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I am trying to prove $|z|\cdot |w|=|z \cdot w|$

$|z\cdot w|=\sqrt{(ac-bd)^{2}+(ad+bc)^{2}}$ and $|z|\cdot|w|=\sqrt{a^{2}+b^{2}}\cdot\sqrt{c^{2}+d^{2}}=\sqrt{(a^{2}+b^{2})\cdot(c^{2}+d^{2})}=\sqrt{(a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2})}$

And I'm stuck
Help?

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Now, $$(ac-bd)^2+(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=$$ $$=a^2c^2+a^2d^2+b^2c^2+b^2d^2=a^2(c^2+d^2)+b^2(c^2+d^2)=(a^2+b^2)(c^2+d^2)$$

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  • $\begingroup$ you just did what i did backwards how dose this help me prove that $a^2c^2+a^2d^2+b^2c^2+b^2d^2=(ac-bd)^2+(ad+bc)^2$ $\endgroup$ – Lendion Oct 30 '17 at 9:56
  • $\begingroup$ @Lendion Sorry, I thought that this step is obvious. I added something. See now. $\endgroup$ – Michael Rozenberg Oct 30 '17 at 10:00
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We have$$|z.w|=|z|.|w|\iff|z.w|^2=\bigl(|z|.|w|\bigr)^2=|z|^2.|w|^2$$and\begin{align}|z.w|^2&=z.w.\overline{z.w}\\&=z.w.\overline z.\overline w\\&=z.\overline z.w.\overline w\\&=|z|^2.|w|^2.\end{align}

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It's easier if you know that you can write $z = r_z e^{i\theta_z}$ where $r_z$ and $\theta_z$ are real.

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  • 1
    $\begingroup$ ... but actually I like the way shown by José Carlos Santos even better. $\endgroup$ – David K Oct 30 '17 at 10:28

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