Complex Numbers

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 $|z + w| \le |z| + |w|$. ∎

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.

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 $|zw| = |z|\,|w|$. ∎