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# Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced

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## 10 Questions MCQ Test Maths 35 Years JEE Main & Advanced Past year Papers | Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced

Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced for JEE 2022 is part of Maths 35 Years JEE Main & Advanced Past year Papers preparation. The Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced questions and answers have been prepared according to the JEE exam syllabus.The Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced MCQs are made for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced below.
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Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 1

### If z1 = a + ib and z2 = c + id are complex  numbers such that |z1| = |z2|=1 and Re=0, then the pair of complex numbers w1 = a + ic and w2 = b + id satisfies – (1985 - 2 Marks)

Detailed Solution for Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 1

(a, b, c) z1 = a + ib   and  z2 = c + id.
ATQ  |zi|2 = |z2|2 = 1
⇒​ a2 + b2 =1   and   c2 + d2 = 1. ....(1)

Also Re = 0 ⇒​ ac + bd=0

= α(say)  ....(2)

From (1) and (2), we get
b2 α2 + b2 = c2α2 + c2 ⇒​ b2 = c2;
Similarly   a2  = d2

∴

and

Also Re = ab + cd = (bα) b + c (-cα)
= α(b2 -c2 )=0

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 2

### Let z1 and z2 be complex numbers such that z1 ≠ z2 and |z1| = |z2| . If z1 has positive real part and z2 has negative imaginary part, then may be (1986 - 2 Marks)

Detailed Solution for Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 2

(a, d)   Let z1 = a + ib, a > 0 and b ∈ R; z2 = c+ id, d < 0, c∈R
then |z1| = | z2|
⇒ a2 + b2 = c2+d2
⇒ a2- c2 = d2-b2 ....(1)

Now ,

[Using (1)]

= purely imaginary number or zero in case a + c = b + d = 0.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 3

### If z1 and z2 are two nonzero complex numbers such that |z1 + z2|=|z1|+ |z2|, then Arg z1 - Argg  z2 is equal to                                  (1987 - 2 Marks)

Detailed Solution for Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 3

(c) Let z1 = r1 (cos θ1 +i sinθ1

and z2 = r2 (cos θ2 +i sinθ2)

where  r1 = |z1|, r2 =|z2|, θ1 = arg(z1), θ2= arg(z2)

∴ z1 + z2 = r1 (cos θ1 + i sinθ1 ) + r2 (cos θ2 +i sinθ2)

= (r1 cos θ1+ r2 cos θ2) + i(r1 sin θ1+r2 sin θ2)

=  r12 cos2 θ1 + r22 cos2 θ2 + 2r1r2 cos θ1 cosθ2

+ r12 sin2 θ1 + r22 sin2 θ2 + 2r1r2 sin θ1 sinθ2

= r12 + r22 + 2r1 r2 cos ( θ12)

and |z1| + |z2| = r1+r2

Since |z1 + z2| = | z1 |+ |z2| (given)

⇒ |z1 + z2|2 = |z1|2 + | z2 |2+ 2| z1| |z2|

⇒ r12 + r22+2r1 r2 cos (θ1 - θ ) = r12 + r22+2r1r2

⇒ cos (θ12) = 1 ⇒ θ1 2 =0

⇒ Arg (z1) = arg (z2)

Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 4

The value of  is (1987 - 2 Marks)

Detailed Solution for Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 4

(d) Let  z=

Then by DeMoivre’s theorem, we have

Now,

=  [Using z7 = cos 2π + i sin 2π = 1]

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 5

If ω is an imaginary cube root of unity, then (1 + ω – ω2)7 equals          (1998 - 2 Marks)

Detailed Solution for Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 5

(d) We have (1 + ω + ω2)7 = (- ω2 2)7

(-2)72)7 = -128 ω14 = -128ω2

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 6

The value of the sum , where i = , equals             (1998 - 2 Marks)

Detailed Solution for Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 6

This forms a G.P.

Sum of G.P.= i(1 + i)

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 7

If  = x + iy, then (1998 - 2 Marks)

Detailed Solution for Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 7

(d) Taking  – 3i common from C2, we get

(∵ C2 ≡ C3)

⇒ x = 0,   y = 0

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 8

Let z1 an d z2 be two distinct complex numbers an d let z = (1 – t) z1 + tz2 for some real number t with 0 < t < 1. If Arg (w) denotes the principal argument of a non-zero complex number w, then (2010)

Detailed Solution for Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 8

(a,c,d) Given that z = (1 – t) z1+ tz2 where 0 < t<1

⇒ z divides the join of z1and zinternally in the ratio t : (1– t).

∴ z1 , z and z2 are collinear

⇒ |z –z1| + |z –z2| = |z1– z2|

Also z = (1 – t) z1+ tz2

= t = purely real number

∴ arg = 0  ⇒ arg(z - z) = arg(z-z)

Also

⇒ (z – z1) (z2 –z1)

⇒  = 0

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 9

Let w =and P = {wn : n  = 1, 2, 3, ...}. Further H1 =

and H2  where c is the set of all complex numbers. If z1∈ P∩H1, z2 ∈ P∩H2 and O represents the origin, then ∠z1Oz2 =    (JEE Adv. 2013)

Detailed Solution for Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 9

(c, d)

and wn

∴ P contains all those points which lie on unit circle and have arguments

and so on.

As z1∈ P ∩ H1 and z2 ∈ P ∩ H2, therefore z1 and z2 can have possible positions as shown in the figure.

∴ ∠z1Oz2 can be

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 10

Let a, b ∈ R and a2 + b2 ≠ 0.

Suppose  where

i =. If z = x+ iy and z ∈ S, then (x, y) lies on (JEE Adv. 2016)

Detailed Solution for Test: MCQs (One or More Correct Option): Complex Numbers | JEE Advanced - Question 10

(a, c, d) z =  = x + iy

⇒  x + iy

⇒

⇒   x2 + y2

⇒   x2 + y2 -

∴ Locus of z is a circle with centre  and radius

irrespective of ‘a’ +ve or –ve

Also for b = 0, a ≠ 0, we get, y = 0

∴ locus is x-axis

and for a = 0, b ≠ 0 we get x = 0

⇒ locus is y-axis.

∴ a, c, d are the correct options.

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## Maths 35 Years JEE Main & Advanced Past year Papers

132 docs|70 tests