Let O be the origin,the point A be z_1 = sqrt | vedclass

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(D) Given $z_1 = \sqrt{3} + 2\sqrt{2}i$. The modulus is $|z_1| = \sqrt{(\sqrt{3})^2 + (2\sqrt{2})^2} = \sqrt{3 + 8} = \sqrt{11}$.Given $\sqrt{3}|z_2|

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$ABO$ is an obtuse angled isosceles triangle

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(D) Given $z_1 = \sqrt{3} + 2\sqrt{2}i$. The modulus is $|z_1| = \sqrt{(\sqrt{3})^2 + (2\sqrt{2})^2} = \sqrt{3 + 8} = \sqrt{11}$.Given $\sqrt{3}|z_2| = |z_1|$,so $|z_2| = \frac{|z_1|}{\sqrt{3}} = \sqrt{\frac{11}{3}}$.Given $\arg(z_2) - \arg(z_1) = \frac{\pi}{6}$,the angle $\angle AOB = \frac{\pi}{6}$.The area of $	riangle ABO = \frac{1}{2} |z_1| |z_2| \sin(\angle AOB) = \frac{1}{2} \cdot \sqrt{11} \cdot \sqrt{\frac{11}{3}} \cdot \sin(\frac{\pi}{6}) = \frac{1}{2} \cdot \frac{11}{\sqrt{3}} \cdot \frac{1}{2} = \frac{11}{4\sqrt{3}}$.Using the Law of Cosines in $	riangle ABO$,$AB^2 = |z_1|^2 + |z_2|^2 - 2|z_1||z_2| \cos(\frac{\pi}{6}) = 11 + \frac{11}{3} - 2 \cdot \sqrt{11} \cdot \sqrt{\frac{11}{3}} \cdot \frac{\sqrt{3}}{2} = \frac{44}{3} - 11 = \frac{11}{3}$.Since $|z_2|^2 = \frac{11}{3}$ and $AB^2 = \frac{11}{3}$,we have $|z_2| = AB$. Thus,$	riangle ABO$ is an isosceles triangle with $\angle OAB = \angle AOB = \frac{\pi}{6}$.The third angle $\angle ABO = \pi - (\frac{\pi}{6} + \frac{\pi}{6}) = \frac{2\pi}{3}$.Since $\angle ABO = \frac{2\pi}{3} > \frac{\pi}{2}$,the triangle is obtuse angled isosceles.

Let $O$ be the origin,the point $A$ be $z_1 = \sqrt{3} + 2\sqrt{2}i$,and the point $B(z_2)$ be such that $\sqrt{3}|z_2| = |z_1|$ and $\arg(z_2) = \arg(z_1) + \frac{\pi}{6}$. Then:

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