JEE Main Complex Number Questions with Solutions PDF

The JEE Main Complex Number questions are based on subtopics such as Definition of Complex Numbers, Complex Plane (Argand Diagram), Polar Form and Euler's Formula, Operations with Complex Numbers (Addition, Subtraction, Multiplication, Division) and many more.

JEE Main is held in 2 sessions in January and April every year. The exam runs for a duration of 3 hours

JEE Main Complex Number Questions with Solutions- Download PDF

The JEE Main Complex Number questions with solutions that are asked in JEE Main are explained in detail and are added to the PDF, and students can directly download the PDF by clicking on the link below.

JEE Main Complex Number Questions with Solutions

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Top 10 JEE Main Complex Number Questions with Solutions

The Top 10 JEE Main complex number questions with solutions are listed below and comprise all the major sub-topics like Roots of Complex Numbers, Complex Conjugate Roots Theorem, and Complex Exponential Form with detailed solutions step by step.

Q1: Let a complex number z, |z| ≠ 1, satisfy log1/√2 [(|z| + 11) / (|z| – 1)2] ≤ 2. Then, the largest value of |z| is equal to?

1. 8
2. 5
3. 6
4. 7

Ans. Option D

Detailed Solution:

Given that |z| ≠ 1.

log1/√2 [(|z| + 11) / (|z| – 1)2] ≤ 2

[(|z| + 11) / (|z| – 1)2] ≥ 1 / 2

 2|z| + 22 ≥ (|z| – 1)2

2|z| + 22 ≥ |z|2 – 2|z| + 1

|z|2 – 4|z| – 21 ≤ 0

(|z| – 7) (|z| + 3) ≤ 0

|z| ≤ 7

The largest value of |z| = 7

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Q2: The least value of |z| where z is a complex number that satisfies the inequality, i = √-1, is equal to

1. 2
2. 3
3. 8
4. √5

Ans: Option B

Detailed Solution:

2[{(|z| + 3) (|z| – 1)} / {|z| + 1}] ≥ log√2(16)

2[{(|z| + 3) (|z| – 1)} / {|z| + 1}] ≥ 23 [{(|z| + 3) (|z| – 1)} / {|z| + 1}] ≥ 3

|z|2 + 2 |z| – 3 ≥ 3 |z| + 3

 |z|2 – |z| – 6 ≥ 0

(|z| – 3) (|z| + 2) ≥ 0

|z|min = 3

The least value of |z| = 3

Q3: If |z1 + z2|2 = |z1|2 + |z2|2 , then z1/z2 is

1. purely imaginary
2. purely real
3. zero of purely imaginary
4. neither real nor imaginary

Ans: Option A

Detailed Solution:

Given that |z1 + z2|2 = |z1|2 + |z2|2 

So, z1/z2 is purely imaginary.

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Q4: If z and w are two complex numbers such that |z| ≤ 1, |w|≤ 1, and, Then z is equal to

1. One or i
2. i or -i
3. i or -1
4. 1 or -1

Ans: Option D

Detailed Solution:

Given |z| ≤ 1 and |w| ≤ 1

Let z = x1 + iy1

w = x2 + iy2

So x12 + y12 ≤ 1

x22 + y22 ≤ 1

Therefore, |z + iw| = |x1 + y1+ i(x2+iy2)| = 2

(x1 – y1)2 + (y2+x2)2 = 4 ….(1)

Also |z + iw| = |x1 + y1 + i(x2+ iy2)| = 2

(x1 – y2)2+(y1 – x2)2 = 4 ….(2)

Subtract (2) – (1)

(y1 – x2)2-(y1 + x2)2 = 0

y12 + x22– 2y1x2– y12– x22– 2y1x2 = 0

y1x2 = 0

y1 = 0

x12 ≤1

 -1≤ x ≤1

So z = 1+ i 0 or -1 + i0

Que 5: If 3/(2+ cos θ+i sin θ) = a+ib, then [(a-2)2+b2] is

1. 0
2. -1
3. 1
4. 2

Ans: Option C

Detailed Solution:

Given that 3/(2+ cos θ+i sin θ) = a+ib

(3/(2+ cos θ+ i sin θ))((2 + cos θ)-i sin θ)/(2 + cos θ)- i sin θ)

((6 + 3 cos θ)- i 3 sin θ)/((2+cos θ)2+sin2 θ)

((6 + 3 cos θ)- i(3 sin θ))/(5 + 4 cos θ)

Comparing with a+ib, we get;

a = (6+3 sin θ)/(5+4 cos θ)

b = -3 sin θ/(5+4 cos θ)

(a – 2)2+ b2 = [(6 + 3 sin θ)/(5 + 4 cos θ)] – 22+ (-3 sin θ/(5+4 cos θ))2

= ((-4 – 5 cos θ)2+ 9 sin2θ)/(5 + 4 cos θ)2

= ((4 +5 cos θ)2+ 9 sin2θ)/(5 + 4 cos θ)2

= (16 + 40 cos θ + 25 cos2θ+ 9 sin2θ)/(5 + 4 cos θ)2

= (16 + 40 cos θ + 16 cos2θ+ 9 (sin2θ + cos2θ))/(5 + 4 cos θ)2

= (16 + 40 cos θ + 16 cos2θ+ 9)/(5 + 4 cos θ)2

= (16 cos2θ+40 cos θ+ 25)/(5 + 4 cos θ)2

= (4 cos θ+5)2/(5+4 cos θ)2

= 1

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Q 6: If it is a real number, then the argument of sin θ + i cos θ is

1. π- tan-1 (4/3)
2. – tan-1 (3/4)
3. π- tan-1 (4/3)
4. tan-1 (4/3)

Ans: Option A

Detailed Solution:

Let z is real.

4 sin θ + 3 cos θ = 0

 tan θ = -3/4 [θ lies in 2nd quadrant]

arg (sin θ+ i cosθ ) = π + tan-1 (cos θ /sin θ)

π – tan-1(4/3)

Q 7: If √(x+iy) = ±(a+ib), then √(-x-iy) is equal to

1. ±(b+ia)
2. ±(a-ib)
3. (ai+b)
4. None of these

Ans: Option D

Detailed Solution:

Given that √(x+iy) = ±(a+ib)

Squaring both sides, we get;

(x+iy) = (a+ib)2

= a2 + 2aib – b2

= a2– b2 + 2aib

Comparing real and imaginary parts, we get the following;

x = a2-b2

y = 2ab

-x – iy = -a2 + b2– i(2ab)

-x – is = (-b + a)2

Taking square root, we get;

√(-x-iy) = ±(-b+ia)

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Q 8: The point in the set { z ∈ C: arg ((z-2)/(z-6i)) = π/2} (where C denotes the set of all complex numbers) lies on the curve, which is

1. hyperbola
2. pair of lines
3. parabola
4. circle

Ans: Option D

Detailed Solution:

Given that arg ((z-2)/(z-6i) = π/2

Arg (z-2) – arg(z-6i) = π/2

z = x+iy

arg ((x-2)+iy) – arg(x+(y-6)i = π/2

We know, tan-1A – tan-1B = tan-1[(A-B)/(1+AB)] 

x(x-2) + y(y-6) = 0

x2 – 2x +1 + y2 – 6y + 9 – 10 = 0

(x – 1)2 + (y – 3)2 = √(10)2

The point lies on a circle.

Q 9: The complex number (-√3+3i)(1-i)/(3+√3i)(i)(√3+√3i), when represented in the argand diagram, lies in

1. in the second quadrant,
2. in the first quadrant,
3. on the Y-axis (imaginary axis)
4. on the X-axis (real axis)

Ans: Option C

Detailed Solution:

Given z = (-√3+3i)(1-i)/(3+√3i)(i)(√3+√3i)

Taking √three outside

= √3(-1+√3i)(1-i)/(√3)2(√3+i)(i)(1+i)

= (1/√3)(-1+√3i)(1-i)/(-1+i√3)(1+i)

= (1-i)/√3(1+i)

Multiply the numerator and denominator with (1-i)

= (1-i)(1-i)/√3(1+i)(1-i)

= (1-i)2/√3(1+1)

= (1-2i-1)/2√3

= -2i/2√3

= -i/√3, purely imaginary

So, z lies on the Y-axis.

Q 10: The complex numbers sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for

1. x = nπ
2. x = (n+½)π
3. x = 0
4. No values of x

Ans: Option D

Detailed Solution:

Given that sin x + i cos 2x and cos x – i sin 2x are conjugate to each other.

sin x – i cos 2x = cos x – i sin 2x

Comparing real and imaginary parts

sin x = cos x ..(i)

cos 2x = sin 2x…(ii)

From equation (i), tan x = 1

From equation (ii) tan 2x = 1

2 tan x/(1-tan2x) = 1, not satisfied by tan x = 1.

Therefore, no value of x is possible.

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