UPSC NDA Math Model Paper 10

Let z=x+ iy lies on |z|=2
⇒√x2+y2=2
⇒x2+y2=4, which is a circle with centre at (0,0) and radius is 2.
Let z=x+ iy, then z is the conjugate of a complex number given by,
z=x−iy
Now,

4

z

=

4

x−iy

⇒

4

z

=

4

x−iy

×

x+iy

x+iy

⇒

4

z

=

4(x+iy)

x2+y2

⇒

4

z

=

4z

4

(∵x2+y2=4)
⇒

4

z

=z
Taking mod both sides, we get
⇒|

4

z

|=|z|
Let

4

z

=z′
Hence, |z′|=2(∵|z|=2)
|z′| lies on the circle.