Lời giải:Ta có \(\left( 1+i \right).z.\left| z \right|-1=\left( i-2 \right)\left| z \right| \Leftrightarrow \left( 1+i \right).z.\left| z \right|=1-2\left| z \right|+i.\left| z \right|\)
\(\Rightarrow \left| \left( 1+i \right).z.\left| z \right| \right|=\left| 1-2\left| z \right|+i.\left| z \right| \right| \Leftrightarrow \sqrt{2}.{{\left| z \right|}^{2}}=\sqrt{{{\left( 1-2\left| z \right| \right)}^{2}}+{{\left| z \right|}^{2}}}\)
\(\Leftrightarrow 2{{\left| z \right|}^{4}}=5{{\left| z \right|}^{2}}-4\left| z \right|+1 \Leftrightarrow \left( \left| z \right|-1 \right)\left( 2{{\left| z \right|}^{3}}+2{{\left| z \right|}^{2}}-3\left| z \right|+1 \right)=0\).
Do \(\left| z \right|\) là một số nguyên nên suy ra \(\left| z \right|=1\).
