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

<p>$$z_1=\sqrt{3}+2 \sqrt{2} i \quad \& \frac{\left|z_2\right|}{\left|z_1\right|}=\frac{1}{\sqrt{3}}$$</p> <p>given $\arg \left(\frac{z_2}{z_1}\right)=\frac{\pi}{6}$</p> <p>$$\begin{aligned} & z_2=\frac{\left|z_2\right|}{\left|z_1\right|} \cdot z_1 \mathrm{e}^{i\left(\frac{\pi}{6}\right)} \\ & z_2=\frac{1}{\sqrt{3}} \cdot \frac{(\sqrt{3}+2 \sqrt{2} i)(\sqrt{3}+i)}{2} \\ & z_2=\frac{(3-2 \sqrt{2})+i(2 \sqrt{6}+\sqrt{3})}{2 \sqrt{3}} \end{aligned}$$</p> <p>Now,</p> <p>$z_1-z_2=\frac{(3+2 \sqrt{2})+i(2 \sqrt{6}-\sqrt{3})}{2 \sqrt{3}}$</p> <p>$\left|z_1-z_2\right|=\left|z_2\right| \Rightarrow \Delta \mathrm{ABO}$ is isosceles with angles $\frac{\pi}{6}, \frac{\pi}{6} \& \frac{2 \pi}{3}$</p>