JEE Main 10 Jan 2019 Shift 1 Solved Paper

3 |z1| = 4|z2|
⇒

|z1|

|z2|

=

4

3

⇒

|3z1|

|2z2|

= 2
Let

3z1

2z2

= a = 2 cos θ + 2i sin θ
z =

3z1

2z2

+

2z2

3z1

= a +

1

a

=

5

2

cos θ +

3

2

i sin θ
Now all options are incorrect
Remark
There is a misprint in the problem actualproblem should be :
"Let z1 and z2 be any non-zero complexnumber such that 3|z1| = 2 |z2|
If z =

3z1

2z2

+

2z2

3z1

, then
Given
3 |z1| = 2|z2|
Now |

3z1

2z2

| = 1
Let

3z1

2z2

= a = cos θ + i sin θ
z =

3z1

2z2

+

2z2

3z1

= a +

1

a

= 2 cos θ
∴ Im (z) = 0
Now option (4) is correct