If $\left|\mathrm{z}-\frac{2}{\mathrm{z}}\right|=2$ then the greatest value of $\left|\mathrm{z}\right|$ is

$$\begin{gathered} \left|\right|z_1|-|z_2|\le |z_1-z_2|\le |z_1|+|z_2\mid \\ \text{ Let }{z}_1=z\text{ and }{z}_2=\frac{2}{z} \end{gathered}$$

$\left|\right|z\left|-\frac{2}{\left|z\right|}\right|\le \left|z-\frac{2}{z}\right|\le \left|z\right|+\frac{2}{\left|z\right|}$

$\left|\right|z\left|-\frac{2}{\left|z\right|}\right|\le 2\le \left|z\right|+\frac{2}{\left|z\right|}$

$-2\le \left|z\right|-\frac{2}{\left|z\right|}\le 2$

$-2\left|z\right|\le |z{|}^2-2\le 2|z|$

${\left|z\right|}^2-2\left|z\right|\le 2$

${\left(\left|z\right|-1\right)}^2\le 3\Rightarrow \left|z\right|\le \sqrt{3}+1$