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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 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.
a) 1
b) 3
c) 2
d) -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. d) -2 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
a) 1
b) -1
c) 2
d) -2
Q3. (cos θ + i sin θ)4 / (sin θ + i cos θ)5 is equal to ____________. 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 / 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)
a) sin(9θ) − i cos (9θ)
b) cos(9θ) − i cos (9θ)
c) cos(9θ) − i sin (9θ)
d) None of these
Q4. Given z = (1 + i√3)100, then find the value of Re (z) / Im (z). 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
a) 100 / √3
b) – 1 / √3
c) 1 / √3
d) 10 / √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? 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
a) (a + b)3
b) a2− ab + b2
c) a2 + b2
d) a3 + b3
Q6. If ω is an imaginary cube root of unity, (1 + ω − ω2)7 equals to ___________. b) −128ω2 Solution: (1 + ω − ω2)7 = (1 + ω + ω2 − 2ω2)7 = (−2ω2)7 = −128ω14 = −128ω12ω2 = −128ω2
a) −64ω2
b) −128ω2
c) 64ω2
d) −112ω2
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