Explanation

(b)
$\log_{\sqrt{3}}{\left( \frac{\left| \mathcal{z} \right|^{2} - \left| \mathcal{z} \right| + 1}{2 + \left| \mathcal{z} \right|} \right) < 2}$
$\Rightarrow \ \frac{\left| \mathcal{z} \right|^{2} - \left| \mathcal{z} \right| + 1}{2 + \left| \mathcal{z} \right|} < \left( \sqrt{3} \right)^{2}$
$\mathcal{\Rightarrow \ \ |\ z}\left. \ \right|^{2} - \left| \mathcal{z} \right| + 1 < 3\left( 2 + \left| \mathcal{z} \right| \right)$
$\Rightarrow \ \ \left| \mathcal{z} \right|^{2} - 4\left| \mathcal{z} \right| - 5 < 0$
$\Rightarrow \ \ (\left| \mathcal{z} \right| + 1)\left( \left| \mathcal{z} \right| - 5 \right) < 0$
$\Rightarrow \ - 1 < \left| \mathcal{z} \right| < 5 \Rightarrow \left| \mathcal{z} \right| < 5\ as\ \left| \mathcal{z} \right| > 0$
$\therefore\$ Locus of $\mathcal{z}$ is |$\mathcal{\ z}| < 5$