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 $$|z| = - \sqrt 2 - 1$$