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

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
$$
R=2\max_{-\pi/2\le \theta \le \pi/2}r_1(\theta)=\max_{1 \le t \le 3}\sqrt{2[t+\sqrt{t^2-1}]}=\sqrt{2(3+2\sqrt{2})}.
$$
