QJEEM22Mabf99: JEE Main 2022 Mathematics - Class 11 JEEM22Mabf99 Solution: If be a complex number such that , then the maximum value of is :

We know,

$\left| {|{z_1}| - |{z_2}|} \right| \le \left| {{z_1} + {z_2}} \right| \le |{z_1}| + |{z_2}|$

$\therefore$ $\left| {|z| - {1 \over {|z|}}} \right| \le \left| {z - {1 \over z}} \right|$

$\Rightarrow \left| {|z| - {1 \over {|z|}}} \right| \le 2$ [Given $\left| {z - {1 \over z}} \right| = 2$ ]

$\Rightarrow \left| {{{|z{|^2} - 1} \over {|z|}}} \right| \le 2$

$\Rightarrow - 2 \le {{|z{|^2} - 1} \over {|z|}} \le 2$

$\therefore$ ${{|z{|^2} - 1} \over {|z|}} \le 2$

$\Rightarrow |z{|^2} - 1 \le 2|z|$

$\Rightarrow |z{|^2} - 2|z| - 1 \le 0$

$\Rightarrow |z{|^2} - 2|z| + 1 - 2 \le 0$

$\Rightarrow {(|z| - 1)^2} - 2 \le 0$

$\Rightarrow - \sqrt 2 \le |z| - 1 \le \sqrt 2$

$\Rightarrow 1 - \sqrt 2 \le |z| \le 1 + \sqrt 2$ ..... (1)

or

$- 2 \le {{|z{|^2} - 1} \over {|z|}}$

$\Rightarrow |z{|^2} - 1 \le - 2|z|$

$\Rightarrow |z{|^2} + 2|z| - 1 \le 0$

$\Rightarrow |z{|^2} + 2|z| + 1 - 2 \le 0$

$\Rightarrow {(|z| + 1)^2} - 2 \le 0$

$\Rightarrow - \sqrt 2 \le |z| + 1 \le + \,\sqrt 2$

$\Rightarrow - \sqrt 2 - 1 \le |z| \le \sqrt 2 - 1$ ...... (2)

From (1) and (2) we get,

Maximum value of $|z| = \sqrt 2 + 1$ and minimum value of $|z| = - \sqrt 2 - 1$