Complex Numbers and Quadratic Equation Test

Result

Complex Numbers and Quadratic Equation Test - 6

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
2 / -0.83

The value of (-1 + √-3)² + (-1 - √-3)² is

A
8

B
4

C
-4

D
-2

Solution

The given expression can be simplified as follows:

-1 + √(-3) = -1 + i √3
-1 - √(-3) = -1 - i √3

Now, (-1 + i √3)² = (-1)² + 2*(-1)*(i √3) + (i √3)²
= 1 - 2i √3 - 3
= -2i √3 - 2

Similarly, (-1 - i √3)² = (-1)² + 2*(-1)*(-i √3) + (-i √3)²
= 1 + 2i √3 - 3
= 2i √3 - 2

Adding the two results together:
(-2i √3 - 2) + (2i √3 - 2) = -4

Therefore, the value of the expression is -4.
Question 2
2 / -0.83

If α is a complex a number such that α² +α+1 = 0 then α³¹ is

A
0

B
1

C
α

D
α²

Solution

Since α³ +1=(α+1)(α² −α+1), using the given information, we get α³ +1 = 0
Question 3
2 / -0.83

A
0

B
-1

C
1

D
none of these
Question 4
2 / -0.83

If n is any integer, then iⁿ is

A
1 ,-1 , i ,-i

B
i

C
1 , -1

D
none of these
Question 5
2 / -0.83

The points z = x + iy which satisfy the equation | z | = 1 lie on

A
the line x = 1

B
the line y = 1

C
the circle whose centre is origin and radius = 1

D
the line x + y = 1

Solution

z = x + iy
|z| = (x² + y² )^1/2
x² + y² = 1
Equation of circle with centre (0,0) and radius 1 unit.
Question 6
2 / -0.83

The equation z² =

A
a unique solution

B
no solution

C
two solutions

D
four solutions

Solution
Question 7
2 / -0.83

The points of the complex plane given by the condition arg. (z) = (2n + 1) π, n ∈ I lie on

A
the imaginary semi axis z = iy , y > 0

B
the imaginary semi axis z = iy , y < 0

C
the positive real semi axis z = x , x > 0

D
the negative real semi axis z = x , x < 0
Question 8
2 / -0.83

If z = x + yi ; x ,y ∈ R, then locus of the equation , where c ∈R and b ∈ C, b ≠ 0 are fixed, is

A
a parabola

B
a straight line

C
a circle

D
none of these

Solution

As b and c are linear constants ,independent of x and y, then by substituting them in the equation given, we get an equation linear in x and y. Thus, the given equation represents a straight line.
Question 9
2 / -0.83

The complex number z which satisfies lies on

A
the y - axis

B
the line x + y =1

C
the x - axis

D
none of these
Question 10
2 / -0.83

(1+i)²ⁿ + (1 −i)²ⁿ ,n ∈N is

A
a purely real number

B
a purely imaginary number

C
a non-real complex number

D
none of these
Question 11
2 / -0.83

A
9 π/10

B
5 π/6

C
6 π/5

D
none of these

Solution

sin(6 π/5) + i(1+cos(6 π/5)) or −sin(π/5) + i(1 −cos(π/5)) lies in the second quadrant of complex plane hence its argument is given as
arg(x+iy) = π−tan^-1 |y/x| (∀x <0,y ≥0)
= π−tan^-1 |1 −cos(π/5)/sin(π/5)|
= π−tan^-1 |2sin2(π/10)/2sin(π/10)cos(π/10)|
= π−tan^-1 |sin(π/10)/cos(π/10)|
= π−tan^-1 |tan(π/10)|
= π−tan^-1 (tan(π/10)) (∵tan(π/10)>0)
= π−π/10(∵−π/2 ≤tan −1(x)≤π/2)
= 9 π/10
Question 12
2 / -0.83

then z⁶⁹ is equal to :

A
i

B
1

C
-i

D
none of these

Solution
Question 13
2 / -0.83

If ω is a cube root of unity, then the linear factors of x³ +y³ in complex numbers are

A
(x+y)(x+y ω)(x+y ω² )

B
(x+y)(x −y ω)(x −y ω² )

C
(x+y)(x+y ω)(x −y ω² )

D
(x −y)(x+y ω)(x+y ω² )
Question 14
2 / -0.83

can be expressed as

A
|z+1|²

B
∣z² ∣+1

C
∣z² ∣+2

D
none of these

Solution

(z+1) (z(bar)+1)
((x+1)+iy) ((x+1)-iy)
= (x+1)² + y²
= |(x+1)² + iy² |
= |(x+iy) + 1|²
= |z+1|²
Question 15
2 / -0.83

If α,βare non-real cube roots of unity then αβ+ α⁵ +β⁵ equals

A
0

B
1

C
3

D
-1