Q 1 :

Let θ1,θ2,,θ10 be positive valued angles (in radian) such that θ1+θ2++θ10=2π. Define the complex numbers z1=eiθ1, zk=zk-1eiθk for k=2,3,,10,
where i=-1. Consider the statements P and Q given below:

P|z2-z1|+|z3-z2|++|z10-z9|+|z1-z10|2π

Q|z22-z12|+|z32-z22|++|z102-z92|+|z12-z102|4π                        [2021]

  • P is TRUE and Q is FALSE  

     

  • Q is TRUE and P is FALSE  

     

  • both P and Q are TRUE  

     

  • both P and Q are FALSE

     

(3)

Since, |z1|=|z2|==|z10|=1

θ2=arc(z1,z2)

|z2-z1|=length of line ABlength of arc AB

|z3-z2|=length of line BClength of arc BC

 Sum of length of these 10 lines  Sum of length of arcs (i.e. 2π)

                                                                   [ θ1+θ2+θ3++θ10=2π]

 P: |z2-z1|+|z3-z2|++|z1-z10|2π

P is true.

Now, |z22-z12|=|z2-z1||z2+z1|

We know that 

|z2+z1||z2|+|z1|2

 |z22-z12|+|z32-z22|++|z12-z102|2{|z2-z1|+|z3-z2|++|z1-z10|}2(2π)Q4π

Q is also true.



Q 2 :

Let S be the set of all complex numbers z satisfying |z-2+i|5. If the complex number z0 is such that 1|z0-1| is the maximum of the set {1|z-1|:zS}, then the principal argument of 4-z0-z0¯z0-z0¯+2i is                     [2019]

  • π4    

     

  • 3π4    

     

  • π2    

     

  • -π2

     

(4)

S: |z-2+i|5 represents boundary and outer region of circle with centre (2,-1) and radius 5 units.

z0S, such that 1|z0-1| is the maximum.

 |z0-1| is minimum.

z0S with |z0-1| as minimum will be a point on boundary of circle of region S  which lies on radius of this circle, which passes through (1,0).

 z0, 1, 2-i are collinear, or (x0,y0), (1,0), (2,-1) are collinear.

 Using slopes of parallel lines,x'

y0x0-1=-12-1  y0=1-x0

Now,   4-z0-z0¯z0-z0¯+2i=4-(z0+z0¯)(z0-z0¯)+2i

=4-2x02iy0+2i=4-2x02i(1-x0)+2i

=2(2-x0)2(2-x0)i=1i=-i

 Arg(4-z0-z0¯z0-z0¯-2i)=Arg(-i)=-π2



Q 3 :

Let complex numbers α and 1α¯ lie on circles (x-x0)2+(y-y0)2=r2 and (x-x0)2+(y-y0)2=4r2 respectively.  If z0=x0+iy0 satisfies the equation 2|z0|2=r2+2, then |α|=                      [2013]

  • 12    

     

  • 12    

     

  • 17    

     

  • 13

     

(3)

Since, α lies on the circle (x-x0)2+(y-y0)2=r2

 |α-z0|2=r2

(α-z0)(α¯-z0¯)=r2

αα¯-αz0¯-α¯z0+z0z0¯=r2

|α|2+|z0|2-αz0¯-α¯z0=r2    ...(i)

Also 1α¯ lies on the circle (x-x0)2+(y-y0)2=4r2

 |1α¯-z0|2=4r2(1α¯-z0)(1α-z0¯)=4r2

1αα¯-z0α-z0¯α¯+z0z0¯=4r2

1|α|2-z0α¯|α|2-z0¯α|α|2+|z0|2=4r2

1+|α|2|z0|2-z0α¯-z0¯α=4r2|α|2              ...(ii)

On subtracting equation (i) from (ii), we get

1-|α|2+|z0|2(|α|2-1)=r2(4|α|2-1)

or (|α|2-1)(|z0|2-1)=r2(4|α|2-1)

Using |z0|2=r2+22, we get

(|α|2-1)r22=r2(4|α|2-1)

|α|2-1=8|α|2-2

|α|2=17

|α|=17



Q 4 :

Let z be a complex number such that the imaginary part of z is non-zero and a=z2+z+1 is real. Then a cannot take the value           [2012]

  • -1

     

  • 13    

     

  • 12    

     

  • 34

     

(4)

 Im(z)0

z is non-real

and equation z2+z+(1-a)=0 will have non-real roots, if D<0

1-4(1-a)<04a<3a<34

 a cannot take the value 34.



Q 5 :

Let z=x+iy be a complex number where x and y are integers. Then the area of the rectangle whose vertices are the roots of the equation: zz¯3+z¯z3=350 is        [2009]

  • 48

     

  • 32

     

  • 40

     

  • 80

     

(1)

Given: z=x+iy, where x and y are integers

Also, zz¯3+z¯z3=350|z|2(z¯2+z2)=350

(x2+y2)(x2-y2)=175

(x2+y2)(x2-y2)=25×7              ...(i)

or (x2+y2)(x2-y2)=35×5             ...(ii)

 x,y are integers

 x2+y2=25 and x2-y2=7           [From eq (i)]

x2=16, y2=9

x=±4, y=±3

 Vertices of rectangle are 

(4,3), (4,-3), (-4,-3), (-4,3)

 Area of rectangle =8×6=48 sq. units

Now from eq. (ii),

x2+y2=35 and x2-y2=5

x2=20, which is not possible for any integral value of x.



Q 6 :

Let z=cosθ+isinθ. Then the value of m=115Im(z2m-1) at θ=2° is                   [2009]

  • 1sin2°    

     

  • 13sin2°    

     

  • 12sin2°    

     

  • 14sin2°

     

(4)

z=cosθ+isinθ

z2m-1=(cosθ+isinθ)2m-1=cos(2m-1)θ+isin(2m-1)θ

[By De Moivre's theorem: (cosθ+isinθ)n=cosnθ+isinnθ]

 Im(z2m-1)=sin(2m-1)θ

 m=115Im(z2m-1)=m=115sin(2m-1)θ

=sinθ+sin3θ+sin5θ+ (upto 15 terms)

=sin[15(2θ2)]·sin[θ+14θ]sinθ

[ sinα+sin(α+β)+sin(α+2β)+ (n terms)=sin(nβ2)·sin(α+(n-1)β2)sin(β2)]

=sin15θ·sin15θsinθ=sin30°·sin30°sin2°=14sin2°



Q 7 :

A particle P starts from the point z0=1+2i, where i=-1. It moves horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point z1. From z1 the particle moves 2 units in the direction of the vector i^+j^ and then it moves through an angle π2 in anticlockwise direction on a circle with centre at origin, to reach a point z2. The point z2 is given by                    [2008]

  • 6+7i    

     

  • -7+6i    

     

  • 7+6i    

     

  • -6+7i

     

(4)

The initial position of point is Z0=1+2i

 Z1=(1+5)+(2+3)i=6+5i

Now Z1 is moved through a distance of 2 units in the direction i^+j^ (i.e. by 1+i)

 It becomes Z1'=Z1+(1+i)=7+6i

Now OZ1' is rotated through an angle π2 in anticlockwise direction,

therefore Z2=iZ1'=-6+7i.



Q 8 :

If |z|=1 and z±1, then all the values of z1-z2 lie on                 [2007]

  • a line not passing through the origin

     

  • |z|=2

     

  • the x-axis

     

  • the y-axis

     

(4)

Given: |z|=1 and z±1

To find the locus of ω=z1-z2

Now, ω=z1-z2=zzz¯-z2            [ |z|=1|z|2=zz¯=1]

=1z¯-z=purely imaginary number

 ω must lie on y-axis.



Q 9 :

A man walks a distance of 3 units from the origin towards the north-east (N 45° E) direction. From there, he walks a distance of 4 units towards the north-west (N 45° W) direction to reach a point P. Then the position of P in the Argand plane is                 [2007]

  • 3eiπ/4+4i    

     

  • (3-4i)eiπ/4    

     

  • (4+3i)eiπ/4    

     

  • (3+4i)eiπ/4

     

(4)

OP=OA+AP

OP=OA+OB

OP=3eiπ/4+4ei(π/2+π/4)

=3eiπ/4+4eiπ/2eiπ/4

=3eiπ/4+4ieiπ/4

=eiπ/4(3+4i)



Q 10 :

If a,b,c are integers, not all simultaneously equal and ω is a cube root of unity (ω1), then minimum value of |a+bω+cω2| is             [2005]

  • 0

     

  • 1

     

  • 32    

     

  • 12

     

(2)

Given that a,b,c are integers not all equal and ω is cube root of unity 1,

then |a+bω+cω2|

=|a+b(-1+i32)+c(-1-i32)|

=|(2a-b-c2)+i(b3-c32)|

=12(2a-b-c)2+3(b-c)2

=12[(a-b)2+(b-c)2+(c-a)2]

R.H.S. will be minimum when a=b=c, but according to the question, we cannot take a=b=c.

 The minimum value is obtained when any two are zero and third is a minimum magnitude integer i.e. 1.

 b=c=0, a=1 gives us the minimum value =1.



Q 11 :

The locus of z which lies in shaded region (excluding the boundaries) is best represented by                  [2005]

  • z:|z+1|>2 and |arg(z+1)|<π4

     

  • z:|z-1|>2 and |arg(z-1)|<π4

     

  • z:|z+1|<2 and |arg(z+1)|<π2

     

  • z:|z-1|<2 and |arg(z+1)|<π2

     

(1)

In the figure, we see that 

AB=AC=AD=2

 BCD is an arc of a circle with centre at A and radius 2. Shaded region is exterior part of this sector ABCDA.

 For any point represented by z on arc BCD we should have 

|z-(-1)|=2

and for shaded region, |z+1|>2               ...(i)

For shaded region, we also have 

-π4<arg(z+1)<π4

or |arg(z+1)|<π4             ...(ii)

From (i) and (ii), we get (1) is the correct option.



Q 12 :

If ω(1) be a cube root of unity and (1+ω2)n=(1+ω4)n, then the least positive value of n is                     [2004]

  • 2

     

  • 3

     

  • 5

     

  • 6

     

(2)

(1+ω2)n=(1+ω4)n

(-ω)n=(1+ω)n=(-ω2)n

ωn=1n=3



Q 13 :

If |z|=1 and ω=z-1z+1 (where z-1), then Re(ω) is                   [2003]

  • 0

     

  • -1|z+1|2    

     

  • |zz+1|·1|z+1|2

     

  • 2|z+1|2

     

(1)

Given that |z|=1 and ω=z-1z+1 (z-1)

Now we know that zz¯=|z|2

zz¯=1                 (for |z|=1)

 ω=(z-1z+1)×(z¯+1z¯+1)

=zz¯+z-z¯-1zz¯+z+z¯+1=2iy2+2x

[ zz¯=1 and taking z=x+iy so that z+z¯=2x and z-z¯=2iy]

Re(ω)=0



Q 14 :

Let ω=-12+i32, then the value of the determinant |1111-1-ω2ω21ω2ω4| is                  [2002]

  • 3ω    

     

  • 3ω(ω-1)    

     

  • 3ω2    

     

  • 3ω(1-ω)

     

(2)

Applying R1R1+R2+R3, we get

|1101-1-ωω21ω2ω4|=|3001-1-ω2ω21ω2ω4|

=3[-ω-1-ω]=3(ω2-ω)



Q 15 :

The complex numbers z1,z2 and z3 satisfying z1-z3z2-z3=1-i32 are the vertices of a triangle which is                   [2001]

  • of area zero  

     

  • right-angled isosceles  

     

  • equilateral  

     

  • obtuse-angled isosceles

     

(3)

z1-z3z2-z3=1-i32

arg(z1-z3z2-z3)=arg(1-i32)

arg(cos(-π3)+isin(-π3))

angle between (z1-z3) and (z2-z3) is 60°.

and  |z1-z3z2-z3|=|1-i32|

|z1-z3z2-z3|=1|z1-z3|=|z2-z3|  (Imp Step)

The  with vertices z1,z2,z3 is isosceles with vertical angle 60°.

Hence rest of the two angles should also be 60° each.

Required triangle is an equilateral triangle.



Q 16 :

Let z1 and z2 be nth roots of unity which subtend a right angle at the origin. Then n must be of the form                      [2001]

  • 4k+1    

     

  • 4k+2    

     

  • 4k+3    

     

  • 4k

     

(4)

Let z=(1)1/n=(cos2kπ+isin2kπ)1/n

z=cos2kπn+isin2kπn,  k=0,1,2,,n-1.

Let z1=cos(2k1πn)+isin(2k1πn)

and z2=cos(2k2πn)+isin(2k2πn)

be the two values of z such that they subtend right angle at origin.

 2k1πn-2k2πn=±π24(k1-k2)=±n

As k1 and k2 are integers and k1k2,

 n=4k,  kI.



Q 17 :

For a complex number z, let Re(z) denote the real part of z. Let S be the set of all complex numbers z satisfying z4-|z|4=4iz2, where i=-1. Then the minimum possible value of |z1-z2|2, where z1,z2S with Re(z1)>0 and Re(z2)<0, is _______              [2020]



(8)

Let z=x+iy

z4-|z|4=4iz2

z4-(zz¯)2=4iz2z2(z2-z¯2)=4iz2

z=0 or z2-z¯2=4i

4ixy=4ixy=1

Locus of z is a rectangular hyperbola xy=1

Given that Re(z1)>0 and Re(z2)<0

 |z1-z2|min=(1+1)2+(1+1)2=8

|z1-z2|min2=8

 



Q 18 :

Let ω1 be a cube root of unity. Then the minimum of the set {|a+bω+cω2|2:a,b,c distinct non-zero integers} equals ______.         [2019]



(3)

a,b,c are distinct non-zero integers

Min. value of |a+bω+cω2|2 is to be found |a+bω+cω2|2

=|a+b(-1+i32)+c(-1-i32)|2

=|12(2a-b-c)+i32(b-c)|2

=14(2a-b-c)2+34(b-c)2

=14(4a2+b2+c2-4ab+2bc-4ac+3b2+3c2-6bc)

=a2+b2+c2-ab-bc-ca

=12[(a-b)2+(b-c)2+(c-a)2]

For minimum value, let us consider a=3, b=2, c=1

  minimum value =12[1+1+4]=62=3



Q 19 :

For a non-zero complex number z, let arg(z) denote the principal argument of z, with -π<arg(z)π. Let ω be the cube root of unity for which 0<arg(ω)<π. Let α=arg(n=12025(-ω)n). Then the value of 3απ is _________.            [2025]



(-2)

α=arg(-ω+ω2-ω3++(-ω)2025)

α=arg(-ω((-ω)2025-1)-ω-1)α=arg(2ω2)

α=-2π3  3απ=-2



Q 20 :

Let  denote the set of all real numbers. Let z1=1+2i and z2=3i be two complex numbers, where i=-1.

Let S={(x,y)×:|x+iy-z1|=2|x+iy-z2|}. Then which of the following statements is (are) TRUE?      [2025]

  • S is a circle with centre (-13,103)

     

  • S is a circle with centre (13,83)

     

  • S is a circle with radius 23

     

  • S is a circle with radius 223

     

Select one or more options

(1, 4)

We have, |x+iy-1-2i|=2|x+iy-3i|

(x-1)2+(y-2)2=4(x2+(y-3)2)

3x2+3y2+2x-20y+31=0

x2+y2+2x3-20y3+313=0

 S is a circle with centre (-13,103) and radius 

=19+1009-313=89=223



Q 21 :

Let a,b and a2+b20. Suppose S={zC:z=1a+ibt, t+, t0}, where i=-1. If z=x+iy and zS, then (x,y) lies on       [2016]

  • the circle with radius 12a and centre (12a,0) for a>0, b0

     

  • the circle with radius -12a and centre (-12a,0) for a<0, b0

     

  • the x-axis for a0, b=0

     

  • the y-axis for a=0, b0

     

Select one or more options

(1, 3, 4)

z=1a+ibt=x+iy

x+iy=a-ibta2+b2t2x=aa2+b2t2,  y=-bta2+b2t2

x2+y2=1a2+b2t2=xax2+y2-xa=0

 Locus of z is a circle with centre (12a,0) and radius 12|a|

irrespective of 'a' being +ve or -ve

Also for b=0, a0, we get y=0

 locus is x-axis

and for a=0, b0, we get x=0

locus is y-axis.

Hence, 1,3,4 are the correct options.



Q 22 :

Let w=3+i2 and P={wn:n=1,2,3,}. Further, H1={z:Re(z)>12}  and  H2={z:Re(z)<-12},

where c is the set of all complex numbers. If z1PH1, z2PH2 and O represents the origin, then z1Oz2=                      [2013]

  • π2

     

  • π6

     

  • 2π3

     

  • 5π6

     

Select one or more options

(3, 4)

We have w=3+i2=cosπ6+isinπ6

wn=cosnπ6+isinnπ6

 P contains all those points which lie on unit circle and have arguments π6,2π6,3π6 and so on.

Since, z1PH1 and z2PH2, therefore z1 and z2 can have possible positions as shown in the figure.

 z1Oz2 can be 2π3 or 5π6



Q 23 :

Match the statements in Column I with those in Column II.                        [2010]

[Note: Here z takes values in the complex plane and Imz and Rez denote, respectively, the imaginary part and the real part of z.]

  Column I   Column II
(A) The set of points z satisfying 
|z-i|z||=|z+i|z||
is contained in or equal to
(p) an ellipse with eccentricity 45
(B) The set of points z satisfying 
|z+4|+|z-4|=10
is contained in or equal to
(q) the set of points z satisfying Imz=0
(C) If |w|=2, then the set of points 
z=w-1w
is contained in or equal to
(r) the set of points z satisfying |Imz|1
(D) If |w|=1, then the set of points 
z=w+1w
is contained in or equal to
(s) the set of points z satisfying |Rez|<2
    (t) the set of points z satisfying |z|3

 

  • (A)(q,r),  B(p),  C(p,s,t),  D(q,r,s,t)

     

  • (A)(q,r,s,t),  B(p),  C(p,s,t),  D(q,r)

     

  • (A)(q,r,s,t),  B(p,s,t),  C(p),  D(q,r)

     

  • (A)(p),  B(p,s,t),  C(q,r,s,t),  D(q,r)

     

(1)

(A)(q,r)

|z-i|z||=|z+i|z||

z is equidistant from two points (0,|z|) and (0,-|z|), which lie on imaginary axis.

 z must lie on real axis Im(z)=0. Also |Im(z)|1

(B)p

Sum of distances of z from two points (-4,0) and (4,0) is 10 which is greater than 8.

z traces an ellipse with 2a=10 and 2ae=8

e=45

(C)(p,s,t)

Let ω=2(cosθ+isinθ), then

z=ω-1ω=2(cosθ+isinθ)-12(cosθ-isinθ)

x+iy=32cosθ+i52sinθ

Here, |z|=9+254=3443 and |Re(z)|2

Also x=32cosθ, y=52sinθ4x29+4y225=1

Which is an ellipse with e=1-925=45

(D)(q,r,s,t)

Let ω=cosθ+isinθ then z=2cosθIm(z)=0

Also |z|3 and |Im(z)|1, |Re(z)|2