Mathematics MCQ on Complex Numbers and Quadratic Equations for JEE and Engineering Exam 2022

MCQ on Complex Numbers and Quadratic Equations: To be an expert in JEE Mathematics, it is absolutely necessary to practice and be familiar will all the concepts as well as the questions of different types. This is essential to gain mastery over the subject. We have also often heard the common saying, “Practice Makes a Man Perfect”, hence students have to practice, practice and practice till they master the subject.

MCQ on Complex Numbers and Quadratic Equation

In this post we are providing you MCQ on Complex Numbers and Quadratic Equations, which will be beneficial for you in upcoming JEE and Engineering Exams.

MCQ on Complex Numbers and Quadratic Equations

Q1. Given z = (1 + i√3)100, then find the value of Re (z) / Im (z).
a) -1 / √3.
b) 1 / √3.
c) 1 
d) √3.

View Answer

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.

Q2. 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?
a) a2 + b2
b) a3 + b3
c) a + b
d) none

View Answer

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

Q3. The real values of x and y for which the equation is (x + iy) (2 − 3i) = 4 + i is satisfied, are __________.
a) x = 5 / 13, y = 14 / 13
b) x = 7 / 13, y = 11 / 13
c) x = 14 / 13, y = 5 / 13
d) x = 5 / 13, y = 10 / 13

View Answer

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

Q4.  If ω is an imaginary cube root of unity, (1 + ω − ω2)7 equals to ___________.
a) 128ω2
b) -128ω2
c) -120ω2
d) 120ω2

View Answer

(1 + ω − ω2)= (1 + ω + ω− 2ω2)7

= (−2ω2)7

= −128ω14

= −128ω12ω2

= −128ω2

Q5. If z is a complex number, then the minimum value of |z| + |z − 1| is ______.
a) -1
b) 0
c) 1
d) None 

View Answer

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.

Q6. If (1 − i) x + (1 + i) y = 1 − 3i, then (x, y) = ______________.
a) (2,-2)
b) (2,-1)
c) (2,1)
d) (2,-1)

View Answer

(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).

Q7.  For any two complex numbers z1 and z2 and any real numbers a and b; |(az1 − bz2)|2 + |(bz1 + az2)|2 = ___________.

View Answer

|(az1 − bz2)|+ |(bz1 + az2)|2

= a2 |z1|2 + b2|z2|2 − 2 Re (ab)|z1\overline{z2}| + b2 |z1|+ a2|z2|2 + 2 Re (ab)|\overline{z1}z2|

= (a2 + b2) (|z1|2 + |z2|2)

Q8. The common roots of the equations x12 − 1 = 0, x4 + x2 + 1 = 0 are __________.  
a) x = ± 1, ± ω
b) x = ± ω, 0
c) x = ± ω2, ± ω
d) x = ± ω3, ± ω2

View Answer

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

Q9. 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.

View Answer

One of the numbers must be a conjugate of z1 = 1 + i√3 i.e. z2 = 1 − i√3 or z3 = zei2π/3 and

z2 = ze−i2π/3 , z= (1 + i√3) [cos (2π / 3) + i sin (2π / 3)] = −2

Q10. The difference between the corresponding roots of x2 + ax + b = 0 and x+ bx + a = 0 is same and a≠b, then what is the relation between a and b?

View Answer

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

Given |α − β| =|γ − δ| ⇒ (α + β)− αβ

= (γ + δ)−4γδ

⇒ a2 − 4b = b2 − 4a

⇒ (a− b2) + 4 (a − b) = 0

⇒ a + b + 4 = 0 (Because a≠b)

 


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