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JEE Questions for Maths Complex Numbers Quiz 15 - MCQExams.com
JEE
Maths
Complex Numbers
Quiz 15
Given that | z − 1 | = 1, where z is a point on the Argand plane. Then
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2)
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none of these
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0
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1
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3
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1
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2
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3
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4
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2)
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none of these
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z lies only on a circle
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z lies only on the real axis
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z lies either on the real axis or on a circle
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none of these
If z satisfies | z − 1| < | z + 3 |, then ω = 2z + 3 − i satisfies
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| ω − 5 − i | < | ω + 3 + i |
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|ω − 5| < |ω + 3| and Re (ω) > 1
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Im (iω) < 1
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|arg (ω −| < π/2
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ω2
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2ω2
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3ω2
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1
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3
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4
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none of these
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none of these
The common roots of the equation z
3
+ 2z
2
+ 2z + 1 = 0 and z
1985
+ z
100
+ 1 = 0 are
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1, ω
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1, ω2
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ω, ω2
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none of these
Let z be a root of x
5
− 1 = 0, with z ≠ 1 Then the value of z
15
+ z
16
+ z
17
+....+ z
50
is equal to :
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1
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2
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ω
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none of these
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The cube roots of unity
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are collinear
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lie on a circle of radius √3
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form an equilateral triangle
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none of these
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0
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1
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−1
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none of these
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| z | = 5
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| z | < 5
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| z | > 5
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none of these
If | z − 3 + 2i | ≤ 4, then the sum of least and greatest value of |z| is
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− 1
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i
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0
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none of these
The solution of the equation 2z = | z | + 2i is
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none of these
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none of these
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none of these
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− 1
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1
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0
0%
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1 or i
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i or − i
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1 or − 1
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i or − 1
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n1 = n2 + 1
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n1 = n2 − 1
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n1 = n2
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n1 > 0, n2 > 0
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i
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i − 1
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− i
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0
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2)
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3ω
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3ω( ω − 1)
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3ω2
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3ω (1 − ω)
For all complex numbers z
1
, z
2
satisfying |z
1
| = 12 and |z
2
− 3 − 4i | = 5, the minimum value of | z
1
− z
2
| is
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0
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2
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7
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0.7
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Clockwise rotation around origin through an angle α.
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Anticlockwise rotation around origin through an angle α
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Reflection in the line through origin with slope tan α
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Reflection in the line through origin with slope tan α/2
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0
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None of these
The centre of a regular polygon of n sides is located at the point z = 0 and one of its vertex z
1
is known . If z
2
be the vertex adjacent to z
1
, then z
2
is equal to
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2)
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None of these
If z
1
, z
2
, z
3
, z
4
are affixes of four points in the Argand plane and z is the affix of a point such that |z - z
1
| = |z - z
2
| = |z - z
3
|=|z - z
4
|, z
1
, z
2
, z
3
, z
4
are
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Concyclic
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Vertices of a parallelogram
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Vertices of a rhombus
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In a straight line
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2)
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1
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–1
PQ and PR are two infinite rays. QAR is an arc. Point lying in the shaded excluding the boundary satisfies
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6 + 7i
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–7 + 6i
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7 + 6i
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–6 + 7i
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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
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