Question

lf ${\log }_{\sqrt{3}}\frac{|z^2|-|z|+1}{2+|z|}<2$, then locus of $z$ is

• A. $|z|<1$
• B. $0<|z|<5$
• C. $|z|>1$
• D. $|z|>5$

Answer 1

The correct option is B $0<|z|<5$

Given, ${\log }_{\sqrt{3}}\frac{|z^2|-|z|+1}{2+|z|}<2$

Therefore, $\frac{|z{|}^2-|z|+1}{2+|z|}<{\sqrt{3}}^2$
$\Rightarrow |z^2|-|z|+1<6+3|z|$

$|z{|}^2-4|z|-5<0$

$-1<|z|<5$

But $|z|<0$ is not possible.