Complex Numbers | Real Analysis

See a complex number as a point in the plane. The real part runs sideways, the imaginary part runs up. The modulus is the distance from the origin to that point. Adding two complex numbers draws the diagonal of a parallelogram. That diagonal is never longer than the two sides added together. This is the triangle inequality. For example, walk 3 steps east with a 🧭 compass, then turn and walk 4 more. The straight-line distance back is shorter than the bent path. The modulus measures exactly that straight line.

Examples & Exercises

Prove completely that for all complex numbers $z, w \in \mathbb{C}$ , the triangle inequality $|z + w| \le |z| + |w|$ holds. Start, without hints, by expanding $|z + w|^2$ through the conjugate.

|z + w| \le |z| + |w| \qquad (z, w \in \mathbb{C})

z = 1.2 + 0.8i

w = -0.6 + 1.1i

|z + w| = \sqrt{3.97} \approx 1.992

while

|z| + |w| \approx 2.695

, a slack of about

0.703

Show proof

First expand $|z + w|^2$ using $|\zeta|^2 = \zeta \bar{\zeta}$ for $\zeta \in \mathbb{C}$ . Since $\overline{z + w} = \bar{z} + \bar{w}$ , $|z + w|^2 = (z + w)(\bar{z} + \bar{w}) = z\bar{z} + z\bar{w} + w\bar{z} + w\bar{w} = |z|^2 + z\bar{w} + \bar{z}w + |w|^2$ . Here $\bar{z}w = \overline{z\bar{w}}$ , because $\overline{z\bar{w}} = \bar{z}\,\overline{\bar{w}} = \bar{z}w$ . For any complex number $\zeta$ , $\zeta + \bar{\zeta} = 2\,\mathrm{Re}\,\zeta$ ; taking $\zeta = z\bar{w}$ gives $|z + w|^2 = |z|^2 + 2\,\mathrm{Re}(z\bar{w}) + |w|^2$ .
Next, a lemma: $\mathrm{Re}\,\zeta \le |\zeta|$ for every $\zeta \in \mathbb{C}$ . Proof: write $\zeta = u + vi$ with $u, v \in \mathbb{R}$ . Then $\mathrm{Re}\,\zeta = u \le |u| = \sqrt{u^2} \le \sqrt{u^2 + v^2} = |\zeta|$ , where the middle step uses $u^2 \le u^2 + v^2$ (as $v^2 \ge 0$ ) and the monotonicity of the square root. Hence $\mathrm{Re}(z\bar{w}) \le |z\bar{w}|$ .
Now compute $|z\bar{w}| = |z|\,|w|$ directly. Both sides are $\ge 0$ , so it suffices to compare squares: $|z\bar{w}|^2 = (z\bar{w})\,\overline{(z\bar{w})} = (z\bar{w})(\bar{z}w) = (z\bar{z})(w\bar{w}) = |z|^2 |w|^2 = (|z|\,|w|)^2$ . The square root is monotone on $[0, \infty)$ (equivalently, nonnegative numbers with equal squares are equal), so $|z\bar{w}| = |z|\,|w|$ .
Combine everything: $|z + w|^2 = |z|^2 + 2\,\mathrm{Re}(z\bar{w}) + |w|^2 \le |z|^2 + 2|z\bar{w}| + |w|^2 = |z|^2 + 2|z|\,|w| + |w|^2 = (|z| + |w|)^2$ . Both sides are $\ge 0$ and the square root is monotone on $[0, \infty)$ , so taking square roots gives $|z + w| \le |z| + |w|$ . ∎

Prove completely that for all $z, w \in \mathbb{C}$ , the parallelogram identity $|z + w|^2 + |z - w|^2 = 2|z|^2 + 2|w|^2$ holds. Expand all four terms in the form $\zeta \bar{\zeta}$ and show explicitly how the cross terms cancel.

|z + w|^2 + |z - w|^2 = 2|z|^2 + 2|w|^2 \qquad (z, w \in \mathbb{C})

z = 1.2 + 0.8i

w = -0.6 + 1.1i

|z + w|^2 + |z - w|^2 = 3.97 + 3.33 = 7.30

and

2|z|^2 + 2|w|^2 = 2(2.08 + 1.57) = 7.30

Show proof

Expand the sum term. Since $\overline{z + w} = \bar{z} + \bar{w}$ , $|z + w|^2 = (z + w)(\bar{z} + \bar{w}) = z\bar{z} + z\bar{w} + w\bar{z} + w\bar{w} = |z|^2 + z\bar{w} + \bar{z}w + |w|^2$ .
Expand the difference term. Since $\overline{z - w} = \bar{z} - \bar{w}$ , $|z - w|^2 = (z - w)(\bar{z} - \bar{w}) = z\bar{z} - z\bar{w} - w\bar{z} + w\bar{w} = |z|^2 - z\bar{w} - \bar{z}w + |w|^2$ . Only the cross terms $z\bar{w}$ and $\bar{z}w$ have changed sign.
Add the two expansions. The cross terms cancel, $+z\bar{w} - z\bar{w} = 0$ and $+\bar{z}w - \bar{z}w = 0$ , while the diagonal terms double: $|z + w|^2 + |z - w|^2 = (|z|^2 + |w|^2) + (|z|^2 + |w|^2) = 2|z|^2 + 2|w|^2$ . ∎
Why this identity is a bridge: by the Jordan-von Neumann theorem, a norm satisfying the parallelogram law for all vectors is exactly a norm coming from an inner product — this computation in $\mathbb{C}$ is the seed of that direction, and it returns in the parallelogram-law scene of the sequences chapter.

Prove completely that for all $z, w \in \mathbb{C}$ : (i) $|zw| = |z|\,|w|$ , (ii) $\overline{z + w} = \bar{z} + \bar{w}$ , (iii) $\overline{zw} = \bar{z}\,\bar{w}$ , and (iv) $z\bar{z} = |z|^2$ . Write $z = a + bi$ and $w = c + di$ with $a, b, c, d \in \mathbb{R}$ and carry out the component computations explicitly.

|zw| = |z|\,|w|, \qquad \overline{z + w} = \bar{z} + \bar{w}, \qquad \overline{zw} = \bar{z}\,\bar{w}, \qquad z\bar{z} = |z|^2

z = 1 + i

w = 1 - 0.5i

zw = 1.5 + 0.5i

, so

|zw| = \sqrt{2.5} \approx 1.581

, matching

|z|\,|w| = \sqrt{2} \cdot \sqrt{1.25} = \sqrt{2.5}

Show proof

Start with (iv). Since $\bar{z} = a - bi$ , $z\bar{z} = (a + bi)(a - bi) = a^2 - abi + abi - b^2 i^2 = a^2 + b^2 = |z|^2$ . The cross terms $-abi + abi$ cancel, and $i^2 = -1$ turns $-b^2 i^2$ into $+b^2$ .
(ii) Conjugate of a sum: $z + w = (a + c) + (b + d)i$ , so $\overline{z + w} = (a + c) - (b + d)i = (a - bi) + (c - di) = \bar{z} + \bar{w}$ .
(iii) Conjugate of a product: first $zw = (a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i$ , so $\overline{zw} = (ac - bd) - (ad + bc)i$ . Meanwhile $\bar{z}\,\bar{w} = (a - bi)(c - di) = ac - adi - bci + bdi^2 = (ac - bd) - (ad + bc)i$ . The two computations agree, so $\overline{zw} = \bar{z}\,\bar{w}$ .
(i) Modulus of a product: using (iii) and (iv), $|zw|^2 = (zw)\,\overline{(zw)} = (zw)(\bar{z}\bar{w}) = (z\bar{z})(w\bar{w}) = |z|^2 |w|^2 = (|z|\,|w|)^2$ . Both sides are $\ge 0$ , so taking square roots gives $|zw| = |z|\,|w|$ . ∎