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 + n^{2}) 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 + n^{2}) = a^{2} + b^{2 }

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

**Solution:**

First, note that |−z|=|z| and |z_{1} + z_{2}| ≤ |z_{1}| + |z_{2}|

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 z_{1} and z_{2} and any real numbers a and b; |(az_{1} − bz_{2})|^{2} + |(bz_{1} + az_{2})|^{2} = ___________.

**Solution:**

|(az_{1} − bz_{2})|^{2 }+ |(bz_{1} + az_{2})|^{2}

= (a_{2} + b_{2}) (|z_{1}|^{2} + |z_{2}|^{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} + 9y^{2} = 25x^{2} + 25 (y − 8)^{2} ….(i) and

⇒ |z − 4| = |z − 8|

|x − 4 + iy| = |x − 8 + iy|

(x − 4)^{2} + y^{2} = (x − 8)^{2} + y^{2}

⇒ x = 6

Putting x = 6 in (i), we get y^{2} − 25y + 136 = 0

y = 17, 8

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

**Question 5:** If z_{1} = 10 + 6i, z_{2} = 4 + 6i and z is a complex number such that

**Solution:**

Given numbers are z_{1} = 10 + 6i, z_{2} = 4 + 6i and z = x + iy

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

12y − y^{2} − 72 + 6y = x^{2} − 14x + 40 …..(i)

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

From (i), (x^{2 }− 14x + 49) + (y^{2} − 18y + 81) = 18

(x − 7)^{2} + (y − 9)^{2} = 18 or

^{2}+ (y − 9)

^{2}]

^{½ }= [18]

^{½ }= 3√2

|(x − 7) + i (y − 9)| = 3√2 or

|z − 7 −9i| = 3√2.

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

**Solution: **

One of the numbers must be a conjugate of z_{1} = 1 + i√3 i.e. z_{2} = 1 − i√3 or z_{3} = z_{1 }e^{i2π/3 }and

z_{2} = z_{1 }e^{−i2π/3 }, z_{3 } = (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 a^{3} + b^{3} + c^{3} = 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, ω^{2}, 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, −2ω^{2}

x = −1, 1 − 2ω, 1 − 2ω^{2}

**Question 9: **If 1, ω, ω^{2}, ω^{3}……., ω^{n−1} are the n, n^{th} roots of unity, then (1 − ω) (1 − ω^{2}) …..

(1 − ω^{n − 1}) = ____________.

**Solution:**

Since 1, ω, ω^{2}, ω^{3}……., ω^{n−1} are the n, n^{th} roots of unity, therefore, we have the identity

= (x − 1) (x − ω) (x − ω^{2}) ….. (x − ω^{n−1}) = x^{n }− 1 or

(x − ω) (x − ω^{2})…..(x − ω^{n−1}) = x^{n}−1 / x−1

= x^{n−1} + x^{n−2} +….. + x + 1

Putting x = 1 on both sides, we get

(1 − ω) (1 − ω^{2})….. (1 − ω^{n−1}) = n

**Question 10:** If a = cos (2π / 7) + i sin (2π / 7), then the quadratic equation whose roots are α = a + a^{2} + a^{4} and β = a^{3} + a^{5} + a^{6} is _____________.

**Solution:**

a = cos (2π / 7) + i sin (2π / 7)

a^{7 }= [cos (2π / 7) + i sin (2π / 7)]^{7}

= cos 2π + i sin 2π = 1 …..(i)

S = α + β = (a + a^{2} + a^{4}) + (a^{3} + a^{5} + a^{6})

S = a + a^{2} + a^{3} + a^{4} + a^{5} + a^{6}

or

P = α * β = (a + a^{2} + a^{4}) (a^{3} + a^{5} + a^{6})

= a^{4} + a^{6} + a^{7} + a^{5} + a^{7} + a^{8} + a^{7} + a^{9} + a^{10}

= a^{4} + a^{6} + 1 + a^{5} + 1 + a + 1 + a^{2} + a^{3} (From eqn (i)]

= 3+(a + a^{2} + a^{3} + a^{4} + a^{5} + a^{6})

= 3 + S = 3 − 1 = 2 [From (ii)]

Required equation is, x^{2 }− Sx + P = 0

x^{2} + x + 2 = 0.

**Question 11:** Let z_{1} and z_{2} 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 z_{1} = [cos 2r_{1}π / n] + i sin [2r_{1}π / n] and z_{2} = [cos 2r_{2}π / n] + i sin [2r_{2}π / n].

Then ∠Z_{1} O Z_{2} = amp (z_{1} / z_{2}) = amp (z_{1}) − amp (z_{2})

= [2 (r_{1} − r_{2})π] / [n]

= π / 2

(Given) n = 4 (r_{1} − r_{2})

= 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} / i^{5 }([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)

z^{100 }= [2 (cos π / 3 + i sin π / 3)]^{100 }

= 2^{100 }(cos 100π / 3 + i sin 100π / 3)

= 2^{100 }(−cos π / 3 − i sin π / 3)

= 2^{100}(−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 + bω^{2}) (aω^{2} + bω),where α = ω and β = ω^{2} = (a + b) (a^{2} + abω^{2} + abω + b^{2})

= (a + b) (a^{2}− ab + b^{2})

= a^{3} + b^{3}

**Question 15:** If ω is an imaginary cube root of unity, (1 + ω − ω^{2})^{7} equals to ___________.

**Solution:**

(1 + ω − ω^{2})^{7 }= (1 + ω + ω^{2 }− 2ω^{2})^{7}

= (−2ω^{2})^{7}

= −128ω^{14}

= −128ω^{12}ω^{2}

= −128ω^{2}

**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^{1/3}ω^{2}

The given expression [x + yω + zω^{2}] / [xω + yω^{2} + z] = (1 / ω) * [xω + yω^{2} + z] / [xω + yω^{2} + z]

= ω^{2}.

**Question 17:** The common roots of the equations x^{12 }− 1 = 0, x^{4} + x^{2} + 1 = 0 are __________.

**Solution:**

x^{12 }− 1 = (x^{6} + 1) (x^{6} − 1)

= (x^{6} + 1) (x^{2} − 1) (x^{4} + x^{2} + 1)

Common roots are given by x^{4} + x^{2} + 1 =0

x^{2} = [−1 ± i √3] / [2] = ω, ω^{2} or ω^{4}, ω^{2} (Because ω^{3} = 1) or

x = ± ω^{2}, ± ω

**Question 18:** Given that the equation z^{2} + (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 z^{2} + (p + iq)z + r + is = 0 ……(i)

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

α^{2} + (p + iq)α + r + is = 0 or

α^{2} + pα + r + i (qα + s) = 0

Equating real and imaginary parts, we have α^{2} + pα + r = 0 and qα + s = 0

Eliminating α, we get

(−s / q)^{2} + p (−s / q) + r = 0 or

s^{2 }− pqs + q^{2}r = 0 or

pqs = s^{2 }+ q^{2}r

**Question 19:** The difference between the corresponding roots of x^{2} + ax + b = 0 and x^{2 }+ 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 x^{2} + ax + b = 0 and x^{2 }+ 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

⇒ a^{2} − 4b = b^{2} − 4a

⇒ (a^{2 }− b^{2}) + 4 (a − b) = 0

⇒ a + b + 4 = 0 (Because a≠b)

**Question 20: **If b_{1 }b_{2 }= 2 (c_{1} + c_{2}), then at least one of the equations x^{2} + b_{1}x + c_{1} = 0 and x^{2} + b_{2}x + c_{2} = 0 has ____________ roots.

**Solution:**

Let D_{1} and D_{2} be discriminants of x^{2} + b_{1}x + c_{1} = 0 and x^{2} + b_{2}x + c_{2} = 0, respectively.

Then,

D_{1} + D_{2} = b_{1}^{2} − 4c_{1} + b_{2}^{2} − 4c_{2}

= (b_{1}^{2 }+ b_{2}^{2}) − 4 (c_{1} + c_{2})

= b_{1}^{2 }+ b_{2}^{2} − 2b_{1}b_{2} [Because b_{1}b_{2} = 2 (c_{1} + c_{2})] = (b_{1} – b_{2})^{2} ≥ 0

⇒ D_{1} ≥ 0 or D_{2} ≥ 0 or D_{1} and D_{2} both are positive.

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

**Question 21:** If the roots of the equation x^{2} + 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 ≤ 4m^{2} or

4a^{2} − 4b ≤ 4m^{2}

⇒ a^{2} − m^{2} ≤ b and discriminant D > 0 or

4a^{2 }− 4b > 0

⇒ a^{2} − m^{2} ≤ b and b < a^{2}.

Hence, b ∈ [a^{2} − m^{2 }, a^{2}).

**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 i^{4 }= 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:**

^{2}θ] / [1 + 4 sin

^{2}θ] = 0

3 − 4 sin^{2} 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

**Also Read**

Quadratic Equation JEE Advanced Previous Year Questions With Solutions

Complex Number JEE Advanced Previous Year Questions With Solutions

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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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