Maximum value of $Z$ How to find maximum value of $| Z| $ if: $$ \Big|    Z-\dfrac{4}{Z}      \Big|=2;      $$
Where $Z$ is a complex mumber
 A: From here,  $$\left|z-\frac4z\right|\ge |z|-\left|\frac4z\right|$$ 
So, $|z|-\frac4{|z|}\le 2$
$\implies |z|^2-2|z|-4\le 0\implies (|z|-1)^2\le 5$
$\implies  -\sqrt 5\le |z|-1\le \sqrt5\implies 1-\sqrt5\le |z|\le 1+\sqrt5$
$\implies 0< |z|\le 1+\sqrt5$ 
A: Write $Z=re^{i\phi}$, then
$$
4 = |Z-\frac{4}{Z}|^2
\\= (re^{i\phi}-\frac{4}{r}e^{-i\phi})(re^{-i\phi}-\frac{4}{r}e^{i\phi})
\\= r^2-8cos(2\phi)+\frac{16}{r^2}
$$
Write $cos(2\phi)=c$ (this parameter varies freely in $[-1,1]$ given an appropriate choice of $\phi$) and maximize $r$ subject to the constraint
$$
r^2-8c+\frac{16}{r^2}=4
$$
now, write $s=r^2$:
$$
s^2-2(4c+2)s+16=0
\\ s\in{4c+2\pm\sqrt{(4c+2)^2-16}}
$$
$s$ is thus maximized for $c=1$, and
$$
|Z|_{max} = r_{max}\\ = \sqrt{s_{c=1}}
\\ = \sqrt{6+2\sqrt{5}}
\\ = \sqrt{1+2\sqrt{5}+5}
\\ = 1+\sqrt5
$$
A: Setting $z=Z/2$ we have
$$
|Z-4/Z|=2 \iff |z-z^{-1}|=1.
$$
So the problem is to find 
$$
R=2\max_{z \in M}|z| \ \text{ where }\  M:=\{z \in \mathbb{C}:\ |z-z^{-1}|=1\}.
$$
We have
$$
M=\{re^{i\theta}:\ r>0, \ -\pi \le \theta \le\pi,\ r^4-(1+2\cos\theta)r^2+1=0\}.
$$
Solving
$$
r^4-(1+2\cos\theta)r^2+1=0,\ r>0,\  -\pi \le \theta \le\pi,
$$
we obtain
\begin{eqnarray}
r_1^2(\theta)&=&\frac{1}{2}\left(1+2\cos\theta+\sqrt{(1+2\cos\theta)^2-1}\right),\  -\frac{\pi}{2} \le \theta \le \frac{\pi}{2},\\
r_2^2(\theta)&=&\frac{1}{2}\left(1+2\cos\theta-\sqrt{(1+2\cos\theta)^2-1}\right),\  -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}.
\end{eqnarray}
Since
$$
r_1^2(\theta)\ge r_2^2(\theta) \quad \forall \theta \in [-\pi/2,\pi/2],
$$
it follows that
$$
R=2\max_{-\pi/2\le \theta \le \pi/2}r_1(\theta)=\max_{1 \le t \le 3}\sqrt{2[t+\sqrt{t^2-1}]}=\sqrt{2(3+2\sqrt{2})}.
$$
