JEE Complex Numbers MCQs | Previous Year Questions(PYQs) Download PDF

MCQs for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers

1. The value of 1 + i² + i⁴ + i⁶ + … + i²ⁿ is

(a) positive

(b) negative

(c) 0

(d) cannot be evaluated

Correct option: (d) cannot be evaluated

Solution:

1 + i² + i⁴ + i⁶ + … + i²ⁿ = 1 – 1 + 1 – 1 + … (–1)ⁿ

This cannot be evaluated unless the value of n is known.

2. If a + ib = c + id, then

(a) a² + c² = 0

(b) b² + c² = 0

(c) b² + d² = 0

(d) a² + b² = c² + d²

Correct option: (d) a² + b² = c² + d²

Solution:

Given,

a + ib = c + id

⇒ |a + ib| = |c + id|

⇒ √(a² + b²) = √(c² + d²)

Squaring on both sides, we get;

a² + b² = c² + d²

3. If a complex number z lies in the interior or on the boundary of a circle of radius 3 units and centre (– 4, 0), the greatest value of |z +1| is

(a) 4

(b) 6

(c) 3

(d) 10

Correct option: (b) 6

Solution:

The distance of the point representing z from the centre of the circle is |z – (-4 + i0)| = |z + 4|

According to the given,

|z + 4| ≤ 3

Now,

|z + 1| = |z + 4 – 3| ≤ |z + 4| + |-3| ≤ 3 + 3 ≤ 6

Hence, the greatest value of |z + 1| is 6.

4. The value of arg (x) when x < 0 is

(a) 0

(b) π/2

(c) π

(d) none of these

Correct option: (c) π

Solution:

Let z = x + 0i and x < 0

Since the point (-x, 0) lies on the negative side of the real axis,

|z| = |x + oi| = √[(-1)² + 0)] = 1

∴ Principal argument (z) = π

Alternative method:

Let x = cos θ + i sin θ

For θ = π, x should be negative.

Thus, x < 0 for θ = π.

5. If 1 – i, is a root of the equation x² + ax + b = 0, where a, b ∈ R, then the value of a – b is

(a) -4

(b) 0

(c) 2

(d) 1

Correct option: (a) -4

Solution:

Given that 1 – i is the root of x² + ax + b = 0.

Thus, 1 + i is also the root of the given equation since non-real complex roots occur in conjugate pairs.

Sum of roots = −a/1 = (1 – i) + (1 + i)

⇒ a = – 2

Product of roots, b/1 = (1 – i)(1 + i)

b = 1 – i²

b = 1 + 1 {since i² = -1}

⇒ b = 2

Now, a – b = -2 – 2 = -4

6. Number of solutions of the equation z² + |z|² = 0 is

(a) 1

(b) 2

(c) 3

(d) infinitely many

Correct option: (d) infinitely many

Solution:

Given,

z² + |z|² = 0, z ≠ 0

⇒ (x + iy)² + [√(x² + y²)]² = 0

⇒ x² – y² + i2xy + x² + y² = 0

⇒ 2x² + i2xy = 0

⇒2x (x + iy) = 0

⇒ x = 0 or x + iy = 0 (not possible)

Therefore, x = 0 and z ≠ 0.

Thus, y can have any real value.

Hence, there exist infinitely many solutions.

7. If [(1 + i)/(1 – i)]^x = 1, then

(a) x = 2n + 1, where n ∈ N

(b) x = 4n, where n ∈ N

(c) x = 2n, where n ∈ N

(d) x = 4n + 1, where n ∈ N

Correct option: (b) x = 4n, where n ∈ N

Solution:

Given,

[(1 + i)/(1 – i)]^x = 1

By rationalising the denominator,

[(1 + i)(1 + i)/ (1 – i)(1 + i)]^x = 1
[(1 + i)²/ (1 – i + i – i²)]^x = 1
[(1 + i² + 2i)/(1 + 1)]^x = 1
[(1 – 1 + 2i)/ 2]^x = 1

i^x = 1

Thus, i^x = i⁴ⁿ, where n is any positive integer.

8. If the complex number z = x + iy satisfies the condition |z + 1| = 1, then z lies on

(a) x-axis

(b) circle with centre (1, 0) and radius 1

(c) circle with centre (–1, 0) and radius 1

(d) y-axis

Correct option: (c) circle with centre (–1, 0) and radius 1

Solution:

Given,

z = x + iy

and

|z + 1| = 1

|x + iy + 1| = 1

⇒ |(x + 1) + iy| = 1

⇒ √[(x +1)² + y²] = 1

Squaring on both sides,

(x + 1)² + y² = 1

This is the equation of a circle with centre (–1, 0) and radius 1.

9. The simplified value of (1 – i)³/(1 – i³) is

(a) 1

(b) -2

(c) -i

(d) 2i

Correct option: (b) -2

Solution:

(1 – i)³/(1 – i³)

= (1 – i)³/(1³ – i³)

= (1 – i)³/ [(1 – i)(1 + i + i²)]

= (1 – i)²/(1 + i – 1)

= (1 – i)²/i

= (1 + i² – 2i)/i

= (1 – 1 – 2i)/i

= -2i/i

= -2

10. sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for:

(a) x = nπ

(b) x = [n + (1/2)] (π/2)

(c) x = 0

(d) No value of x

Correct option: (d) No value of x

Solution:

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

So, sin x – i cos 2x = cos x – i sin 2x

On comparing real and imaginary parts of both sides, we get

⇒ sin x = cos x and cos 2x = sin 2x

⇒ sin x/cos x = 1 and (cos 2x/sin 2x) = 1

⇒ tan x = 1 and tan 2x = 1

Now, consider tan 2x = 1

Using the formula tan 2A = 2 tan A/(1 – tan²A),

(2 tan x)/(1 – tan²x) = 1

However, this is not possible for tan x = 1.

Therefore, for no value of x, sinx + i cos 2x and cos x – i sin 2x are conjugate to each other

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