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Test: 35 Year JEE Previous Year Questions: Complex Numbers


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Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 1

z and w are two nonzero complex numbers such that | z | = | w| and Arg z + Arg w = π then z equals [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 1

(b) Let | z | = | ω | = r

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 2

If | z – 4 | < | z – 2 |, its solution is given by [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 2

(c) Given | z – 4 | < | z – 2 |
Let z = x + iy
⇒ | (x – 4) + iy) | < | (x – 2) + iy |
⇒ (x – 4)2 + y2 < (x – 2)2 + y2
⇒ x2 – 8x + 16 < x2 – 4x + 4
⇒ 12 < 4x
⇒ x > 3
⇒ Re(z) > 3

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 3

The locus of the centre of a circle which touches the circle | z – z1 | = a and | z – z2 | = b externally (z , z1 & z2 are complex numbers) will be [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 3

(b) Let the circle be |z – z0| = r. Then according to given conditions |z0 – z1| = r + a and |z0 – z2|= r + b.
Eliminating r, we get |z0 – z1| –|z0 – z2| = a – b.
∴ Locus of centre z0 is |z – z1| –|z – z2| = a – b,
which represents a hyperbola.

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 4

If z and ω are two non-zero complex numbers such that zω = 1 and Arg(z) - Arg(ω) = , then is equal to [2003]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 4

(a) | |=| || w|=| z ||ω|=|zω|= 1
Arg()= arg() + arg(ω)= - arg(z) + argω  
= -

=-1

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 5

Let Z1 and Z 2 be two roots of the equation Z2 + aZ +b= 0 , Z being complex. Further , assume that the origin, Z1 and Z2 form an equilateral triangle. Then [2003]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 5

z 2 + az +b= 0 ;  z1 + z2 = - a & z1 z2=b 0, z1,z2 form an equilateral Δ 
∴ 02 + z12 + z22 = 0.z1 + z1.z2+z2 .0 (for an equilateral triangle, 
z12 + z22 + z32 = z1z2+ z2 z3+ z3z1 )
⇒ z12 + z22= z1z2
⇒ ( z1 + z2 )2=3 z1z2
∴a2= 3b

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 6

If   = 1 then         [2003]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 6

= ⇒ (i)x = 1; 

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 7

Let z and w be complex numbers such that  = 0 and arg zw = π. Then arg z equals [2004] 

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 7

arg zw = π ⇒ arg z + argw = π...(1)




Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 8

If z = x- iy and = p+ iq, then  is equal to [2004]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 8

= p+ iq ⇒ z = p3 + (iq)3 + 3 p (iq)( p+ iq)

⇒ x - iy = p3 - 3pq2 + i(3p2q-q3)

∴ x = p3 - 3pq2 ⇒ 

y = q3 - 3p2q ⇒ 

 

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 9

If | z2 - 1 |=|z |2+1, then z lies on              [2004]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 9

|z2 - 1|=|z|2+1 ⇒|z2 - 1|2

⇒  

⇒  

⇒  ⇒ 

⇒ z is purely imaginary

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 10

If the cube roots of unity are 1,ω ,ω2 then the roots of the equation ( x – 1)3 + 8 = 0, are [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 10

(x - 1)3 + 8=0  
⇒ (x - 1) = (-2) (1)1 /3  
⇒ x – 1 = – 2 or -2ω or - 2ω2
or x = – 1 or 1 – 2ω or 1 – 2ω2 .

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 11

If z1 and z2 are two non- zero complex numbers such that | z1 +z2| = | z1| + | z2| , then arg z1 – arg z2 is equal to [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 11

|z1 +z2| = |z1| + |z2| ⇒ z1 and z2 are collinear and are to the same side of origin; hence arg z1 – arg z2 = 0.

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 12

If  and |w| = 1, then z lies on [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 12

As given w  ⇒ |w| = = 1

 ⇒ 

 ⇒ distance of z from origin and point  is

same hence z lies on bisector of the line joining points (0, 0) and (0, 1/3).
Hence z lies on a straight line.

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 13

The value of is                  [2006]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 13

= i × 0 – i                [∵ e-2πi = 1] = – i

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 14

If z2 +z + 1=0 , where z is complex number, then the value of

[2006]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 14

 z2 +z + 1=0 ⇒ z = ω or ω2

So,

and

∴ The given sum = 1+1 + 4 + 1 + 1 + 4 = 12

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 15

If | z + 4 | ≤ 3, then the maximum value of | z + 1 | is[2007]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 15

z lies on or inside the circle with centre (–4, 0) and radius 3 units.

From the Argand diagram maximum value of | z + 1| is  6

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 16

The conjugate of a complex n umber  then that complex number is [2008] 

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 16

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 17

Let R be the real line. Consider the following subsets of the plane R × R:
S ={(x, y): y = x + 1 and 0 < x < 2}
T ={(x, y): x – y is an integer},
Which one of  the following is true? [2008]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 17

Given S = {(x , y) : y = x + 1 and 0 < x < 2}
∵ x ≠ x  + 1 for any x ∈(0, 2) ⇒ (x, x) ∉ S
∴ S is not reflexive.
Hence S in not an equivalence relation.
Also T  ={x, y): x – y is an integer}
​∵ x – x = 0 is an integer " x ∈ R
∴ T is reflexive.
If  x – y is an integer then y – x is also an integer
∴T is symmetric If  x – y is an integer and y – z is an integer then (x – y) + (y– z) = x – z is also an integer.
∴ T is transitive Hence T is an equivalence relation.

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 18

The number of complex numbers z such that |z – 1| = |z + 1| = |z – i| equals [2010]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 18

Let z = x+ iy
|z -1| = |z +1| (x -1)2 + y2 = (x +1)2+y2
⇒ Re z =0 ⇒ x = 0
|z -1| = |z-i| (x-1)2+ y2= x2 +(y -1)2
⇒ x =y
|z+1| = |z-i| (x+1)2 + y2 = x2 +(y-1)2
Only (0, 0) will satisfy all conditions.
⇒ Number of complex number z  = 1

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 19

Let α, β be real and z be a complex number. If z2 + αz + β = 0 has two distinct roots on the line Re z =1, then it is necessary that :[2011]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 19

∵ Real part of roots is 1 Let roots are 1 + pi, 1 + q
∴ sum of roots = 1 + pi + 1 + qi = – a which is real
⇒  q = – p or root are 1 + pi and 1 – pi product of roots = 1 + p2 = β∈ (1,)
p ≠ 0 as roots are distinct.

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 20

If ω(≠1) is a cube root of unity, and (1 + ω)7 = A +Bω.Then (A, B) equals [2011]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 20

(1 + ω)7 = A + Bω;
​ (–ω2)7 = A + Bω – ω2 = A + Bω;
1 + ω = A + Bω
⇒ A = 1, B = 1.

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 21

If z ≠ 1 and is real, then the point represented by the complex number z lies : [2012]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 21

Now   is real ⇒  Im  = 0

⇒  

⇒​  Im [(x2 – y2 + 2ixy) (x – 1) – iy)] = 0

⇒  2xy (x – 1) – y (x2 – y2) = 0

⇒ y(x2 + y2 – 2x) = 0 ⇒  y = 0;  x2 + y2 – 2x = 0

∴ z lies either on real axis or on a circle through origin.

 

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 22

If z is a complex number of unit modulus and argument θ, then arg       equals: [JEE M 2013]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 22

Given | z | = 1, arg z = θ

As we know,  

∴   = arg (z) = θ

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 23

If z is a complex number such that z ≥ 2, then the minimum value of 

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 23

We know minimum value of 

Thus minimum value of 

Since, | Z |≥2 therefore 

⇒  

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 24

A complex number z is said to be unimodular if |z| = 1.
Suppose z1 and z2 are complex numbers such that  is unimodular and z2 is not unimodular. Then the point z1 lies on a:     [JEE M 2015]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 24

⇒  

⇒  

⇒ 

⇒ 

⇒  

⇒   

⇒ Point z1 lies on circle of radius 2.

Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 25

A value of θ for which    is purely imaginary, is: [JEE M 2016]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Complex Numbers - Question 25

Rationalizing the given expression

For the given expression to be purely imaginary, real part of the above expression should be equal to zero.

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