JEE Main 2025 — 28 January, Morning Shift — Question 18

$z_{1}=\sqrt{3}+2 \sqrt{2} i \& \frac{\left|z_{2}\right|}{\left|z_{1}\right|}=\frac{1}{\sqrt{3}}$

given $\arg \left(\frac{z_{2}}{z_{1}}\right)=\frac{\pi}{6}$

$\mathrm{z}_{2}=\frac{\left|\mathrm{z}_{2}\right|}{\left|\mathrm{z}_{1}\right|} \cdot \mathrm{z}_{1} \mathrm{e}^{\mathrm{i}\left(\frac{\pi}{6}\right)}$

$\mathrm{z}_{2}=\frac{1}{\sqrt{3}} \cdot \frac{(\sqrt{3}+2 \sqrt{2} \mathrm{i})(\sqrt{3}+\mathrm{i})}{2}$

$z_{2}=\frac{(3-2 \sqrt{2})+i(2 \sqrt{6}+\sqrt{3})}{2 \sqrt{3}}$

Now,

$\mathrm{z}_{1}-\mathrm{z}_{2}=\frac{(3+2 \sqrt{2})+\mathrm{i}(2 \sqrt{6}-\sqrt{3})}{2 \sqrt{3}}$

$\left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|=\left|\mathrm{z}_{2}\right| \Rightarrow \triangle \mathrm{ABO}$ is isosceles with angles

$\frac{\pi}{6}, \frac{\pi}{6} \& \frac{2 \pi}{3}$