complex no.

We have
\left|z+\dfrac{2}{z}\right|=4\quad\Rightarrow\ \left|z+\dfrac{2}{z}\right|^2=16
which gives us
|z|^2+\dfrac{4}{|z|^2}+2\left(\dfrac{z}{\overline{z}}+\dfrac{\overline{z}}{z}\right)=16
\Rightarrow\ |z|^2+\dfrac{4}{|z|^2}+2\left(\dfrac{z^2+\overline{z}^2}{|z|^2}\right)=16

\Rightarrow\ |z|^4 + 4 + 2(z^2+\overline{z}^2)=16|z|^2

\Rightarrow\ |z|^4+4+2\big((z+\overline{z})^2-2z\overline{z}\big)=16|z|^2

\Rightarrow\ |z|^4+4+2(z+\overline{z})^2-4|z|^2=16|z|^2

\Rightarrow\ |z|^4-20|z|^2+4=-2(z+\overline{z})^2=-8(\mathrm{Re}(z))^2\leq 0

Let t=|z|^2. So that the above inequality reduces to
t^2-20t+4\leq 0
which gives us
2(5-2\sqrt{6})\leq t \leq 2(5+2\sqrt{6})
Hence, we get
\sqrt{2(5-2\sqrt{6})}\leq |z| \leq \sqrt{2(5+2\sqrt{6})}

Thanks Kaymant Sir(which clear my concept) and decoder.