If log _{frac{1}{sqrt{3}}}left{frac{|z|^2-|z| | vedclass

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(C) Given that,$\log _{\frac{1}{\sqrt{3}}}\left\{\frac{|z|^2-|z|+1}{2+|z|}\right\}>-2$. Since the base $a = \frac{1}{\sqrt{3}}$ satisfies $0 < a < 1$,

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a circle

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(C) Given that,$\log _{\frac{1}{\sqrt{3}}}\left\{\frac{|z|^2-|z|+1}{2+|z|}\right\}>-2$. Since the base $a = \frac{1}{\sqrt{3}}$ satisfies $0 $\frac{|z|^2-|z|+1}{2+|z|} Simplifying the right side: $\left(\frac{1}{\sqrt{3}}\right)^{-2} = (\sqrt{3})^2 = 3$. Thus,$\frac{|z|^2-|z|+1}{2+|z|} Multiplying both sides by $(2+|z|)$ (since $|z| \ge 0$,$2+|z| > 0$): $|z|^2-|z|+1 $|z|^2-|z|+1 $|z|^2-4|z|-5 Factoring the quadratic: $(|z|-5)(|z|+1) Since $|z| \ge 0$,$|z|+1$ is always positive,so we must have $|z|-5 This represents the interior of a circle with radius $5$ centered at the origin.

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

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