TITLE: How to show that $\overline{zw}=\overline{z}\,\overline{w}$?
QUESTION [9 upvotes]: I thought about first multiplying two complex which aren't in the conjugate form:
$$zw=a c+i a d+i b c-b d$$
Then multiply two complex conjugates:
$$\overline{z}\,\overline{w}=a c\color{red}{-}i a d\color{red}{-}i b c-b d$$
And then note the differences between them, which are those red signals. Then I thought about expanding $\overline{zw}$ without negating the proper terms:
$$\overline{zw}=\overline{a c+i a d+i b c-b d}$$
And then apply the overline as if it were an operation to negate those two signals:
$$\overline{zw}=a c-i a d-i b c-b d$$
Is my reasoning correct? I'm thinking it is but I'm feeling that something is missing.

REPLY [7 votes]: Yes. More concisely,
\begin{align*}
\overline{zw} & =\overline{\left(a+ib\right)\left(c+id\right)}\\
 & =\overline{ac+iad+ibc-bd}\\
 & =ac-iad-ibc-bd\\
 & =\left(a-ib\right)\left(c-id\right)\\
 & =\left(\overline{a+ib}\right)\left(\overline{c+id}\right)\\
 & =\overline{z}\overline{w}
\end{align*}