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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.

Red Banana
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2 Answers2

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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*}

parsiad
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That's a fine way to proceed. You could make it more explicit by combining like factors, so you have the standard $x + yi$ form, and then conjugating, as you suggest, but the end result would be the same.

amWhy
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  • Thanks for the answer. I'm feeling something weird: Multiplying $\overline{z},\overline{w}$ gives me a guarantee that $\overline{z},\overline{w}=a c-i a d-i b c-b d$ but I feel I can't extend that guarantee to $\overline{zw}$. – Red Banana May 11 '13 at 01:29
  • You can see this works by expanding $(zw)$, conjugating, factoring the conjugation of $zw$, to obtain the conjugation of $(z)(w)$, then showing this is equal to $\overline z \overline w$ – amWhy May 11 '13 at 01:32
  • @amWhy: Nice feedback too +1 – Amzoti May 11 '13 at 06:25