Exercise 832Algebra › Complex numbersHigh schoolAdded 3 Nov 2026

Triangle inequality for complex numbers

#triangle inequality#modulus#complex

Statement

Prove that for all $z,w\in\mathbb{C}$ :
$|z+w|\le|z|+|w|.$

Hint

Use $|z+w|^2=(z+w)(\bar z+\bar w)$ and $\operatorname{Re}(z\bar w)\le|z\bar w|$ .

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

Bound. $\operatorname{Re}(z\bar w)\le|z\bar w|=|z||w|$ , so
$|z+w|^2\le|z|^2+|w|^2+2|z||w|=(|z|+|w|)^2.$

Conclusion. Taking square roots gives the triangle inequality.

Equality case. $z$ and $w$ are nonnegatively proportional (same argument, one is a nonneg real multiple of the other).