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JEE Questions for Maths Complex Numbers Quiz 5 - MCQExams.com
JEE
Maths
Complex Numbers
Quiz 5
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Hyperbola
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Parabola
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Ellipse
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Straight line
If z = x + iy and |z - 2 + i| = |z - 3 - i|, then locus of z is
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2x + 4y - 5 = 0
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2x - 4y - 5 = 0
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x + 2y = 0
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x - 2y + 5 = 0
If |z
2
– 1| = |z|
2
+ 1, then z lies on
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An ellipse
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The imaginary axis
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A circle
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The real axis
The equation |z – 5i| ÷ |z + 5i| = 12, where z = x + iy, represents a/an
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Circle
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Ellipse
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Parabola
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None of these
The number of solutions for the equations |z – 1| = |z – 2| = |z – i| is
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One solution
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3 solutions
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2 solutions
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No solution
If a = cos θ + i sin θ, then
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i cot θ/2
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i tan θ/2
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i cos θ/2
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i cosec θ/2
If z
1
, z
2
,.., z
n
are complex numbers such that |Z
1
|= |z
2
|=......=|z
n
|= 1 , then |z
1
+ z
2
+......+ z
n
| is equal to
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|z1+ z2 +z3......+ zn|
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|z1|+| z2| +|z3|......+ |zn|
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n
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√n
The conjugate of a complex number is
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2)
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In which quadrant of the complex plane, the point
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Fourth
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First
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Second
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Third
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A = 4 B = 5 C=3 D = 2
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A = 4 B =1 C=2 D = 6
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A = 6 B = 5 C=2 D=3
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A = 4 B=1 C=2 D =3
If z = x + iy is a variable complex number such that,
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x2 - y2 - 2x = 1
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x2 + y2 - 2x = 1
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x2 + y2 - 2x = 1
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x2 + y2 + 2x = 1
If ω = e
i2π/3
and a, b, c, x, y, z are non-zero complex numbers such that a + bω + cω
2
= y, a + bω
2
+ cω = z.Then, the value of
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2
0%
3
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1
0%
4
If (cos θ + i sinθ )(cos 2θ + i sin 2θ)...(cos nθ + i sin nθ) = 1, then the value of is
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4mπ
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0%
The value of the expression
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2)
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0%
None of these
The value of
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1
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- 1
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- i
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i
If z ≠ 1 and z
2
/(z - 1)is real, then the point represented by the complex number z lies
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either on the real axis or on a circle passing through the origin
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on a circle with centre at the origin
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either on the real axis or on a circle not passing through the origin
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on the imaginary axis
The shaded region, where P = (-1, 0), Q = (-1 + √2, √R = (-1 + √2, -√2), S = (1,is represented by
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|z + 1| > 2, arg (z +< π/4
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|z + 1| < 2, arg (z +< π/4
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|z - 1| > 2, arg (z +> π/4
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|z - 1| < 2, arg (z +> π/4
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π/2
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π/6
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2π/3
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5π/6
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10π/3
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20π/3
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16π/3
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32π/3
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1
0%
–1
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i
0%
–i
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0
0%
1
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2
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3
The cube roots of unity lie on the circle
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| z | = 1
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| z − 1 | = 1
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| z + 1 | = 1
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| z − ω| = 1
If ω( ≠is a cube root of unity and (1 + ω)
7
= A + Bω, then A and B are respectively
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0, 1
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1, 1
0%
1, 0
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−1, 1
Let z and ω be two non−zero complex numbers such that | z | = | ω | and Arg z + Arg ω = π, then z equals :
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ω
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− ω
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0%
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0
0%
1
0%
i
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ω
If ω is an imaginary cube root of unity, then (1 + ω − ω
2
)
7
equals
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128 ω
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− 128 ω
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128 ω2
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− 128 ω2
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− 1
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1
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− i
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i
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equal to 1
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less than 1
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greater than 3
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equal to 3
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of area zero
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right angled isosceles
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equilateral
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obtuse angled isosceles
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2)
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0%
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4
0%
–4
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5
0%
–5
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–1
0%
0
0%
1
0%
None
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2)
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0%
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(3,1)
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(3, –1)
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(–3, 1)
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(–3, –1)
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2)
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0%
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1
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–1
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2
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–2
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[–2, 13]
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[0, 13]
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[2, 13]
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[–13, 2]
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1
0%
0
0%
–1
0%
w
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0%
2)
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0%
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0%
2
0%
–1
0%
1
0%
–2
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0%
0%
2)
0%
0%
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0%
Circle
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An ellipse
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Parabola
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A straight line
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18
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54
0%
6
0%
12
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0%
2
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0%
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2)
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0%
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4
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8
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2
0%
12
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2)
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None of these
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0
0%
1
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2
0%
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2)
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–2i
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2
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0
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