Question

The number of complex numbers z such that |z – 1| = |z + 1| = |z – i| equals

(a) 0 

(b) 1 

(c) 2 

(d) ∞

Answer 1

(b) : Ist solution :
|z – 1| = |z + 1| = |z – i| reads that the distance of desired complex number z is same from three points in the complex plane –1, 1 and i. These points are non-collinear, hence the desired number is the centre of the (unique) circle passing through these three non-collinear points.

2nd solution :
We resort to definition of modulus.
|z – 1| = |z + 1| ⇒ |z – 1|2 = |z + 1|²
⇒ (z − 1)(bar z − 1) = (z + 1)(bar z + 1)
⇒ z bar z − z − bar z + 1 = z bar z + z + bar z + 1
⇒ z +bar z = 0 (z being purely imaginary)
Thus x = 0
Again, |z – 1|² = |z – i|²
⇒ (x – 1)² + y² = x² + (y – 1)²
⇒ 1 + y² = (y – 1)² (because x = 0)
⇒ 1 + y² = y² – 2y + 1
∴ y = 0
Thus, (0, 0) is the desired point.