For $z, w \in \mathbb{C}$, $z \overline{w} = \overline{w \overline{z}}$.

Question

The standard proof of the triangle inequality in $\mathbb{C}$ invokes the fact that $$z \overline{w} = \overline{w \overline{z}}.$$ I am trying to understand why this is the case. I know that for any $z$, we have $$\overline{\overline{z}} = z.$$ I can say that: $$z \overline{w} = \overline{\overline{z \overline{w}}}.$$ For any $z_1, z_2,$, $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$. $$z \overline{w} = \overline{(\overline{z \overline{w}})} = \overline{\overline{z} \cdot \overline{\overline{w}}} = \overline{\overline{z} \cdot w} = \overline{w \overline{z}}.$$ I am silently invoking what is essentially an "associative law" to apply the "outer" conjugate to the inner complex number, which also contains a conjugate. Is this valid? Is there a different way to write this proof?

Answer 1

As far as I can tell, your proof is valid and correct. As for other proof methods, although it's quite a basic technique, you can just use the real & imaginary parts of $z$ and $w$, including in their conjugates, to show the $2$ sides are equal. In particular, let

$$z = x + yi \tag{1}\label{eq1A}$$ $$w = a + bi \tag{2}\label{eq2A}$$

where $x$, $y$, $a$ and $b$ are real. Then you have

$$\begin{equation}\begin{aligned} z \overline{w} & = (x + yi)(a - bi) \\ & = xa -bxi + ayi + by \\ & = (xa + by) + (-bx + ay)i \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

$$\begin{equation}\begin{aligned} w \overline{z} & = (x - yi)(a + bi) \\ & = xa + bxi - ayi + by \\ & = (xa + by) + (bx - ay)i \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

Thus, you get

$$\begin{equation}\begin{aligned} \overline{w \overline{z}} & = (xa + by) + (-bx + ay)i \\ & = z \overline{w} \end{aligned}\end{equation}\tag{5}\label{eq5A}$$