Question

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

• A. a circle
• B. a straight line
• C. interior of the circle
• D. ellipse

Answer 1

The correct option is C interior of the circle

$\left|\frac{|z{|}^2-|z|+1}{|z|+2}\right|<(\sqrt{3}{)}^2=3$
As the denominator on LHS is positive we can multiply by it on both the sides.

$\Rightarrow ||z{|}^2-|z|+1|<3|z|+6$
Now, $|z{|}^2-|z|+1>0$ as the discriminant is negative and the coefficient of $|z{|}^2$ is positive.
$\Rightarrow |z{|}^2-|z|+1<3|z|+6$
$\Rightarrow |z{|}^2-4|z|-5<0$
$\Rightarrow (|z|-5)(|z|+1)<0$
By definition $|z|\ge 0$

Hence, $|z|<5$.

This represents the interior of a circle.