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