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My textbook asked me to prove that a complex number $r\operatorname{cis}(x)$, denoted by $z$, when multiplied by its conjugate is equal to its modulus squared. I realise that the second half of my 'proof' was probably unnecessary as the modulus is simply $r$, but I decided to include it anyway.

$$z = r \operatorname{cis}\left({\theta}\right) \quad \& \quad \bar{z}= r \operatorname{cis}\left({-\theta}\right) $$

$$z\bar{z} = r\times r \operatorname{cis}\left({\theta + \left({-\theta}\right)}\right)$$

$$= r^2 \operatorname{cis} \left({0}\right)$$

$$ = r^2 \left({\cos(0)+ i \sin(0)}\right)$$

$$= r^2 (1 + 0i)$$

$$\boxed{z \bar{z} = r^2}$$

$$\vert z \vert = \sqrt{(r\cos \theta)^2+(r\sin \theta)^2}$$

$$ = \sqrt{r^2 \cos^2 \theta + r^2 \sin^2 \theta}$$

$$ = \sqrt{r^2(\sin^2 \theta + \cos^2 \theta)}$$

$$ = \sqrt{r^2(1)}$$

$$=\sqrt{r^2}$$

$$\vert z \vert = r$$

$$\boxed{\vert z \vert^2 = r^2}$$

$$\text{Hence} \quad z\bar{z} \equiv \vert z \vert ^2$$

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    $\begingroup$ The second half is not necessary, the first half looks good to me. $\endgroup$ – Laars Helenius Feb 27 '15 at 23:58
  • $\begingroup$ But the second half may contain insights that are worthwhile that are not in the first half. $\endgroup$ – Michael Hardy Feb 28 '15 at 1:16
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It's correct. As far as $|z|$ is concerned, $|z| = r $ by definition.

Thus the second part is not needed.

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  • $\begingroup$ It's not true by definition, it's true because $|cis(\theta)|=1$ for every $\theta$ and applying multiplicativeness $\endgroup$ – Stella Biderman Dec 29 '16 at 3:25

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