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 ei2π/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, nth roots of unity, then (1 − ω) (1 − ω2) …..
(1 − ωn − 1) = ____________.
Solution:
Since 1, ω, ω2, ω3……., ωn−1 are the n, nth 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:
11/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, p1/3 = −q1/3(1)1/3.
Hence α = −q1/3, β = −q1/3 ω and γ = −q1/3ω2
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