For the complex number $\mathrm{z}$ satisfying the condition $\left|\mathrm{ }\mathrm{z}+\frac{2}{\mathrm{z}}\mathrm{ }\right|=2$, the maximum value of $\left|\mathrm{ }\mathrm{z}\mathrm{ }\right|$ is

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

Now, for maximum or minimum value of $\left|z\right|,$ LHS inequality will hold true as it corresponds to the two complex numbers being on opposite sides of origin.

Hence, $\left|z\right|-\frac{2}{\left|z\right|}\le 2$ (without loss of generality let LHS > 0)

${\Rightarrow \left|z\right|}^2-2\left|z\right|-2\le 0$

$\Rightarrow \left|z\right|\le \frac{2+\sqrt{4+8}}{2}$

$\Rightarrow \left|z\right|\le 1+\sqrt{3}$