Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.3 AIME Prove it Ireland

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

Solution

If $|z| \ge 1$ , then
$|z|^2 + 2|z - 1| \ge 1 + 2|z - 1| \ge 1,$
with equality iff $z = 1$ .

If $|z| < 1$ , then
$|z - 1| \ge |z| - 1 = 1 - |z|,$
and so
$|z|^2 + 2|z - 1| - 1 \ge |z|^2 + 2(1 - |z|) - 1 = (|z| - 1)^2 > 0.$

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