Q 1 :

If w-w¯z1-z is purely real where w=α+iβ, β≠0 and z≠1, then the set of the values of z is           [2006]

  • {z:|z|=1}  

     

  • {z:z=z¯}

     

  • {z:z≠1}  

     

  • {z:|z|=1, z≠1}

     

(4)

∵w-w¯z1-z is purely real

∴ (w-w¯z1-z)¯=(w-w¯z1-z)⇒w¯-wz¯1-z¯=w-w¯z1-z

⇒w¯-wz¯-w¯z+wzz¯=w-w¯z-wz¯+w¯zz¯

⇒w-w¯=(w-w¯)|z|2

⇒|z|2=1  (∵w=α+iβ and β≠0)

⇒|z|=1 and also given that z≠1

∴The required set is {z:|z|=1,z≠1}



Q 2 :

If z1,z2 and z3 are complex numbers such that

|z1|=|z2|=|z3|=|1z1+1z2+1z3|=1, then |z1+z2+z3| is                [2000]

  • equal to 1

     

  • less than 1

     

  • greater than 3

     

  • equal to 3

     

(1)

Given : |z1|=|z2|=|z3|=1

Now, |z1|=1⇒|z1|2=1⇒z1z1¯=1

Similarly z2z2¯=1,  z3z3¯=1

Now, |1z1+1z2+1z3|=1⇒|z1¯+z2¯+z3¯|=1

⇒|z1+z2+z3¯|=1⇒|z1+z2+z3|=1



Q 3 :

For all complex numbers z1,z2 satisfying |z1|=12 and |z2-3-4i|=5, the minimum value of |z1-z2| is         [2002]

  • 0

     

  • 2

     

  • 7

     

  • 17

     

(2)

|z1|=12 ⇒ z1 lies on a circle with centre (0,0) and radius 12 units.

And |z2-3-4i|=5 ⇒ z2 lies on a circle with centre (3,4) and radius 5 units.

From figure, it is clear that |z1-z2| i.e., distance between z1 and z2 will be minimum when they lie at A and B respectively,

i.e., O,C,B,A are collinear as shown.

Then |z1-z2|=AB=OA-OB=12-2(5)=2.

As above is the minimum value, we must have |z1-z2|≥2.



Q 4 :

If arg(z)<0, then arg(-z)-arg(z)=               [2000]

  • π

     

  • -π

     

  • -π2

     

  • π2

     

(1)

Given: arg(z)<0 (given) ⇒ arg(z)=-θ

Now, z=rcos(-θ)+isin(-θ)=r[cosθ-isinθ]

Again, -z=-r[cosθ-isinθ]=r[cos(π-θ)+isin(π-θ)]

∴ arg(-z)=π-θ

Thus arg(-z)-arg(z)=(π-θ)-(-θ)=π-θ+θ=π



Q 5 :

Let A={1967+1686 isinθ7-3icosθ:θ∈ℝ}. If A contains exactly one positive integer n, then the value of n is                     [2023]



(281)

281(49+18sinθcosθ+i(21cosθ+42sinθ))49+9cos2θ

is a positive integer. For positive integer Im(z)=0

⇒21cosθ+42sinθ=0

⇒tanθ=-12,  sin2θ=-45,  cos2θ=45

Now Re(2)=281(49-9sin2θ)49+9cos2θ

=281(49-9×(-45))49+9×45

=281(49+365)49+365

= 281



Q 6 :

For any integer k, let αk=cos(kπ7)+isin(kπ7), where i=-1. The value of the expression ∑k=112|αk+1-αk|∑k=13|α4k-1-α4k-2| is            [2015]



(4)

 Given: αk=coskπ7+isinkπ7=eiπk7

αk+1-αk=eiπ(k+1)7-eiπk7=eiπk7(eiπ7-1)

|αk+1-αk|=|eiπ7-1|

⇒∑k=112|αk+1-αk|=12|eiπ7-1|

Similarly, ∑k=13|α4k-1-α4k-2|=3|eiπ7-1|

∴∑k=112|αk+1-αk|∑k=13|α4k-1-α4k-2|=4



Q 7 :

If z is any complex number satisfying |z-3-2i|≤2, then the minimum value of |2z-6+5i| is               [2011]



(5)

Given: |z-3-2i|<2,

which represents a circular region with centre (3,2) and radius 2.

Now, |2z-6+5i|=2|z-(3-52i)|

=2×distance of z from P  (where z lies in or on the circle)

Also min distance of z from P=52

∴ Minimum value of |2z-6+5i|=5



Q 8 :

Let z be a complex number with non-zero imaginary part. If 2+3z+4z22-3z+4z2 is a real number, then the value of |z|2 is _______.                    [2022]



(0.50)

 Let X=4z2+3z+24z2-3z+2

It can be written as=1+6z4z2-3z+2  ;  Now X=1+62(2z+1z)-3

∵ X∈ℝ, then 2z+1z∈ℝ

⇒2z+1z=2z¯+1z¯⇒2(z-z¯)-z-z¯|z|2=0

∵ (z-z¯)(2-1|z|2)=0

∵ z≠z¯ (given), So, |z|2=12



Q 9 :

Let z¯ denote the complex conjugate of a complex number z and let i=-1. In the set of complex numbers, the number of distinct roots of the equation z¯-z2=i(z¯+z2)is _______ .         [2022]



(4)

 Given, z¯-z2=i(z¯+z2)

It can be written as z¯(1-i)=z2(1+i)

So |z¯||1-i|=|z|2|1+i|

|z|=|z|2 ⇒ |z|=0 or |z|=1

Let arg(z)=α. So from (i), we get

2nπ-α-π4=2α+π4

⇒α=13(4n-12)π=(4n-1)π6

So we will get 3 distinct values of α. Hence there will be total 4 possible values of complex number z.



Q 10 :

Let S={a+b2:a,b∈ℤ},   T1={(-1+2)n:n∈ℕ} and T2={(1+2)n:n∈ℕ}. Then which of the following statements is (are) TRUE?         [2024]

  • ℤ∪T1∪T2⊂S

     

  • T1∩(0,12024)=ϕ, where ϕ denotes the empty set.

     

  • T2∩(2024,∞)≠ϕ

     

  • For any given a,b∈ℤ, cos(π(a+b2))+isin(π(a+b2))∈ℤ if and only if b=0, where i=-1.

     

Select one or more options

(1, 3, 4)

(a) S={a+b2:a,b∈ℤ}

     For b=0; ℤ⊂S

     T1=(-1+2)n=m+2 n, m,n∈ℤ

     T2=(1+2)n=m1+2n1, m1,n1∈ℤ

      For n∈ℕ, elements of T1 and T2 are of the form a+b2

      Hence ℤ∪T1∪T2⊂S

(b) Now, -1+2<1 and its higher powers decrease

      ⇒(-1+2)n<1 and can be made in (0,12024) for some higher n

(c) 1+2>1 and its higher power increases

      ⇒(1+2)n can be made in (2024,∞) for some higher n

(d) cosπ(a+b2)+isinπ(a+b2)∈ℤ

      a+b2 is an integer ⇒b=0

 



Q 11 :

Let z¯ denote the complex conjugate of a complex number z. If z is a non-zero complex number for which both real and imaginary parts of (z¯)2+1z2 are integers, then which of the following is/are possible value(s) of |z|?                      [2022]

  • (43+32052)14

     

  • (7+334)14

     

  • (9+654)14

     

  • (7+136)14

     

(1)

Let z=r.eiθ⇒z¯=re-iθ

∴ (z¯)2+1z2=r2e-2iθ+1r2e2iθ=(r2+1r2)e-2iθ=a+ib (say), 

where a,b∈ℤ

So, (r2+1r2)2=a2+b2⇒r8-(a2+b2-2)r4+1=0

⇒r4=(a2+b2-2)±(a2+b2-2)2-42

For option (a): |z|4=43+32052

⇒a2+b2=45 i.e. (a,b)=(±6,±3) or (±3,±6)

For option (b): |z|4=7+334⇒a2+b2=112

For option (c): a2+b2=132

For option (d): a2+b2=133



Q 12 :

Let S be the set of all complex numbers z satisfying |z2+z+1|=1. Then which of the following statements is/are TRUE?             [2020]

  • |z+12|≤12 for all z∈S

     

  • |z|≤2 for all z∈S

     

  • |z+12|≥12 for all z∈S

     

  • The set S has exactly four elements

     

Select one or more options

(2, 3)

|z2+z+1|=1

⇒|(z+12)2+34|≥1⇒|(z+12)|2≥14⇒|z+12|≥12

also |(z2+z)+1|=1

⇒|z2+z|-1≤1⇒|z2+z|≤2

⇒||z2|-|z||≤|z2+z|≤2⇒|r2-r|≤2⇒r=|z|≤2; ∀z∈S

Hence, set S is infinite



Q 13 :

Let s, t, r be non-zero complex numbers and L be the set of solutions z=x+iy (x,y∈ℝ, i=-1) of the equation

 sz+tz¯+r=0,  where z¯=x-iy. Then, which of the following statement(s) is (are) TRUE?               [2018]

  • If L has exactly one element, then |s|≠|t|

     

  • If |s|=|t|, then L has infinitely many elements

     

  • The number of elements in L∩{z:|z-1+i|=5} is at most 2

     

  • If L has more than one element, then L has infinitely many elements

     

Select one or more options

(1, 3, 4)

We have, 

       sz+tz¯+r=0                 ⋯(i)

On taking conjugate   

       s¯z¯+tz¯+r¯=0                   ⋯(ii)

On solving Eqs. (i) and (ii), we get

z=r¯t-rs¯|s|2-|t|2

(a) For unique solutions of z

       |s|2-|t|2≠0⇒|s|≠|t|

       It is true.

(b) If |s|=|t|, then r¯t-rs¯ may or may not be zero.

       So, z may have no solution.

        ∴ L may be an empty set.

        It is false.

(c) If elements of set L represents line, then this line and given circle intersect at maximum two points.

        Hence, it is true.

(d) In the case locus of z is a line, so L has infinite elements.

        Hence, it is true.



Q 14 :

For a non-zero complex number z, let arg(z) denote the principal argument with -π<arg(z)≤π. Then, which of the following statement(s) is (are) FALSE?     [2018]

  • arg(-1-i)=π4, where i=-1

     

  • The function f:ℝ→(-π,π], defined by f(t)=arg(-1+it) for all t∈ℝ, is continuous at all points of ℝ, where i=-1

     

  • For any two non-zero complex numbers z1 and z2, arg(z1z2)-arg(z1)+arg(z2) is an integer multiple of 2π

     

  • For any three given distinct complex numbers z1,z2 and z3, the locus of the point z satisfying the condition arg((z-z1)(z2-z3)(z-z3)(z2-z1))=π, lies on a straight line

     

Select one or more options

(1, 2, 4)

(a) arg(-1-i)=-3π4

      ∴ (a) is false

(b) f(t)=arg(-1+it)=[π-tan-1(t),t≥0-π+tan-1(t),t<0

      limt→0-f(t)=-π  and  limt→0+f(t)=π

       LHL≠RHL⇒f is discontinuous at t=0

      ∴ (b) is false

(c) arg(z1z2)-argz1+argz2

     =2nπ+argz1-argz2-argz1+argz2

     =2nπ, multiple of 2π

      ∴ (c) is true

(d)  arg((z-z1)(z2-z3)(z-z3)(z2-z1))=π

      ⇒(z-z1)(z2-z3)(z-z3)(z2-z1)=k,  k∈ℝ

     ⇒(z-z1z-z3)=k(z2-z1z2-z3)

     ⇒z,z1,z2,z3 are concyclic, i.e. z lies on a circle.

      ∴ (d) is false



Q 15 :

Let a, b, x and y be real numbers such that a-b=1 and y≠0. If the complex number z=x+iy satisfies Im(az+bz+1)=y, then which of the following is(are) possible value(s) of x?          [2017]

  • -1+1-y2

     

  • -1-1-y2

     

  • 1+1+y2

     

  • 1-1+y2

     

Select one or more options

(1, 2)

a-b=1, y≠0

Im(az+bz+1)=y

⇒Im[(a(x+iy)+b(x+1)+iy)·(x+1)-iy(x+1)-iy]=y

⇒-(ax+b)y+ay(x+1)(x+1)2+y2=y

⇒-axy-by+axy+ay(x+1)2+y2=y

⇒a-b=(x+1)2+y2

⇒1=(x+1)2+y2

∴ x=-1±1-y2

 



Q 16 :

Let z1 and z2 be two distinct complex numbers and 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]

  • |z-z1|+|z-z2|=|z1-z2|

     

  • Arg(z-z1)=Arg(z-z2)

     

  • |z-z1z¯-z1¯z2-z1z2¯-z1¯|

     

  • Arg(z-z1)=Arg(z2-z1)

     

Select one or more options

(1, 3, 4)

 Given: z=(1-t)z1+tz2, where 0<t<1

⇒z=(1-t)z1+tz2(1-t)+t

⇒z divides the join of z1 and z2 internally in the ratio t:(1-t)

∴ z1,z,z2 are collinear           

        

⇒|z-z1|+|z-z2|=|z1-z2|

Also z=(1-t)z1+tz2

⇒z-z1z2-z1=t, which is purely real number

∴ arg(z-z1z2-z1)=0 ⇒arg(z-z1)=arg(z2-z1)

Also z-z1z2-z1=t⇒z¯-z¯1z¯2-z¯1=t

⇒z-z1z2-z1=z¯-z¯1z¯2-z¯1

⇒(z-z1)(z¯2-z¯1)=(z¯-z¯1)(z2-z1)

⇒ |z-z1z¯-z¯1z2-z1z¯2-z¯1|=0



Q 17 :

Let z be a complex number satisfying |z|3+2z2+4z¯-8=0, where z¯ denotes the complex conjugate of z. Let the imaginary part of z be non-zero.

Match each entry in List-I to the correct entries in List-II.

  List - I   List - II
(P) |z|2 is equal to (1) 12
(Q) |z-z¯|2 is equal to (2) 4
(R) |z|2+|z+z¯|2 is equal to (3) 8
(S) |z+1|2 is equal to (4) 10
    (5) 7

 

The correct option is:                                                          [2023]

  • (P) → (1), (Q) → (3), (R) → (5), (S) → (4)

     

  • (P) → (2), (Q) → (1), (R) → (3), (S) → (5)

     

  • (P) → (2), (Q) → (4), (R) → (5), (S) → (1)

     

  • (P) → (2), (Q) → (3), (R) → (5), (S) → (4)

     

(2)

Given, |z|3+2z2+4z¯-8=0                ⋯(i)

|z¯|3+2z¯2+4z-8=0  [Conjugate both sides]

⇒2(z2-z¯2)+4(z¯-z)=0

⇒2(z-z¯)[z+z¯-2]=0

∵ z=z¯ (Not possible)  or  z+z¯=2

∴ z=1+bi (b≠0)⇒z¯=1-bi

(1+b2)3/2+2(1-b2+2bi)+4(1-bi)-8=0  [from (i)]

(1+b2)3/2-2(1+b2)=0

⇒(1+b2)(1+b2-2)=0

∵ 1+b2≠0⇒1+b2-2=0⇒b2=3

(P) |z|2=1+b2=1+3=4

(Q) |z-z¯|2=|1+ib-1+ib|2=4b2=12

(R) |z|2+|z+z¯|2=4+|1+ib+1-ib|2=4+4=8

(S) |z+1|2=|1+1+ib|2=4+b2=4+3=7



Q 18 :

Let S=S1∩S2∩S3, where

S1={z∈ℂ:|z|<4},  S2={z∈ℂ:Im[z-1+3 i1-3 i]>0}, and S3={z∈ℂ:Re(z)>0}.                [2013]

Q.  Area of S=

  • 10π3

     

  • 20π3

     

  • 16π3

     

  • 32π3

     

(2)

S1:x2+y2<16

S2:Im[(x-1)+i(y+3)1-i3]>0

⇒3(x-1)+(y+3)>0 ⇒y+3x>0

S3:x>0

Then S:S1∩S2∩S3 is as shown in the figure given below.

Area of shaded region

=π4×42+π×42×60°360°

=4π+8π3=20π3



Q 19 :

Let S=S1∩S2∩S3, where

S1={z∈ℂ:|z|<4},  S2={z∈ℂ:Im[z-1+3 i1-3 i]>0}, and S3={z∈ℂ:Re(z)>0}.             [2013]

Q.  minz∈S|1-3i-z|=

  • 2-32

     

  • 2+32

     

  • 3-32

     

  • 3+32

     

(3)

S1:x2+y2<16

S2:Im[(x-1)+i(y+3)1-i3]>0

⇒3(x-1)+(y+3)>0 ⇒y+3x>0

S3:x>0

Then S:S1∩S2∩S3 is as shown in the figure given below.

minz∈S|1-3i-z|= minimum distance between z and (1,-3)

Clearly (from figure) minimum distance between z∈S and (1,-3)

from line y+x3=0 i.e. |3-33+1|=3-32



Q 20 :

Let A,B,C be three sets of complex numbers as defined below                       [2008]

A={z:Im(z)≥1}

B={z:|z-2-i|=3}

C={z:Re((1-i)z)=2}

Q. The number of elements in the set A∩B∩C is

  • 0

     

  • 1

     

  • 2

     

  • ∞

     

(2)

Given : A={z:Im(z)≥1}={(x,y):y≥1}

Clearly A is the set of all points lying on or above the line y=1 in cartesian plane.

B={z:|z-2-i|=3}={(x,y):(x-2)2+(y-1)2=9}

⇒B is the set of all points lying on the boundary of the circle with centre (2,1) and radius 3.

C={z:Re[(1-i)z]=2}={(x,y):x+y=2}

⇒ C is the set of all points lying on the straight line represented by x+y=2.

Graphically, the three sets are represented as shown below :

From graph A∩B∩C consists of only one point P [the common point of the region y≥1, (x-2)2+(y-1)2=9 and x+y=2

∴ n(A∩B∩C)=1



Q 21 :

Let zk=cos(2kπ10)+isin(2kπ10), k=1,2,…,9.                                [2014]

  List - I   List - II
P. For each zk there exists zj such that zk.zj=1 1. True
Q. There exists a k∈{1,2,…,9} such that z1.z=zk has no solution z in the set of complex numbers 2. False
R. |1-z1||1-z2|⋯|1-z9|10 equals 3. 1
S. 1-∑k=19cos(2kπ10) equals 4. 2

 

  • P→1,Q→2,R→4,S→3

     

  • P→2,Q→1,R→3,S→4

     

  • P→1,Q→2,R→3,S→4

     

  • P→2,Q→1,R→4,S→3

     

(3)

(P)→(1):  zk=cos2kπ10+isin2kπ10, k=1 to 9

∴  zk=ei2kπ10

Now zk.zj=1⇒zj=1zk=e-i2kπ10=zk¯

We know if zk is 10th root of unity so will be zk¯

∴ For every zk there exist zi=zk¯

Such that zk.zj=zk.zk¯=1

Hence the statement is true.

(Q)→(2):  z1=zk⇒z=zkz1, z1≠0

∴ We can always find a solution of z1z=zk

Hence the statement is false.

(R)→(3):  We know z10-1=(z-1)(z-z1)⋯(z-z9)

⇒(z-z1)(z-z2)⋯(z-z9)=z10-1z-1=1+z+z2+⋯+z9

For z=1, we get (1-z1)(1-z2)⋯(1-z9)=10

∴ |1-z1||1-z2|⋯|1-z9|10=1

(S)→(4):  1,z1,z2,…,z9 are 10th roots of unity.

∴ z10-1=0

From equation 1+z1+z2+⋯+z9=0,

Re(1)+Re(z1)+Re(z2)+⋯+Re(z9)=0

⇒Re(z1)+Re(z2)+⋯+Re(z9)=-1

⇒∑k=19cos2kπ10=-1⇒1-∑k=19cos2kπ10=2

Hence (3) is the correct option.



Q 22 :

Let A,B,C be three sets of complex numbers as defined below

A={z:Im(z)≥1}

B={z:|z-2-i|=3}

C={z:Re((1-i)z)=2}

 

Q.   Let z be any point in A∩B∩C.  Then, |z+1-i|2+|z-5-i|2 lies between                   [2008]

  • 25 and 29  

     

  • 30 and 34  

     

  • 35 and 39  

     

  • 40 and 44

     

(3)

Given:  A={z:Im(z)≥1}={(x,y):y≥1}

Clearly A is the set of all points lying on or above the line y=1 in Cartesian plane.

B={z:|z-2-i|=3}={(x,y):(x-2)2+(y-1)2=9}

⇒ B is the set of all points lying on the boundary of the circle with centre (2, 1) and radius 3.

C={z:Re[(1-i)z]=2}={(x,y):x+y=2}

⇒ C is the set of all points lying on the straight line represented by x+y=2.

Graphically, the three sets are represented as shown below:

Since, z is a point of A∩B∩C⇒z represents the point P

∴ |z+1-i|2+|z-5-i|2

⇒|z-(-1+i)|2+|z-(5+i)|2

⇒PQ2+PR2=QR2=62=36, which lies between 35 and 39

∴ (3) is correct option.



Q 23 :

Let A,B,C be three sets of complex numbers as defined below

A={z:Im(z)≥1}

B={z:|z-2-i|=3}

C={z:Re((1-i)z)=2}

 

Q.   Let z be any point A∩B∩C and let w be any point satisfying |w-2-i|<3. Then, |z|-|w|+3 lies between         [2008]

  • - 6 and 3  

     

  • - 3 and 6  

     

  • - 6 and 6  

     

  • - 3 and 9

     

(4)

Given:  A={z:Im(z)≥1}={(x,y):y≥1}

Clearly A is the set of all points lying on or above the line y=1 in Cartesian plane.

B={z:|z-2-i|=3}={(x,y):(x-2)2+(y-1)2=9}

⇒ B is the set of all points lying on the boundary of the circle with centre (2, 1) and radius 3.

C={z:Re[(1-i)z]=2}={(x,y):x+y=2}

⇒ C is the set of all points lying on the straight line represented by x+y=2.

Graphically, the three sets are represented as shown below:

Given |w-2-i|<3

⇒Distance between w and (2+i), i.e. S is smaller than 3.

⇒w is a point lying inside the circle with centre S and radius 3.

⇒Distance between z (i.e. the point P) and w should be smaller than 6  (the diameter of the circle)

i.e. |z-w|<6

But we know that ||z|-|w||≤|z-w|

⇒||z|-|w||<6⇒-6<|z|-|w|<6

⇒-3<|z|-|w|+3<9