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I'm trying to prove that if $z,w \in \mathbb{C}$ then $\displaystyle\left|\frac{z}{w}\right| = \frac{|z|}{|w|}$ if $w \neq 0$.

Here is what I have: $$\left|\frac{z}{w}\right|^2=|zw^{-1}|^2 = (zw^{-1})\overline{(zw^{-1})} = \left(\frac{z}{w}\right)\left(\frac{\overline{z}}{\overline{w}}\right)=\frac{z\overline{z}}{w\overline{w}}=\frac{|z|^2}{|w|^2} \blacksquare$$

The text says that a proof can be obtained by applying (a): $$|zw|^2 = (zw)\overline{zw} = (zw)(\overline{z}\cdot\overline{w}) = z\overline{z}w\overline{w}=|z|^2|w|^2$$ to the product $\displaystyle\left(\frac{z}{w}\right)w$.

Is my proof okay and what exactly is the text asking me to do? I feel like the above mentioned product cancels out $w$ and leaves me with $z$ only.

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  • $\begingroup$ Your proof is okay (missing a square after first equal sign) but I dont know what other assumptions you're allowed to use. You probably could have skipped the whole negative-one power step. $\endgroup$ – David Peterson Aug 28 '16 at 23:10
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The argument intended by the text is this:

You get $|z|^2 = |\frac{z}{w}w|^2 = |\frac{z}{w}|^2 \ |w|^2$, and dividing by $|w|^2$ you get $\frac{|z|^2}{|w|^2} = |\frac{z}{w}|^2$.

But your argument is alright too (except for the missing square pointed out in a comment), if you can use those properties of complex conjugation.

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  • $\begingroup$ For this one I'm allowed to use those properties. $\endgroup$ – Trevor Mason Aug 28 '16 at 23:25
  • $\begingroup$ Then it's fine. I assume the author of the text just gave this solution as it leverages the earlier obtained result for the product, rather than to redo a similar argument from scratch. Yet both are valid approaches. $\endgroup$ – quid Aug 28 '16 at 23:28

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