Useful Modulus Identities

Expand |z₁+z₂|² = (z₁+z₂)(z̄₁+z̄₂) and use z₁z̄₁ = |z₁|². The cross term 2Re(z₁z̄₂) is the complex analogue of the 2ab term in (a+b)². This identity is essential for proving the triangle inequalities.

This identity is the complex analogue of the expansion $(a+b)^2 = a^2 + 2ab + b^2$. The cross term $2ab$ becomes $2\operatorname{Re}(z_1\bar{z}_2)$ — a real number that carries the geometric relationship between $z_1$ and $z_2$.

Derivation

Use $|w|^2 = w\bar{w}$:

$$|z_1 + z_2|^2 = (z_1 + z_2)\overline{(z_1 + z_2)} = (z_1 + z_2)(\bar{z}_1 + \bar{z}_2)$$

Expanding:

$$= z_1\bar{z}_1 + z_1\bar{z}_2 + z_2\bar{z}_1 + z_2\bar{z}_2$$ $$= |z_1|^2 + z_1\bar{z}_2 + \overline{z_1\bar{z}_2} + |z_2|^2$$

Since $w + \bar{w} = 2\operatorname{Re}(w)$ for any complex number $w$:

$$\boxed{|z_1 + z_2|^2 = |z_1|^2 + 2\operatorname{Re}(z_1\bar{z}_2) + |z_2|^2}$$

The Companion Identity

Replacing $z_2$ with $-z_2$:

$$|z_1 - z_2|^2 = |z_1|^2 - 2\operatorname{Re}(z_1\bar{z}_2) + |z_2|^2$$

Adding and subtracting the two identities:

$$|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2)$$ $$|z_1 + z_2|^2 - |z_1 - z_2|^2 = 4\operatorname{Re}(z_1\bar{z}_2)$$

The first is the parallelogram law — the sum of squares of diagonals equals twice the sum of squares of sides. It holds for complex numbers exactly as it does for vectors.

Why This Is the Key Identity

The cross term $\operatorname{Re}(z_1\bar{z}_2)$ satisfies $\operatorname{Re}(z_1\bar{z}_2) \leq |z_1\bar{z}_2| = |z_1||z_2|$.

Substituting into the expansion:

$$|z_1 + z_2|^2 \leq |z_1|^2 + 2|z_1||z_2| + |z_2|^2 = (|z_1| + |z_2|)^2$$

Taking square roots gives the triangle inequality $|z_1 + z_2| \leq |z_1| + |z_2|$ in one line. The identity is the engine; the inequality is a corollary.