Question

If ${\log }_{\sqrt{3}}\left(\frac{{\left|z\right|}^2-\left|z\right|+1}{2+\left|z\right|}\right)<2$, then the locus of $z$ is

• A. $\left|z\right|<5$
• B. $\left|z\right|=5$
• C. $\left|z\right|>5$
• D. None of these

Answer 1

Answer: (A)

$\left|z\right|<5$

${\log }_{\sqrt{3}}\left(\frac{{\left|z\right|}^2-\left|z\right|+1}{2+\left|z\right|}\right)<2$

$\Rightarrow \frac{{\left|z\right|}^2-\left|z\right|+1}{2+\left|z\right|}<{\left(\sqrt{3}\right)}^2$

$\Rightarrow {\left|z\right|}^2-\left|z\right|+1<6+3\left|z\right|$

$\Rightarrow {\left|z\right|}^2-4\left|z\right|-5<0$

$\Rightarrow (\left|z\right|-5)(\left|z\right|+1)<0$

$\Rightarrow \left|z\right|-5<0,$

since $\left|z\right|+1>0\Rightarrow \left|z\right|<5.$