# Complex number times conjugate equals square of modulus (proof check)

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.

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

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

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

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

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

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

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

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

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

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

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

$$\vert z \vert = r$$

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

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

• The second half is not necessary, the first half looks good to me. – Laars Helenius Feb 27 '15 at 23:58
• But the second half may contain insights that are worthwhile that are not in the first half. – Michael Hardy Feb 28 '15 at 1:16

It's correct. As far as $|z|$ is concerned, $|z| = r$ by definition.
• It's not true by definition, it's true because $|cis(\theta)|=1$ for every $\theta$ and applying multiplicativeness – Stella Biderman Dec 29 '16 at 3:25