Question

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

• A. $|z|<\frac{1}{3}$
• B. $|z|=1$
• C. $|z|=5$
• D. $1<|z|<5$

Answer 1

Answer: (A), (D)

The correct options are
A $|z|<\frac{1}{3}$
D $1<|z|<5$

$\left|lo{g}_{\sqrt{3}}\right(\frac{|z{|}^2-|z|+1}{|z|+2}\left)\right|<2$
$\Rightarrow -2<{\log }_{\sqrt{3}}\left(\frac{|z{|}^2-|z|+1}{|z|+2}\right)<2$
$\Rightarrow \frac{1}{3}<\frac{|z{|}^2-|z|+1}{|z|+2}<3$
$\Rightarrow |z|+2<3|z{|}^2-3|z|+3$
$\Rightarrow 3|z{|}^2-4|z|+1>0$
$\Rightarrow (3|z|-1)(|z|-1)>0$
Hence
$|z|<\frac{1}{3}$ and $|z|>1$ ...(i)
Similarly
$\frac{|z{|}^2-|z|+1}{|z|+2}<3$
$\Rightarrow |z{|}^2-|z|+1<3|z|+6$
$\Rightarrow |z{|}^2-4|z|-5<0$
$\Rightarrow (|z|+1)(|z|-5)<0$
$\Rightarrow 1<|z|<5$ ...(ii)

From (i) and (ii), we get

$1<|z|<5$