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.