Let $a$ be a positive real number and let
$$M_a = \left\{z \in \mathbb{C^*}: \left|z + \frac{1}{z}\right| = a\right\}$$
Find the minimum and maximum value of $|z|$ when $z\in M_a$. ($\mathbb{C^*}$ denotes the set of non-zero complex numbers).
I did the following:
$$\left|z + \frac{1}{z}\right| = \left(z + \frac{1}{z}\right)\overline{\left(z + \frac{1}{z}\right)}$$
$$=\left(z + \frac{1}{z}\right)\left(\overline{z} + \frac{1}{\overline{z}}\right) = \frac{|z|^4 + 1 + \overline{z}^2 + z^2}{|z|^2} = \frac{|z|^4 + 1 + (\overline{z} + z)^2 - 2|z|^2}{|z|^2} = a^2$$
Rearranging, we get the quadratic:
$$|z|^4 - |z|^2(a^2 + 2) + 1 = -(z+\overline{z})^2 \leq 0 \tag1$$
We can easily find the interval $|z|^2$ must be for the given quadratic to be less than or equal to zero by finding the roots.
$$|z|^2 \in \left[\frac{a^2 + 2 - \sqrt{a^4 + 4a^2}}{2}, \frac{a^2 + 2 + \sqrt{a^4 + 4a^2}}{2}\right]$$
Taking the square root we get
$$\max(|z|) = \frac{a + \sqrt{a^2 + 4}}{2}, \min(|z|) = \frac{-a + \sqrt{a^2 + 2}}{2}$$
However, in this proof, I was not comfortable about one part I skipped over in $(1)$. I tacitly assumed that the quadratic would be equal to $-(z+\overline{z})^2$ for all $z$ such that the quadratic satisfies the inequality. I don't see how I can prove this.