Ireland_2017

Question

Prove for all complex numbers z that |z|^2 + 2|z - 1| 1, with equality iff z = 1.

Accepted Answer

If |z| 1, then |z|^2 + 2|z - 1| 1 + 2|z - 1| 1, with equality iff z = 1. If |z| < 1, then |z - 1| |z| - 1 = 1 - |z|, and so |z|^2 + 2|z - 1| - 1 |z|^2 + 2(1 - |z|) - 1 = (|z| - 1)^2 > 0.