Exercise 1567Algebra › Complex numbersOlympiadAdded 3 Nov 2026

An inequality involving arguments

#argument#inequality#complex

Statement

Let $z,w\in\mathbb{C}^*$ with $\operatorname{Re}(z\bar w)>0$ . Show that $|z+w|>|z-w|$ .

Hint

Square both sides and compare.

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Squaring.
$|z+w|^2=|z|^2+|w|^2+2\operatorname{Re}(z\bar w),$
$|z-w|^2=|z|^2+|w|^2-2\operatorname{Re}(z\bar w).$

Comparison. $|z+w|^2-|z-w|^2=4\operatorname{Re}(z\bar w)>0$ .

Conclusion. $|z+w|>|z-w|$ .

Interpretation. The angle between $z$ and $w$ is acute (cosine positive).