If z ≠ 0 be a complex number such that |z - (1)/(z)| = 2 , then the...

Given $\left|z - \dfrac{1}{z}\right| = 2$

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

i.e. \left|r - \dfrac{1}{r}\right| \le 2$and </p><p>$r + \dfrac{1}{r} \ge 2$which is always true.</p><p>Now \left|r - \dfrac{1}{r}\right| \le 2$

$⇒ r - \dfrac{1}{r} \ge - 2 \text{\text{ and } } r - \dfrac{1}{r} \le 2$

$⇒ {r}^{2} - 1 \le 2r$

$⇒ {\left(r - 1\right)}^{2} \le 2$

$⇒ r - 1 \le \sqrt{2}$

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