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.
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.
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.
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) a^{2} + b^{2}
b) a^{3} + b^{3}
c) a + b
d) none
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}
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
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 − ω^{2})^{7} equals to ___________.
a) 128ω^{2}
b) -128ω^{2}
c) -120ω^{2}
d) 120ω^{2}
(1 + ω − ω^{2})^{7 }= (1 + ω + ω^{2 }− 2ω^{2})^{7}
= (−2ω^{2})^{7}
= −128ω^{14}
= −128ω^{12}ω^{2}
= −128ω^{2}
Q5. If z is a complex number, then the minimum value of |z| + |z − 1| is ______.
a) -1
b) 0
c) 1
d) None
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.
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)
(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 z_{1} and z_{2} and any real numbers a and b; |(az_{1} − bz_{2})|^{2} + |(bz_{1} + az_{2})|^{2} = ___________.
|(az_{1} − bz_{2})|^{2 }+ |(bz_{1} + az_{2})|^{2}
= a_{2} |z_{1}|^{2} + b_{2}|z_{2}|^{2} − 2 Re (ab)|z_{1}$z$| + b_{2} |z_{1}|^{2 }+ a_{2}|z_{2}|^{2} + 2 Re (ab)|$z$z_{2}|
= (a_{2} + b_{2}) (|z_{1}|^{2} + |z_{2}|^{2})
Q8. The common roots of the equations x^{12 }− 1 = 0, x^{4} + x^{2} + 1 = 0 are __________.
a) x = ± 1, ± ω
b) x = ± ω, 0
c) x = ± ω^{2}, ± ω
d) x = ± ω^{3}, ± ω^{2}
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
Q9. 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}.
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
Q10. 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?
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 }− αβ
= (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)
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