<s>[INST] If $z + \frac{1}{z} = r$, where $r$ be a real number and $|r| < 2$. Find $|z|$. Let $r$ be a real number, $|r| < 2,$ and let $z$ be a complex number such that
$$z + \frac{1}{z} = r.$$Find $|z|.$ I am actually stuck on this problem. Any help will be appreciated. Thank you
 [/INST] Multiply the equation by $z$, subtract $rz$, you get
$$
z^2 -rz + 1 = 0.
$$
Can you take it from here?
Hint: $ax^2 + bx + c = 0$ has solutions $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Edit
There are a bunch of heavy answers now. In my opinion the following solution is much more elementary. Just note that that $|z| = z \bar z$ for any complex number. Then use the quadratic formula to obtain $z = \frac{r - \sqrt{r^2 - 4}}{2}$ and $\bar z = \frac{r + \sqrt{r^2 - 4}}{2}$. Then
$$
z \bar z = \left( \frac{r - \sqrt{r^2 - 4}}{2}\right)\left( \frac{r + \sqrt{r^2 - 4}}{2}\right) = \frac{r^2 - r^2 + 4}{4} = 1.
$$
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