If log_{sqrt{3}} left( frac{|z|^2 - |z| + 1}{ | vedclass

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(B) Given the inequality: $\log_{\sqrt{3}} \left( \frac{|z|^2 - |z| + 1}{2 + |z|} \right)  1$,the inequality remains in

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$|z| < 5$

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(B) Given the inequality: $\log_{\sqrt{3}} \left( \frac{|z|^2 - |z| + 1}{2 + |z|} \right) Since the base $\sqrt{3} > 1$,the inequality remains in the same direction:$\frac{|z|^2 - |z| + 1}{2 + |z|} $\frac{|z|^2 - |z| + 1}{2 + |z|} $|z|^2 - |z| + 1 $|z|^2 - |z| + 1 $|z|^2 - 4|z| - 5 Let $t = |z|$,where $t \ge 0$. Then $t^2 - 4t - 5 $(t - 5)(t + 1) This implies $-1 Since $t = |z| \ge 0$,we have $0 \le |z| Thus,the locus of $z$ is $|z| < 5$.

If $\log_{\sqrt{3}} \left( \frac{|z|^2 - |z| + 1}{2 + |z|} \right) < 2$,then the locus of $z$ is

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