JEE Main Maths Complex Numbers previous year questions with solutions are given on this page. Complex numbers is an important topic for the JEE Main exam. Students can definitely expect 1-2 questions on this topic. We also provide Complex Numbers JEE Main questions with solutions PDF. Students can easily download the PDF of Complex Numbers JEE Main previous year questions with solutions.

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Question 1:

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

(a) 8

(b) 5

(c) 6

(d) 7

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

Hence, option (d) is the answer.

Question 2:

The least value of |z| where z is a complex number which satisfies the inequality

\(\begin{array}{l}exp\left ( \frac{(\left| z\right|+3)(\left| z\right|-1)}{\left| \left| z\right|\right|+1} log_{e}2\right )\geq log_{\sqrt{2}}\left|5\sqrt{7}+9i \right|\end{array} \)
, i = √-1, is equal to

(a) 2

(b) 3

(c) 8

(d) √5

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

Hence, option (b) is the answer.

Question 3:

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

(a) purely imaginary

(b) purely real

(c) zero of purely imaginary

(d) neither real nor imaginary

Solution:

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

\(\begin{array}{l}\Rightarrow (z_{1}+z_{2})+(\bar{z_{1}}+\bar{z_{2}})=\left | z_{1} \right |^{2}+\left | z_{2} \right |^{2}\end{array} \)
\(\begin{array}{l}\Rightarrow \left | z_{1} \right |^{2}+\left | z_{2} \right |^{2}=\left | z_{1} \right |^{2}+\left | z_{2} \right |^{2}+z_{1}\bar{z_{2}}+z_{2}\bar{z_{1}}\end{array} \)
\(\begin{array}{l}\Rightarrow 0=z_{1}\bar{z_{2}}+z_{2}\bar{z_{1}}\end{array} \)
\(\begin{array}{l}\Rightarrow z_{1}\bar{z_{2}}=-z_{2}\bar{z_{1}}\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{\bar{z_{1}}}{\bar{z_{2}}}=-\frac{z_{1}}{z_{2}}\end{array} \)
\(\begin{array}{l}\Rightarrow \bar{(\frac{z_{1}}{z_{2}})}=-\frac{z_{1}}{z_{2}}\end{array} \)

So z1/z2 is purely imaginary.

Hence, option (a) is the answer.

Question 4:

If z and w are two complex numbers such that |z| ≤ 1, |w|≤ 1 and

\(\begin{array}{l}|z + iw| =\left | z-i\bar{w} \right |=2\end{array} \)
, Then, z is equal to

(a) 1 or i

(b) i or -i

(c) i or -1

(d) 1 or -1

Solution:

Given |z| ≤ 1and |w| ≤ 1

\(\begin{array}{l}|z + iw| = \left | z-i\bar{w} \right |=2\end{array} \)

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+ i0 or -1 + i0

Hence, option (d) is the answer.

Question 5:

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

(a) 0

(b) -1

(c) 1

(d) 2

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

Hence, option (c) is the answer.

Question 6:

If

\(\begin{array}{l}\frac{3+i\sin \theta }{4-i\cos \theta }, \theta \in [0,2\pi ]\end{array} \)
is a real number, then the argument of sin θ + i cos θ is

(a) π- tan-1 (4/3)

(b) – tan-1 (3/4)

(c) π- tan-1 (4/3)

(d) tan-1 (4/3)

Solution:

Let

\(\begin{array}{l}z=\frac{3+i\sin \theta }{4-i\cos \theta }\times \frac{4+i\cos \theta }{4+i\cos \theta }\end{array} \)
\(\begin{array}{l}= \frac{12-\sin \theta \cos \theta +i(4\sin \theta +3\cos \theta )}{16+\cos ^{2}\theta }\end{array} \)

Given that 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)

Hence, option (a) is the answer.

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

(a) ±(b+ia)

(b) ±(a-ib)

(c) (ai+b)

(d) None of these

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 part, we get;

x = a2-b2

y = 2ab

-x – iy = -a2 + b2– i(2ab)

-x – iy = (-b + ia)2

Taking square root, we get;

√(-x-iy) = ±(-b+ia)

Hence, option (d) is the answer.

Question 8:

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

(a) hyperbola

(b) pair of lines

(c) parabola

(d) circle

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

\(\begin{array}{l}\Rightarrow tan^{-1}\left ( \frac{y}{(x-2)} \right )-tan^{-1}\left ( \frac{(y-6)}{x} \right )=\frac{\pi }{2}\end{array} \)

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

\(\begin{array}{l}\Rightarrow \left ( \frac{xy-(x-2)(y-6)}{x(x-2)+y(y-6)} \right )=tan\frac{\pi }{2}\\\end{array} \)
\(\begin{array}{l}\Rightarrow \left ( \frac{xy-(x-2)(y-6)}{x(x-2)+y(y-6)} \right )=\frac{1 }{0}\\\end{array} \)

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.

Hence, option (d) is the answer.

Question 9:

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

(a) in the second quadrant

(b) in the first quadrant

(c) on the Y-axis (imaginary axis)

(d) on the X-axis (real axis)

Solution:

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

Taking √3 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.

Hence, option (c) is the answer.

Question 10:

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

(a) x = nπ

(b) x = (n+½)π

(c) x = 0

(d) No values of x

Solution:

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

\(\begin{array}{l}\overline{sin x + i cos 2x}= cos x – i sin 2x\end{array} \)

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

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

Therefore, no value of x is possible.

Hence, option (d) is the answer.

Question 11:

The square roots of -7- 24√(-1) are

(a) ±(4+3√(-1))

(b) ±(3+4√(-1))

(c) ±(3-4√(-1))

(d) ±(4-3√(-1))

Solution:

-7- 24√(-1) = -7-24i (since √(-1) = i)

Let √(-7-24i ) = x+iy

Squaring both sides

(-7-24i ) = (x+iy)2

= x2-y2+2xiy

Comparing real and imaginary part, we get

x2-y2 = -7

2xy = -24

solving above 2 equations, x2 = 9 and y2 = 16

x = ±3 and y = ±4

If x = 3, y = -4

If x = -3, y = 4

So, the roots are (-3+4i) and (3-4i).

Hence, option (c) is the answer.

Question 12:

If 1, ω and ω2 are the cube roots of unity, then (1 – ω + ω2)(1 – ω2 + ω4)(1 – ω4 + ω8) (1 – ω8 + ω16) … up to 2n factors is

(a) 2n

(b) 22n

(c) 1

(d) -22n

Solution:

Use 1+ω = -ω2 and 1+ω2 = -ω

(1 – ω + ω2) = -2ω

(1 – ω2 + ω4) = (1 – ω2 + ω3 ω) = (1 – ω2 + ω) = -2ω2

(1 – ω4 + ω8) = 1-ω+ω2 = -2ω

(1 – ω8 + ω16) = (1 – ω2 + ω) = -2ω2

-2ω and -2ω2 are the successive factors.

So the product is (-2ω) (-2ω2 ) (-2ω)(-2ω2 )….(-2ω2 )

= (-2ω)n (-2ω2)n

= 22n (ω3)2n

= 22n

Hence, option (b) is the answer.

Question 13:

Let z and w be two complex numbers such that

\(\begin{array}{l}w = z\overline{z}-2z+2,\end{array} \)
|(z + i) / (z – 3i)| = 1 and Re (w) has a minimum value. Then, the minimum value of n ∈ N, for which wn is real, is equal to

Solution:

Consider z = x + iy

|z + i| = |z-3i|

y = 1
Now w = x2 + y2 – 2x – 2iy + 2
w = x2 + 1 – 2x – 2i + 2
Re(w) = x2 – 2x + 3
Re(w) = (x–1)2 + 2
Re(w)min at x = 1
z = 1 + i
Now w = 1 + 1 –2 – 2i + 2
w = 2(1–i) = 2√2 ei(-π/4)
 wn = 2√2 ei(-π/4)
 If wn is real, n = 4

Question 14:

Let a complex number be w = 1 – √3i. Let another complex number z be such that |zw| = 1 and arg (z) – arg (w) = π / 2. Then the area of the triangle with vertices origin, z and w, is equal to:

(a) 1/2

(b) 4

(c) 2

(d) 1 / 4

Solution:

w = 1 – √3i

|w| = 2

|zw| = 1

⇒ |z| = 1/|w|

= 1/2

arg (z) – arg (w) = π / 2

Area of triangle = (1/2) × (1/2) ×2

= 1/2

Hence option a is the answer.

Question 15:

If (z + i)/(z + 2i) is purely real, then the locus of z is

(a) x – axis

(b) y – axis

(c) y = x

(d) y = x/2

Solution:

Given that (z + i)/(z + 2i) is purely real.

Let z = x+iy

Then (z + i)/(z + 2i) = (x+iy + i)/(x+iy + 2i)

Multiply and divide by the conjugate, and equate the imaginary part to zero.

[(-x(y+2)i) + ix(1+y)]/[(x2 + (y+2)2] = 0

⇒ (-x(y+2)i) + ix(1+y) = 0
⇒ x(y+2) = x(1+y)

⇒ xy + 2x – x – xy = 0

⇒ x = 0

⇒ The locus of z is the y axis.

Hence, option b is the answer.

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