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.
$$z = r \operatorname{cis}\left({\theta}\right) \quad \& \quad \bar{z}= r \operatorname{cis}\left({-\theta}\right) $$
$$z\bar{z} = r\times r \operatorname{cis}\left({\theta + \left({-\theta}\right)}\right)$$
$$= r^2 \operatorname{cis} \left({0}\right)$$
$$ = r^2 \left({\cos(0)+ i \sin(0)}\right)$$
$$= r^2 (1 + 0i)$$
$$\boxed{z \bar{z} = r^2}$$
$$\vert z \vert = \sqrt{(r\cos \theta)^2+(r\sin \theta)^2}$$
$$ = \sqrt{r^2 \cos^2 \theta + r^2 \sin^2 \theta}$$
$$ = \sqrt{r^2(\sin^2 \theta + \cos^2 \theta)}$$
$$ = \sqrt{r^2(1)}$$
$$=\sqrt{r^2}$$
$$\vert z \vert = r$$
$$\boxed{\vert z \vert^2 = r^2}$$
$$\text{Hence} \quad z\bar{z} \equiv \vert z \vert ^2$$
Best Answer
It's correct. As far as $|z|$ is concerned, $|z| = r $ by definition.
Thus the second part is not needed.