Question 430 of 455MediumMCQ
If \[z \neq 0\] be a complex number such that \[\left|z-\frac{1}{z}\right|=2\], then the maximum value of ∣z∣ is :
Solutions · 1
Official solution
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 \[|z| = - \sqrt 2 - 1\]
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