If log_{tan 30^{circ}} left( frac{2|z|^2 + 2| | vedclass

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(C) Given the inequality: $\log_{	an 30^{\circ}} \left( \frac{2|z|^2 + 2|z| - 3}{|z| + 1} ight) < -2$. Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$

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$|z| > 2$

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(C) Given the inequality: $\log_{	an 30^{\circ}} \left( \frac{2|z|^2 + 2|z| - 3}{|z| + 1} ight) Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$,the base is $0 When the base of a logarithm is between $0$ and $1$,the inequality sign reverses when removing the logarithm: $\frac{2|z|^2 + 2|z| - 3}{|z| + 1} > (\frac{1}{\sqrt{3}})^{-2}$. $\frac{2|z|^2 + 2|z| - 3}{|z| + 1} > 3$. $2|z|^2 + 2|z| - 3 > 3(|z| + 1)$. $2|z|^2 + 2|z| - 3 > 3|z| + 3$. $2|z|^2 - |z| - 6 > 0$. Factoring the quadratic expression: $(2|z| + 3)(|z| - 2) > 0$. Since $|z| \geq 0$,$2|z| + 3$ is always positive. Therefore,$|z| - 2 > 0$,which implies $|z| > 2$.

If $\log_{\tan 30^{\circ}} \left( \frac{2|z|^2 + 2|z| - 3}{|z| + 1} \right) < -2$,then:

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