Let $z$ be a complex number such that $\left|Z\right| = 2$ then maximum possible value of $\left|z+\frac{2}{z}\right|$ is

 

Let $z=2(\cos \theta +i\sin \theta )$.Now

$z+\frac{2}{z}=2(\cos \theta +i\sin \theta )+(\cos \theta -i\sin \theta )$

$=3\cos \theta +i\sin \theta$

$\Rightarrow {\left|z+\frac{2}{z}\right|}^2=9{\cos }^2\theta +{\sin }^2\theta$

$=1+4(1+\cos 2\theta )\le 9$

$\therefore$ maximum possible value of $\left|z+\frac{2}{z}\right|\text{ is }3$ which

is attained when z = 1