CircleEasy
Question
A complex number z is said to be unimodular if |z| = 1. Suppose z1 and z2 are complex numbers such that 
is unimodular and z2 is not unimodular. Then the point z1 lies on a :
is unimodular and z2 is not unimodular. Then the point z1 lies on a :Options
A.circle of radius 2
B.circle of radius √2
C.straight line parallel to x-axis
D.straight line parallel to y-axis
Solution
= 1 ⇒ |z1 − 2z2|2 = |2 − z1
|2⇒ (z1 − 2z2)(
− 
) = (2 − z1)(2 − z2)⇒ |z1|2 + 4|z2|2 − 4 − |z1|2 |z2|2 = 0
⇒ 4(|z2|2 − 1) − |z1|2 (|z2|2 − 1) = 0
⇒ |z1|2 − 4 = 0⇒|z1| = 2 is a circle of radius 2 and centre at origin.
Create a free account to view solution
View Solution FreeMore Circle Questions
The equation of the normal to the circle x2 + y2 − 8x − 2y + 12 = 0 at the points whose ordinate is −1...The equation of the tangent to the circle x2 + y2 = 25 which is inclined at 60o angle with x-axis, will be-...The tangent to the hyperbola, x2 - 3y2 = 3 at the point (√3, 0) when associated with two asymptotes constitutes -...Lines L1 : x + y = 2 and L2 : y -(2 + √3)x + (1 + √3) = 0 intersect circle S1 = 0 (radius r) at A,B and A,C ...An equilateral triangle OAB is inscribed in the parabola $y^{2} = 4x$ with the vertex O at the vertex of the parabola. T...