If $S=\{z\in\mathbb C:z\neq0\text{ and }\left|z+\frac{1}{z}\right|=a\}$ find $\min\limits_{z\in S}|z|$ 
If $a$ is a positive real number and $S=\{z\in \mathbb C : z\neq 0\ \text{ and } \vert z+\frac{1}{z}\vert = a\}$, Find $\min\limits_{z\in S}\vert z\vert$ and $\min\limits_{z\in S}\vert\frac{1}{z}\vert$

I first tried using the triangle inequality to get:
$$\vert z\vert^2-a\vert z\vert+1\ge 0$$
$$\vert z\vert\in\big(-\infty,\frac{a-\sqrt{a^2-4}}{2}\big)\cup\big(\frac{a+\sqrt{a^2-4}}{2},\infty\big)$$
since $\vert z\vert\ge 0$ and $a^2>a^2-4$,
$$\vert z\vert\in\big(0,\frac{a-\sqrt{a^2-4}}{2}\big)\cup\big(\frac{a+\sqrt{a^2-4}}{2},\infty\big)$$
So $\min\vert z \vert=0$. But the answer given is $\min\vert z\vert=\frac{\sqrt{a^2-4}-a}{2}$

Let $z=re^{i\phi}$.$\,\,\,$  If $|z+z^{-1}|=a$, then 
$$r^2+r^{-2}+2\cos(2\phi)=a^2 \tag 1$$
We can write $(1)$ in a more convenient form as
$$\left(r+\frac1r\right)^2=a^2-4\sin^2(\phi) \tag 2$$
Note that $|a|\ge 2$ since the minimum of $r+\frac1r$ is $2$.
Now, differentiating both sides of $(2)$ with respect to $\phi$ yields
$$\left(2r-\frac{2}{r^3}\right)\frac{dr}{d\phi}=-4\sin(2\phi)$$
whereupon setting $\frac{dr}{d\phi}$ to $0$ reveals that $2\phi = \ell \pi$ for integer $\ell$.  
For $r\le 1$, the minimum for $r$ occurs when $2\phi =0$.  Substituting $2\phi=0$ into $(2)$ reveals
$$r_{min}+\frac1{r_{min}}=|a| \implies r_{min}=\frac{|a|-\sqrt{a^2-4}}{2}$$
