If z ≠ 0 be a complex number such that z - 1 z = 2 , then the maximum value of z is

Given $\left|z-\frac{1}{z}\right|=2$

We know that $\left|\left|z\right|-\frac{1}{\left|z\right|}\right|\le \left|z-\frac{1}{z}\right|\le \left|z\right|+\frac{1}{\left|z\right|}$ Let $\left|z\right|=r$

i.e. $\left|r-\frac{1}{r}\right|\le 2$ and 

$r+\frac{1}{r}\ge 2$ which is always true.

Now $\left|r-\frac{1}{r}\right|\le 2$

$\Rightarrow r-\frac{1}{r}\ge -2 \& r-\frac{1}{r}\le 2$

$\Rightarrow r^2-1\le 2r$

$\Rightarrow {\left(r-1\right)}^2\le 2$

$\Rightarrow r-1\le \sqrt{2}$

$\therefore {\left|z\right|}_{\max }=1+\sqrt{2}$