Complex Numbers and Quadratic Equations Previous Year Questions With Solutions

Get Complex Numbers and Quadratic Equations previous year questions with solutions here. All solutions are prepared by subject matter experts of Mathematics at BYJU’S. The questions in the article enable the students to predict the difficulty level of the questions in the upcoming JEE Main and JEE Advanced exams. The chapter contains important concepts such as, algebra of complex numbers, modulus and argument, complex conjugate, properties of complex numbers, square root of complex numbers and complex equations, De-moivre’s theorem, Vector representation and rotation of complex numbers and many more.

Other Important Concepts in Complex Numbers and Quadratic Equations

• Complex Numbers
• Quadratic Equations
• Quadratic Inequalities
• De-Moviers Theorem
• Geometrical Representation of a Complex Number
• Modulus and Conjugate of a Complex Number

JEE Main Maths Complex Numbers & Quadratic Equations Previous Year Questions With Solutions

Question 1: If (1 + i) (1 + 2i) (1 + 3i) ….. (1 + ni) = a + ib, then what is 2 * 5 * 10….(1 + n²) is equal to?

Solution:

We have (1 + i) (1 + 2i) (1 + 3i) ….. (1 + ni) = a + ib …..(i)

(1 − i) (1 − 2i) (1 − 3i) ….. (1 − ni) = a − ib …..(ii)

Multiplying (i) and (ii),

we get 2 * 5 * 10 ….. (1 + n²) = a² + b²

Question 2: If z is a complex number, then the minimum value of |z| + |z − 1| is ______.

Solution:

First, note that |−z|=|z| and |z₁ + z₂| ≤ |z₁| + |z₂|

Now |z| + |z − 1| = |z| + |1 − z| ≥ |z + (1 − z)|

= |1|

= 1

Hence, minimum value of |z| + |z − 1| is 1.

Question 3: For any two complex numbers z₁ and z₂ and any real numbers a and b; |(az₁ − bz₂)|² + |(bz₁ + az₂)|² = ___________.

Solution:

|(az₁ − bz₂)|² + |(bz₁ + az₂)|²

= (a₂ + b₂) (|z₁|² + |z₂|²)

Question 4: Find the complex number z satisfying the equations

Solution:

We have

Let z = x + iy, then

⇒ 3|z − 12| = 5 |z − 8i|

3 |(x − 12) + iy| = 5 |x + (y − 8) i|

9 (x − 12)² + 9y² = 25x² + 25 (y − 8)² ….(i) and

⇒ |z − 4| = |z − 8|

|x − 4 + iy| = |x − 8 + iy|

(x − 4)² + y² = (x − 8)² + y²

⇒ x = 6

Putting x = 6 in (i), we get y² − 25y + 136 = 0

y = 17, 8

Hence, z = 6 + 17i or z = 6 + 8i

Question 5: If z₁ = 10 + 6i, z₂ = 4 + 6i and z is a complex number such that

Solution:

Given numbers are z₁ = 10 + 6i, z₂ = 4 + 6i and z = x + iy

amp [(x − 10) + i (y − 6) (x − 4) + i (y − 6)] = π / 4

12y − y² − 72 + 6y = x² − 14x + 40 …..(i)

Now |z − 7 −9i| = |(x − 7) + i (y − 9)|

From (i), (x² − 14x + 49) + (y² − 18y + 81) = 18

(x − 7)² + (y − 9)² = 18 or

|(x − 7) + i (y − 9)| = 3√2 or

|z − 7 −9i| = 3√2.

Question 6: Suppose z₁, z₂, z₃ are the vertices of an equilateral triangle inscribed in the circle |z| = 2. If z₁ = 1 + i√3, then find the values of z₃ and z₂.

Solution:

One of the numbers must be a conjugate of z₁ = 1 + i√3 i.e. z₂ = 1 − i√3 or z₃ = z₁ e^i2π/3 and

z₂ = z₁ e^−i2π/3 , z₃ = (1 + i√3) [cos (2π / 3) + i sin (2π / 3)] = −2

Question 7: If cosα + cos β + cos γ = sin α + sin β + sin γ = 0 then what is the value of cos 3α + cos 3β + cos 3γ?

Solution:

cos α + cos β + cos γ = 0 and sin α + sin β + sin γ = 0

Let a = cos α + i sin α; b = cos β + i sin β and c = cos γ + i sin γ.

Therefore, a + b + c = (cosα + cosβ + cosγ) + i (sinα + sinβ + sinγ) = 0 + i0 = 0

If a + b + c = 0, then a³ + b³ + c³ = 3abc or

(cosα + isina)³ + (cosβ + isinβ)³ + (cosγ + isinγ)³

= 3 (cosα + isinα) (cosβ + isinβ) (cosγ + isinγ)

⇒ (cos3α + isin3α) + (cos3β + isin3β) + (cos3γ + isin3γ)

= 3 [cos (α + β + γ) + i sin (α + β + γ)] or cos 3α + cos 3β + cos 3γ

= 3 cos (α + β + γ).

Question 8: If the cube roots of unity are 1, ω, ω², then find the roots of the equation (x − 1)³ + 8 = 0.

Solution:

(x − 1)³ = −8 ⇒ x − 1 = (−8)^1/3

x − 1 = −2, −2ω, −2ω²

x = −1, 1 − 2ω, 1 − 2ω²

Question 9: If 1, ω, ω², ω³……., ωⁿ⁻¹ are the n, n^th roots of unity, then (1 − ω) (1 − ω²) …..

(1 − ω^{n − 1}) = ____________.

Solution:

Since 1, ω, ω², ω³……., ωⁿ⁻¹ are the n, n^th roots of unity, therefore, we have the identity

= (x − 1) (x − ω) (x − ω²) ….. (x − ωⁿ⁻¹) = xⁿ − 1 or

(x − ω) (x − ω²)…..(x − ωⁿ⁻¹) = xⁿ−1 / x−1

= xⁿ⁻¹ + xⁿ⁻² +….. + x + 1

Putting x = 1 on both sides, we get

(1 − ω) (1 − ω²)….. (1 − ωⁿ⁻¹) = n

Question 10: If a = cos (2π / 7) + i sin (2π / 7), then the quadratic equation whose roots are α = a + a² + a⁴ and β = a³ + a⁵ + a⁶ is _____________.

Solution:

a = cos (2π / 7) + i sin (2π / 7)

a⁷ = [cos (2π / 7) + i sin (2π / 7)]⁷

= cos 2π + i sin 2π = 1 …..(i)

S = α + β = (a + a² + a⁴) + (a³ + a⁵ + a⁶)

S = a + a² + a³ + a⁴ + a⁵ + a⁶

or

P = α * β = (a + a² + a⁴) (a³ + a⁵ + a⁶)

= a⁴ + a⁶ + a⁷ + a⁵ + a⁷ + a⁸ + a⁷ + a⁹ + a¹⁰

= a⁴ + a⁶ + 1 + a⁵ + 1 + a + 1 + a² + a³ (From eqn (i)]

= 3+(a + a² + a³ + a⁴ + a⁵ + a⁶)

= 3 + S = 3 − 1 = 2 [From (ii)]

Required equation is, x² − Sx + P = 0

x² + x + 2 = 0.

Question 11: Let z₁ and z₂ be 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 ____________.

Solution:

1^1/n = cos [2rπ / n] + i sin [2r π / n]

Let z₁ = [cos 2r₁π / n] + i sin [2r₁π / n] and z₂ = [cos 2r₂π / n] + i sin [2r₂π / n].

Then ∠Z₁ O Z₂ = amp (z₁ / z₂) = amp (z₁) − amp (z₂)

= [2 (r₁ − r₂)π] / [n]

= π / 2

(Given) n = 4 (r₁ − r₂)

= 4 × integer, so n is of the form 4k.

Question 12: (cos θ + i sin θ)⁴ / (sin θ + i cos θ)⁵ is equal to ____________.

Solution:

(cos θ + i sin θ)⁴ / (sin θ + i cos θ)⁵

= (cos θ + i sin θ)⁴ / i⁵ ([1 / i] sin θ + cos θ)⁵

= (cosθ + i sin θ)⁴ / i (cos θ − i sin θ)⁵

= (cos θ + i sin θ)⁴ / i (cos θ + i sin θ)⁻⁵ (By property) = 1 / i (cos θ + i sin θ)⁹

= sin(9θ) − i cos (9θ).

Question 13: Given z = (1 + i√3)¹⁰⁰, then find the value of Re (z) / Im (z).

Solution:

Let z = (1 + i√3)

r = √[3 + 1] = 2 and r cosθ = 1, r sinθ = √3, tanθ = √3 = tan π / 3 ⇒ θ = π / 3.

z = 2 (cos π / 3 + i sin π / 3)

z¹⁰⁰ = [2 (cos π / 3 + i sin π / 3)]¹⁰⁰

= 2¹⁰⁰ (cos 100π / 3 + i sin 100π / 3)

= 2¹⁰⁰ (−cos π / 3 − i sin π / 3)

= 2¹⁰⁰(−1 / 2 −i √3 / 2)

Re(z) / Im(z) = [−1/2] / [−√3 / 2] = 1 / √3.

Question 14: If x = a + b, y = aα + bβ and z = aβ + bα, where α and β are complex cube roots of unity, then what is the value of xyz?

Solution:

If x = a + b, y = aα + bβ and z = aβ + bα, then xyz = (a + b) (aω + bω²) (aω² + bω),where α = ω and β = ω² = (a + b) (a² + abω² + abω + b²)

= (a + b) (a²− ab + b²)

= a³ + b³

Question 15: If ω is an imaginary cube root of unity, (1 + ω − ω²)⁷ equals to ___________.

Solution:

(1 + ω − ω²)⁷ = (1 + ω + ω² − 2ω²)⁷

= (−2ω²)⁷

= −128ω¹⁴

= −128ω¹²ω²

= −128ω²

Question 16: If α, β, γ are the cube roots of p (p < 0), then for any x, y and z, find the value of [xα + yβ + zγ] / [xβ + yγ + zα].

Solution:

Since p < 0.

Let p = −q, where q is positive.

Therefore, p^1/3 = −q1^/3(1)^1/3.

Hence α = −q^1/3, β = −q^1/3 ω and γ = −q^1/3ω²

The given expression [x + yω + zω²] / [xω + yω² + z] = (1 / ω) * [xω + yω² + z] / [xω + yω² + z]

= ω².

Question 17: The common roots of the equations x¹² − 1 = 0, x⁴ + x² + 1 = 0 are __________.

Solution:

x¹² − 1 = (x⁶ + 1) (x⁶ − 1)

= (x⁶ + 1) (x² − 1) (x⁴ + x² + 1)

Common roots are given by x⁴ + x² + 1 =0

x² = [−1 ± i √3] / [2] = ω, ω² or ω⁴, ω² (Because ω³ = 1) or

x = ± ω², ± ω

Question 18: Given that the equation z² + (p + iq)z + r + is = 0, where p, q, r, s are real and non-zero has a real root, then how are p, q, r and s related?

Solution:

Given that z² + (p + iq)z + r + is = 0 ……(i)

Let z = α (where α is real) be a root of (i), then

α² + (p + iq)α + r + is = 0 or

α² + pα + r + i (qα + s) = 0

Equating real and imaginary parts, we have α² + pα + r = 0 and qα + s = 0

Eliminating α, we get

(−s / q)² + p (−s / q) + r = 0 or

s² − pqs + q²r = 0 or

pqs = s² + q²r

Question 19: The difference between the corresponding roots of x² + ax + b = 0 and x² + bx + a = 0 is same and a≠b, then what is the relation between a and b?

Solution:

Let α, β and γ,δ be the roots of the equations x² + ax + b = 0 and x² + bx + a = 0, respectively therefore, α + β = −a, αβ = b and δ + γ = −b, γδ = a.

Given |α − β| =|γ − δ| ⇒ (α + β)² − 4αβ

= (γ + δ)² −4γδ

⇒ a² − 4b = b² − 4a

⇒ (a² − b²) + 4 (a − b) = 0

⇒ a + b + 4 = 0 (Because a≠b)

Question 20: If b₁ b₂ = 2 (c₁ + c₂), then at least one of the equations x² + b₁x + c₁ = 0 and x² + b₂x + c₂ = 0 has ____________ roots.

Solution:

Let D₁ and D₂ be discriminants of x² + b₁x + c₁ = 0 and x² + b₂x + c₂ = 0, respectively.

Then,

D₁ + D₂ = b₁² − 4c₁ + b₂² − 4c₂

= (b₁² + b₂²) − 4 (c₁ + c₂)

= b₁² + b₂² − 2b₁b₂ [Because b₁b₂ = 2 (c₁ + c₂)] = (b₁ – b₂)² ≥ 0

⇒ D₁ ≥ 0 or D₂ ≥ 0 or D₁ and D₂ both are positive.

Hence, at least one of the equations has real roots.

Question 21: If the roots of the equation x² + 2ax + b = 0 are real and distinct and they differ by at most 2m then b lies in what interval?

Solution:

Let the roots be α, β

α + β = −2a and αβ = b

Given, |α − β| ≤ 2m

or |α − β|² ≤ (2m)² or

(α + β)²− 4αβ ≤ 4m² or

4a² − 4b ≤ 4m²

⇒ a² − m² ≤ b and discriminant D > 0 or

4a² − 4b > 0

⇒ a² − m² ≤ b and b < a².

Hence, b ∈ [a² − m² , a²).

Question 22: If ([1 + i] / [1 − i])^m = 1, then what is the least integral value of m?

Solution:

= [(1 + i)²] / [2]

= 2i / 2

= i

([1 + i] / [1 − i])^m = 1 (as given)

So, the least value of m = 4 {Because i⁴ = 1}

Question 23: If (1 − i) x + (1 + i) y = 1 − 3i, then (x, y) = ______________.

Solution:

(1 − i) x + (1 + i) y = 1 − 3i

⇒ (x + y) + i (−x + y) = 1 − 3i

Equating real and imaginary parts, we get x + y = 1 and −x + y = −3;

So, x = 2, y = −1.

Thus, the point is (2, −1).

Question 24: [3 + 2i sinθ] / [1 − 2i sinθ] will be purely imaginary if θ = ___________.

Solution:

3 − 4 sin² θ (only if θ be real)

sinθ = ±√3 / 2

= sin(± π / 3)

θ = nπ + (−1)ⁿ (± π / 3)

= nπ ± π / 3

Question 25: The real values of x and y for which the equation is (x + iy) (2 − 3i) = 4 + i is satisfied, are __________.

Solution:

Equation (x + iy) (2 − 3i) = 4 + i

(2x + 3y) + i (−3x + 2y) = 4 + i

Equating real and imaginary parts, we get

2x + 3y = 4 ……(i)

−3x + 2y = 1 ……(ii)

From (i) and (ii), we get

x = 5 / 13, y = 14 / 13

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How Do IIT JEE Mathematics Complex Numbers & Quadratic Equations Question Papers Help Students?

Complex numbers and quadratic equations are one of the most important and fundamental chapters in the preparation of competitive entrance exams. Every year, at least 1-3 questions are covered in JEE Main and other exams, directly and indirectly. The concept of this chapter is incorporated into many other chapters such as functions and coordinate geometry. The chapter itself will help you to score some marks in JEE Main exam as it gets about 7% weight. Students are advised to practice all listed questions to master the concept.

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