JEE Complex Numbers PYQs-Previous Year Questions With Solutions pdf free download

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

\(\begin{array}{l}= (az_{1}-bz_{2})(a\overline{z_{1}}-b\overline{z_{2}})+(bz_{1}+az_{2})(b\overline{z_{1}}+a\overline{z_{2}})\end{array} \)

= (a₂ + b₂) (|z₁|² + |z₂|²)

Question 4: Find the complex number z satisfying the equations

\(\begin{array}{l}|\frac{z-12}{z-8i}| = \frac{5}{3},\ |\frac{z-4}{z-8}| = 1.\end{array} \)

Solution:

We have

\(\begin{array}{l}|\frac{z-12}{z-8i}| = \frac{5}{3}\end{array} \)

,

\(\begin{array}{l}|\frac{z-4}{z-8}| = 1\end{array} \)

Let z = x + iy, then

\(\begin{array}{l}|\frac{z-12}{z-8i}| = \frac{5}{3}\end{array} \)

⇒ 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

\(\begin{array}{l}|\frac{z-4}{z-8}| = 1\end{array} \)

⇒ |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

\(\begin{array}{l}amp\ \frac{z-z_1}{z−z_2} = \frac{\pi}{4}\end{array} \)

, then the value of |z − 7 − 9i| is equal to _________.

Solution:

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

\(\begin{array}{l}amp\ \frac{z-z_1}{z−z_2} = \frac{\pi}{4}\end{array} \)

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

\(\begin{array}{l}\frac{(x − 4) (y − 6) − (y − 6) (x − 10)}{(x − 4) (x − 10) + (y − 6)^2}= 1\end{array} \)

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)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 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

\(\begin{array}{l}= \frac{a(1-a^6)}{1-a}\end{array} \)

or

\(\begin{array}{l}S = \frac{a-1}{1-a}= −1 …..(ii)\end{array} \)

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] / [1 − i] = ([1 + i] / [1 − i]) × [1 + i] / [1 + i]

= [(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 + 2i sinθ] / [1 − 2i sinθ] will be purely imaginary, if the real part vanishes, i.e.,
[3 − 4 sin² θ] / [1 + 4 sin²θ] = 0

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