Maths Quiz on Complex Numbers and Quadratic Equations for IIT JEE & Engineering Exam

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JEE Maths Quiz on Complex Numbers and Quadratic Equation

JEE Maths Quiz on Complex Numbers and Quadratic Equation: In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you maths quiz on Complex Numbers and Quadratic Equations.

Maths Quiz on Complex Numbers and Quadratic Equation

 

Q1. If z is a complex number, then the minimum value of |z| + |z − 1| is ______.
a) 1
b) 3 
c) 2 
d) -1

View Answer

a) 1

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.

Q2. 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.
a) 1
b) -1 
c) 2
d) -2

View Answer

d) -2

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

Q3.  (cos θ + i sin θ)4 / (sin θ + i cos θ)5 is equal to ____________.
a) sin(9θ) − i cos (9θ)
b) cos(9θ) − i cos (9θ)
c) cos(9θ) − i sin (9θ)
d) None of these

View Answer

a) sin(9θ) − i cos (9θ)

Solution:

(cos θ + i sin θ)4 / (sin θ + i cos θ)5

= (cos θ + i sin θ)4 / i([1 / i] sin θ + cos θ)5

= (cosθ + i sin θ)4 / i (cos θ − i sin θ)5

= (cos θ + i sin θ)/ i (cos θ + i sin θ)−5 (By property) = 1 / i (cos θ + i sin θ)9

= sin(9θ) − i cos (9θ)

Q4. Given z = (1 + i√3)100, then find the value of Re (z) / Im (z).
a)  100 / √3
b) – 1 / √3
c) 1 / √3
d) 10 / √3

View Answer

c) 1 / √3

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

Q5. 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 + b)3
b) a2− ab + b2
c) a2 + b2
d) a3 + b3

View Answer

d) a3 + b3

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

Q6. If ω is an imaginary cube root of unity, (1 + ω − ω2)7 equals to ___________.
a) −64ω2
b) −128ω2
c) 64ω2
d) −112ω2

View Answer

b) −128ω2

Solution:

(1 + ω − ω2)= (1 + ω + ω− 2ω2)7

= (−2ω2)7

= −128ω14

= −128ω12ω2

= −128ω2

 

Syllabus and Previous Year Papers

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Download NEET Previous Year Question Papers with Solution Click Here

 

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