i2

=1+i-i2-i+i2+12-i2

=2+i-3i-12=4+2i-3i-1

=-(4+2i)(3i-1)(3i)2-(1)2

=12i-4+6i2-2i10

=10i-1010=i-1

Argz1=3π4

∴12πargz1=12π×3π4=9

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Let z1, z2 and z3 are the points on the argand plane which lie on the circle with equation z-z0=43 (where z0 is the centre of the circle). If z1=0, z2=-4 and z3=4+3z0, then argâ¡z0 is equal to (where argZâ-Ï,Ï)

Options:
A π6
B 5π12
C 5π6
D 2π3

The correct answer is:
5π6

Solution:
Here z1+z2+z33=z0⇒z3=4+3z0
Therefore, center coincides with the circumcentre
⇒ Triangle is equilateral ⇒z1-z2=4
Clearly, z3 either lie in the second or third quadrant

So the centre z0 also lies in the second or third quadrant.
∴z0 can be =-2+23i,-2-23i

⇒argz0=5π6,7π6

Topic: Mathematics, Complex Number, JEE Main
Let z1,z2 be the roots of the equation z2+az+12=0 and z1, z2 form an equilateral triangle with origin. Then, the value of a is

Options:

The correct answer is:
6

Solution:
z2+az+12=0

z1+z2=-a and z1z2=12

If 0, z1, z2 are vertices of equilateral triangles

z2=z1eiπ3

z2=z112+i32

2z2-z1=3iz1

squaring both sides

⇒4z22+z12-4z1z2=-3z12

⇒z1+z22=3z1z2

⇒a2=3×12

⇒a=6

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Let z=32+i25+32-i25. If Rz and Iz respectively denote the real and imaginary parts of z, then

Options:
A Iz=0
B Rz<0 and Iz>0
C Rz>0 and Iz>0
D Rz=-3

The correct answer is:
Iz=0

Solution:
Let α=32+i2, then z=α5+α¯5

We know z¯n=z¯n ,

then, z=2Reα5

Hence, Iz=0.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Let α, β, γ, δ∈Z and let Aα, β, B1, 0, Cγ, δ and D1, 2 be the vertices of a parallelogram ABCD. If AB=10 and the points A and C lie on the line 3y=2x+1, then 2α+β+γ+δ is equal to

Options:
A 10
B 5
C 12
D 8

The correct answer is:
8

Solution:
Plotting the diagram of the given data we get,

We know that diagonals of a parallelogram bisect each other.

⇒Mid-point of AC=Mid-point of BD

⇒α+γ2,β+δ2=1+12,0+22

⇒α+γ2,β+δ2=1,1

⇒α+γ=2, β+δ=2

⇒2α+γ+ β+δ=8

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Let $Z$ and $W$ be complex numbers such that $|Z|=|W|$, and $\arg Z$ denotes the principal argument of $Z$.
Statement 1:If $\arg Z+\arg W=\pi$, then $Z=-\bar{W}$.
Statement 2: $|Z|=|W|$, implies arg $Z-\arg \bar{W}=\pi$.

Options:
A
Statement 1 is true, Statement 2 is false.

B
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

D
Statement 1 is false, Statement 2 is true.

The correct answer is:
Statement 1 is true, Statement 2 is false.

Solution:
Let $|Z|=|\mathrm{W}|=r$
$$
\Rightarrow Z=r e^{i \theta}, \mathrm{W}=r e^{i \phi}
$$
where $\theta+\phi=\pi$
$$
\therefore \quad \bar{W}=r e^{-i \phi}
$$
Now, $Z=r e^{i(\pi-\phi)}=r e^{i \pi} \times e^{-i \phi}=-r e^{-i \phi}$
$$
=-\bar{W}
$$
Thus, statement-1 is true but statement- 2 is false.

Topic: Mathematics, Complex Number, JEE Main
Let $Z$ and $W$ be complex numbers such that $|Z|=|W|$, and $\arg Z$ denotes the principal argument of $Z$.
Statement 1:If $\arg Z+\arg W=\pi$, then $Z=-\bar{W}$.
Statement 2: $|Z|=|W|$, implies arg $Z-\arg \bar{W}=\pi$.

Options:
A
Statement 1 is true, Statement 2 is false.

B
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

D
Statement 1 is false, Statement 2 is true.

The correct answer is:
Statement 1 is true, Statement 2 is false.

Solution:
Let $|Z|=|\mathrm{W}|=r$
$$
\Rightarrow Z=r e^{i \theta}, \mathrm{W}=r e^{i \phi}
$$
where $\theta+\phi=\pi$
$$
\therefore \quad \bar{W}=r e^{-i \phi}
$$
Now, $Z=r e^{i(\pi-\phi)}=r e^{i \pi} \times e^{-i \phi}=-r e^{-i \phi}$
$$
=-\bar{W}
$$
Thus, statement-1 is true but statement- 2 is false.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2012 (19 May Online)
Let z and w be non-zero complex numbers such that zw=z2 and z-z-+w+w-=4. If w varies, then the perimeter of the locus of z is

Options:
A 82 units
B 42 units
C 8 units
D 4 units

The correct answer is:
82 units

Solution:
Given, zw=z2âzw=zz-
âw=z- zâ 0
Now, z-z-+w+w-=4
âz-z-+z+z-=4
Let, z=x+iy, then we get,
x+y=2
which represents a square of side length equal to 22
â The perimeter of the locus is 82 units

Topic: Mathematics, Complex Number, JEE Main
Let Z be a complex number satisfying the relation Z3+4Z¯2Z=0. If the least possible argument of Z is –kπ, then k is equal to (here, arg⁡Z∈-π,π)

Options:

The correct answer is:
0.6

Solution:
Let, Z=reiθ
⇒r3eiθ+4re-i2θ=0
⇒r2ei5θ=-4
⇒r2=4 and ei5θ=-1
⇒r=2 and θ=-3π5,-π5,π5,3π5,π
⇒ least arg⁡Z is -3π5

⇒k=0.6

Topic: Mathematics, Complex Number, JEE Main
Let $z$ be a complex number such that the real part of $\frac{z-2 i}{z+2 i}$ is zero. Then, the maximum value of $|z-(6+8 i)|$ is equal to

Options:
A 12
B 10
C 8
D $\infty$

The correct answer is:
12

Solution:
$\begin{aligned} & \frac{\mathrm{z}-2 \mathrm{i}}{\mathrm{z}+2 \mathrm{i}}+\frac{\overline{\mathrm{z}}+2 \mathrm{i}}{\overline{\mathrm{z}}-2 \mathrm{i}}=0 \\ & \mathrm{z} \overline{\mathrm{z}}-2 \mathrm{i} \overline{\mathrm{z}}-2 \mathrm{i} \mathrm{z}+4(-1) \\ & +\mathrm{z} \overline{\mathrm{z}}+2 \mathrm{z} i+2 \overline{\mathrm{z} i}+4(-1)=0 \\ & \Rightarrow 2|\mathrm{z}|^2=8 \Rightarrow|\mathrm{z}|=2 \\ & |\mathrm{z}-(6+8 \mathrm{i})|_{\text {maximum }}=10+2=12\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Let $z$ be a complex number such that $|z+2|=1$ and $\operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$. Then the value of $|\operatorname{Re}(\overline{z+2})|$ is

Options:
A $\frac{2 \sqrt{6}}{5}$
B $\frac{24}{5}$
C $\frac{1+\sqrt{6}}{5}$
D $\frac{\sqrt{6}}{5}$

The correct answer is:
$\frac{2 \sqrt{6}}{5}$

Solution:
$|z+2|=1, \operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$
Let $z+2=\cos \theta+i \sin \theta$
$\begin{aligned}
& \frac{1}{z+2}=\cos \theta-i \sin \theta \\
& \Rightarrow \frac{z+1}{z+2}=1-\frac{1}{z+2}=1-(\cos \theta-i \sin \theta) \\
& =(1-\cos \theta)+\operatorname{isin} \theta \\
& \operatorname{Im}\left(\frac{z+1}{z+2}\right)=\sin \theta, \sin \theta=\frac{1}{5} \\
& \cos \theta= \pm \sqrt{1-\frac{1}{25}}= \pm \frac{2 \sqrt{6}}{5} \\
& |\operatorname{Re}(\overline{z+2})|=\frac{2 \sqrt{6}}{5}
\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Let z be a complex number such that z-iz+2i=1 and z=52 . Then, the value of z+3i is

Options:
A 10
B 72
C 154
D 23

The correct answer is:
72

Solution:
Given, z-i=z+2i

⇒x2+y-12=x2+y+22
⇒-2y+1=4y+4
⇒6y=-3⇒y=-12

From, z=52, we get,
x2+y2=254⇒x2=244=6⇒z=±6-i2
Hence, z+3i=6+254=494
z+3i=72

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2020 (09 Jan Shift 1)
Let z be a complex number such that z-iz+2i=1 and z=52 . Then the value of z+3i is

Options:
A 10
B 72
C 154
D 23

The correct answer is:
72

Solution:
Given, z-i=z+2i

⇒x2+y-12=x2+y+22
⇒-2y+1=4y+4
⇒6y=-3⇒y=-12

From, z=52, we get,
x2+y2=254⇒x2=244=6⇒z=±6-i2
Hence, z+3i=6+254=494
z+3i=72

Topic: Mathematics, Complex Number, JEE Main
Let $z$ be a complex number such that $|z|+z=3+i$

$($ where $i=\sqrt{-1})$

Then $|\mathrm{z}|$ is equal to :

Options:
A $\frac{\sqrt{34}}{3}$
B $\frac{5}{3}$
C $\frac{\sqrt{41}}{4}$
D $\frac{5}{4}$

The correct answer is:
$\frac{5}{3}$

Solution:
Since, $|z|+z=3+i$

Let $z=a+i b,$ then

$$

|z|+z=3+i \Rightarrow \sqrt{a^{2}+b^{2}}+a+i b=3+i

$$

Compare real and imaginary coefficients on both sides

$$

\begin{array}{l}

b=1, \sqrt{a^{2}+b^{2}}+a=3 \\

\sqrt{a^{2}+1}=3-a \\

a^{2}+1=a^{2}+9-6 a \\

6 a=8 \Rightarrow a=\frac{4}{3}

\end{array}

$$

Then,

$$

|z|=\sqrt{\left(\frac{4}{3}\right)^{2}+1}=\sqrt{\frac{16}{9}+1}=\frac{5}{3}

$$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Let $\mathrm{Z}$ be the point of intersection of the axis and the directrix of the parabola $4 x^2-12 x+4 y+5=0$. If $S$ is its focus, then the point which divides SZ in the ratio $2: 1$ is

Options:
A $\left(\frac{3}{2}, \frac{13}{12}\right)$
B $\left(1, \frac{13}{12}\righ
t)$
C $\left(\frac{3}{4}, \frac{13}{4}\right)$
D $\left(\frac{3}{2}, \frac{13}{4}\right)$

The correct answer is:
$\left(\frac{3}{2}, \frac{13}{12}\right)$

Solution:
Given equation of parabola is $4 x^2-12 x+4 y+5=0$
$\begin{aligned} & \text { i.e }\left(x-\frac{3}{2}\right)^2=-(y-1) \Rightarrow \text { vertex }=\left(\frac{3}{2}, 1\right) \\ & \therefore Z=\left(0+\frac{3}{2}, \frac{1}{4}+1\right)=\left(\frac{3}{2}, \frac{5}{4}\right) \\ & \text { and focus }(s)=\left(0+\frac{3}{2},-\frac{1}{4}+1\right)=\left(\frac{3}{2}, \frac{3}{4}\right) \\ & \text { Now, required point }=\left(\frac{6}{\frac{2}{2}+\frac{3}{2}}, \frac{10}{4}+\frac{3}{4}\right)=\left(\frac{3}{2}, \frac{13}{12}\right)\end{aligned}$

Topic: Mathematics, Parabola, JEE Main
Let Z be the set of integers. If A=x∈Z : 2x+2x2-5x+6=1 and B={x∈Z  :- 3< 2x-1<9}, then the number of subsets of the set A×B, is :

Options:
A 212
B 210
C 218
D 215

The correct answer is:
215

Solution:
Given 2x+2x-2x-3=1=20⇒x=-2, 2, 3.

∴nA=3

Again, -3<2x-1<9⇒-1<x<5

⇒x=0, 1, 2, 3, 4

∴nB=5

∴n A×B=3×5=15

∴Number of subset of A×B=215

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2019 (12 Jan Shift 2)
Let Z be the set of integers, if A=xâZ: x-3x2-5x+6=1 and B=xâZ: 10<3x+1<22, then the number of subsets of the set AÃB is

Options:
A 26
B 28
C 215
D 29

The correct answer is:
26

Solution:
For A,

x-3=1⇒x=2,4 or
x2-5x+6=0⇒x=2,3 but x≠3
∴A=2,4

For B,

B=4,5,6
nA×B=6

∴ number of subsets =26

Topic: Mathematics, Sets and Relations, JEE Main
Let z be those complex numbers which satisfy z+5≤4 and z1+i+z¯1-i⩾-10,i=-1. If the maximum value of |z+1|2 is α+β2, then the value of α+β is

Options:

The correct answer is:
48

Solution:
Let z=x+iy

Given |z+5|≤4

x+52+y2≤16 ...1

z1+i+z¯1-i≥-10

z+z¯+iz-z¯≥-10

x-y+5≥0 ...2

|z+1|2=|z--1|2

Let P-1,0

|z+1|Max.2=PB2 (where B is in 3rd quadrant)

for point of intersection

x+52+y2=16x-y+5=0y=±22

A22-5,22 B-22-5,-22

PB2=+22+42+222

|z+1|2=8+16+162+8

α+β2=32+162

α=32,β=16⇒α+β=48

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Let z∈C be such that z<1. If ω=5+3z51-z, then:

Options:
A 5Reω>1
B 5Imω<1
C 5Reω>4
D 4Imω>5

The correct answer is:
5Reω>1

Solution:
Given z<1 and ω=5+3z51-z

⇒5ω1-z=5+3z

⇒5ω-5ωz=5+3z

⇒z=5ω-53+5ω

Using the given condition z=5ω-13+5ω<1

⇒5ω-1<3+5ω

⇒5ω-1<5ω+35

⇒ω-1<ω--35

We know that the locus of a complex number z1 satisfying z1-a=z2-b is the perpendicular bisector of the line segment joining the points a and b.

Hence, the locus of the point ω satisfying ω-1=ω--35 is the line x=1-352=15

Hence, Reω>15

⇒5Reω>1.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Let z∈C, the set of complex numbers. Then the equation, 2z+3i-z-i=0 represents:

Options:
A A circle with radius 83
B An ellipse with length of minor axis 169
C An ellipse with length of major axis 163
D A circle with diameter 103

The correct answer is:
A circle with radius 83

Solution:
Let z=x+iy so that
2x+iy+3=x+iy-1
⇒2x2+y+32 = x2+y-12
⇒4x2+4y+32=x2+y-12
⇒3x2+3y2+26y+35=0
⇒x2+y2+263y+353=0
⇒r= 0+1699-353
⇒r=649=83

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2017 (08 Apr Online)
Let z∈C with Im(z)=10 and it satisfies 2 z-n2 z+n=2i-1 for some natural number n. Then

Options:
A n=20 and Re(z)=10
B n=40 and Re(z)=10
C n=20 and Re(z)=-10
D n=40 and Re(z)=-10

The correct answer is:
n=40 and Re(z)=-10

Solution:
Put z=x+10i
∴ 2x+10i-n=2i-1⋅2x+10i+n
compare real and imaginary coefficients
x=-10, n=40

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Let $z=\cos \theta+i \sin \theta$. Then, the value of $\sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}\right)$ at $\theta=2^{\circ}$ is

Options:
A
$\frac{1}{\sin 2^{\circ}}$

B
$\frac{1}{3 \sin 2^{\circ}}$

C
$\frac{1}{2 \sin 2^{\circ}}$

D
$\frac{1}{4 \sin 2^{\circ}}$

The correct answer is:
$\frac{1}{4 \sin 2^{\circ}}$

Solution:
Given that $z=\cos \theta+i \sin \theta=e^{i \theta}$
$$
\begin{aligned}
\therefore \sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}\right) & =\sum_{m=1}^{15} \operatorname{Im}\left(e^{i \theta}\right)^{2 m-1}=\sum_{m=1}^{15} \operatorname{Im} e^{i(2 m-1) \theta} \\
& =\sin \theta+\sin 3 \theta+\sin 5 \theta+\ldots+\sin 29 \theta \\
& =\frac{\sin \left(\frac{\theta+29 \theta}{2}\right) \sin \left(\frac{15 \times 2 \theta}{2}\right)}{\sin \left(\frac{2 \theta}{2}\right)}=\frac{\sin (15 \theta) \sin (15 \theta)}{\sin \theta}=\frac{1}{4 \sin 2^{\circ}}
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Main
Let $z \neq-i$ be any complex number such that $\frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+\mathrm{i}}$ is a purely imaginary number.
Then $\mathrm{z}+\frac{1}{\mathrm{z}}$ is:

Options:
A
0

B
any non-zero real number other than 1 .

C
any non-zero real number.

D
a purely imaginary number.

The correct answer is:
any non-zero real number.

Solution:
Let $z=x+i y$
$\frac{z-i}{z+i}$ is purely imaginary means its real part is zero.
$$
\begin{aligned}
&\frac{x+i y-i}{x+i y+i}=\frac{x+i(y-1)}{x+i(y+1)} \times \frac{x-i(y+1)}{x-i(y+1)} \\
&=\frac{x^2-2 i x(y+1)+x i(y-1)+y^2-1}{x^2+(y+1)^2} \\
&=\frac{x^2+y^2-1}{x^2+(y+1)^2}-\frac{2 x i}{x^2+(y+1)^2}
\end{aligned}
$$
for pure imaginary, we have
$$
\begin{aligned}
& \frac{x^2+y^2-1}{x^2+(y+1)^2}=0 \\
\Rightarrow & x^2+y^2=1 \\
\Rightarrow &(x+i y)(x-i y)=1 \\
\Rightarrow & x+i y=\frac{1}{x-i y}=z
\end{aligned}
$$
and $\frac{1}{z}=x-i y$
$$
z+\frac{1}{z}=(x+i y)+(x-i y)=2 x
$$
$\left(z+\frac{1}{z}\right)$ is any non-zero real number

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2014 (12 Apr Online)
Let Z=reiθ ( r>0 and π<θ≤3π ) is a root of the equation Z8-Z7+Z6-Z5+Z4-Z3+Z2-Z+1=0. If the sum of all values of θ is kπ, then k is equal to

Options:

The correct answer is:
16

Solution:
Given equation is -Z9-1-Z-1=0
⇒Z9=-1 where Z≠-1
⇒Z=ei2kπ9, where k=5,6,7,8,10,11,12,13
⇒ θ=10π9,12π9,14π9,16π9,20π9,22π9,24π9,26π9
⇒ the sum of all values of θ is equal to 16π
⇒ k=16

Topic: Mathematics, Complex Number, JEE Main