Question

If $Z_1$, $Z_2$ are two complex numbers satisfying $\left|\frac{Z_1-3Z_2}{3-Z_1\overline{Z_2}}\right|=1,|Z_1|\ne 3$, then $|Z_2|is$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

1

$\left|\frac{Z_1-3Z_2}{3-Z_1\overline{Z_2}}\right|=1$

$\Rightarrow |Z_1-3Z_2|=|3-Z_1\overline{Z_2}|\Rightarrow {|Z_1-3Z_2|}^2={|3-Z_1\overline{Z_2}|}^2$
$\Rightarrow (Z_1-3Z_2)(\overline{Z_1}-3\overline{Z_2})=(3-Z_1\overline{Z_2})(3-\overline{Z_1}{Z}_2)$

$\Rightarrow {\left|Z_1\right|}^2-3Z_1\overline{Z_2}-3Z_2\overline{Z_1}+9{\left|Z_2\right|}^2=9-3\overline{Z_1}{Z}_2-3Z_1\overline{Z_2}+{\left|Z_1\right|}^2{\left|Z_2\right|}^2$

$\Rightarrow {\left|Z_1\right|}^2-{\left|Z_1\right|}^2{\left|Z_2\right|}^2-9+9{\left|Z_2\right|}^2=0$

$\Rightarrow {\left|Z_1\right|}^2(1-{\left|Z_2\right|}^2)-9(1-{\left|Z_2\right|}^2)=0$

$\Rightarrow (1-{\left|Z_2\right|}^2)({\left|Z_1\right|}^2-9)=0$
$\Rightarrow {\left|Z_2\right|}^{}=1(\because \left|Z_1\right|\ne 3given)$