$z$ is a complex number, by the way.
I've tried a lot of things and always end up with a huge algebraic mess and I've wondered if anyone of you has any idea on how to approach this problem.
One of the most "satisfactory" things I've done so far is to expand both the real and imaginary parts for the equation \vert z + 1/z \vert^2 = a^2. These parts yield the constraints:
$2xy(1+1/(x^2+y^2)^2)=0$ ;
$x^2-y^2+\frac{2(x^2-y^2)}{x^2+y^2}+\frac{(x^2-y^2)}{(x^2+y^2)^2}=a^2 $
I've graphed the equations and supposedly the intersection should be my set of points, but this method is far from analytical.
What can I do? Any hints on how to approach this?
Short Solution:
By the Triangle Inequality, $$a=\left|z+\frac{1}{z}\right|\geq \left||z|-\frac{1}{|z|}\right|\,,$$ giving $$-a\leq |z|-\frac{1}{|z|}\leq +a\,,$$ which means $$\frac{\sqrt{a^2+4}-a}{2} \leq |z| \leq \frac{\sqrt{a^2+4}+a}{2}\,.$$ Since $z=\pm \frac{\sqrt{a^2+4}+a}{2}$ and $z=\pm\frac{\sqrt{a^2+4}-a}{2}$ are solutions, both bounds are sharp.
Long Solution: I like this one better, though, because it tells you all possible values of $|z|$.
Suppose that $z=r\exp(\text{i}\theta)$ with $r>0$ and $\theta\in\mathbb{R}$ is a solution to $\left|z+\frac{1}{z}\right|=a$. Then, $$a^2=\left|z+\frac{1}{z}\right|^2=r^2+\frac{1}{r^2}+2\cos(2\theta)\,.$$ Hence, $$a^2-2\leq r^2+\frac{1}{r^2} \leq a^2+2\,.\tag{*}$$
If $a>2$, then, $r$ must satisfy $\left(r+\frac{1}{r}\right)^2\geq a^2$, or $r+\frac{1}{r}\geq a$, as well as$\left(r-\frac{1}{r}\right)^2\leq a^2$, or $-a\leq r-\frac{1}{r}\leq +a$. This means $$0\leq r\leq \frac{a-\sqrt{a^2-4}}{2}\text{ or }r\geq \frac{a+\sqrt{a^2-4}}{2}\,,$$ along with $$\frac{\sqrt{a^2+4}-a}{2} \leq r \leq \frac{\sqrt{a^2+4}+a}{2}\,.$$ Observe that $\frac{a-\sqrt{a^2-4}}{2}> \frac{\sqrt{a^2+4}-a}{2}$ for all $a>2$. Thence, $$\frac{\sqrt{a^2+4}-a}{2}\leq r \leq \frac{a-\sqrt{a^2-4}}{2}\text{ or }\frac{a+\sqrt{a^2-4}}{2}\leq r\leq \frac{\sqrt{a^2+4}+a}{2}\,.$$
If $0 \leq a\leq 2$, then we have by AM-GM that $ r^2+\frac{1}{r^2}\geq 2\geq a^2-2$. Thus, only the inequality on the right-hand side of (*) is relevant. We have $\left(r-\frac{1}{r}\right)^2\leq a^2$, or $-a\leq r-\frac{1}{r}\leq +a$. Therefore, $$\frac{\sqrt{a^2+4}-a}{2} \leq r \leq \frac{\sqrt{a^2+4}+a}{2}\,.$$
Consequently, in both cases, we have $\frac{\sqrt{a^2+4}-a}{2} \leq r \leq \frac{\sqrt{a^2+4}+a}{2}$. The equality on the left side is satisfied when $z=\pm\frac{\sqrt{a^2+4}-a}{2}$, and the equality on the right is satisfied iff $z=\pm\frac{\sqrt{a^2+4}+a}{2}$. Ergo, the maximum value of $|z|=r$ is $\frac{\sqrt{a^2+4}+a}{2}$, whereas the minimum value is $\frac{\sqrt{a^2+4}-a}{2}$.