Question

If ${\log }_{\frac{1}{\sqrt{3}}}\left\{\frac{|z{|}^2-|z|+1}{2+|z|}\right\}>-2$, then $z$ lies inside

• A. a triangle
• B. an ellipse
• C. a circle
• D. a square

Answer 1

Answer: (C)

We have
${\log }_{\frac{1}{\sqrt{3}}}\left\{\frac{|z{|}^2-|z|+1}{2+|z|}\right\}>-2$
$\Rightarrow \frac{|z{|}^2-|z|+1}{2+|z|}<{\left(\frac{1}{\sqrt{3}}\right)}^{-2}$
$[\because {\log }_{a}b>c,b<a^c$ if $0<a<1]$
$\Rightarrow |z{|}^2-|z|+1<(\sqrt{3}{)}^2(2+|z|)$
$\Rightarrow |z{|}^2-|z|+1\times 6+3|z|$
$\Rightarrow |z{|}^2-4|z|+1<6$
$\Rightarrow (|z|-2{)}^2<9$
$\Rightarrow -3<|z|-2<3$
$\Rightarrow -1<|z|<5$
$\Rightarrow 0<|z|<5$
Hence, $z$ lies inside the circle.