Question

|z | =1, then the least value of 1+z/1+z is

Answer 1

Answer:

z∣+∣1−z∣≥∣1+z+1−z∣ (triangle inequality)

⇒ ∣1+z∣+∣1−z∣≥2

∴ minimum value of (∣1+z∣+∣1−z∣)=2

Geometrically ∣z+1∣+∣1−z∣=∣z+1∣+∣z−1∣

which represents sum of distances of z from 1 and −1

it can be seen easily that minimum (PA+PB)=AB=2