Q 1 :

If w-w¯z1-z is purely real where w=α+iβ, β0 and z1, then the set of the values of z is           [2006]

  • {z:|z|=1}  

     

  • {z:z=z¯}

     

  • {z:z1}  

     

  • {z:|z|=1, z1}

     

(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 β0)

|z|=1 and also given that z1

The required set is {z:|z|=1,z1}



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=1z1z1¯=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.

[IMAGE 5]

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

[IMAGE 6]

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+1686isinθ7-3icosθ:θ}. 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)

[IMAGE 7]

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, then 2z+1z

2z+1z=2z¯+1z¯2(z-z¯)-z-z¯|z|2=0

 (z-z¯)(2-1|z|2)=0

 zz¯ (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},   T1={(-1+2)n:n} and T2={(1+2)n:n}. Then which of the following statements is (are) TRUE?         [2024]

  • T1T2S

     

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

     

  • T2(2024,)ϕ

     

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

     

Select one or more options

(1, 3, 4)

(a) S={a+b2:a,b}

     For b=0; S

     T1=(-1+2)n=m+2n, m,n

     T2=(1+2)n=m1+2n1, m1,n1

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

      Hence T1T2S

(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)

      a+b2 is an integer b=0