If $\left| {z - \dfrac{4}{z}} \right| = 2$ , then the maximum value of$\left| z \right|$ is

$\max |Z|=\sqrt{2(3+\sqrt{5})}=1+\sqrt{5}$

Explanation:

Assuming that $Z \in \mathbb{C}$ we have

$Z=\rho e^{i \phi}$ and then knowing that

$|Z|=\sqrt{Z \bar{Z}}$ follows

$\begin{array}{l} \left(\rho e^{i \phi}-\frac{4}{\rho} e^{-i \phi}\right)\left(\rho e^{-i \phi}-\frac{4}{\rho} e^{i \phi}\right)=4 \text { or } \\ \rho^{2}-8 \cos (2 \phi)+\frac{16}{\rho^{2}}=4 \text { or } \\ \rho^{4}-2(2+4 \cos (2 \phi)) \rho^{2}+16=0 \end{array}$

Now solving for $\rho^{2}$

$\rho^{2}=2(1+2 \cos (2 \phi) \pm \sqrt{4 \cos (2 \phi)+2 \cos (4 \phi)-1})$

for $\phi=0$ we have

$\rho_{\max }^{2}=2(1+2+\sqrt{4+2-1})=2(3+\sqrt{5})$

then

$\max |Z|=\sqrt{2(3+\sqrt{5})}=1+\sqrt{5}$