JEE Main 2024 — 6 April, Shift 2 — Question 9

$\frac{\mathrm{z}_{1}-2 \mathrm{z}_{2}}{\frac{1}{2}-\mathrm{z}_{1} \overline{\mathrm{z}}_{2}} \times \frac{\overline{\mathrm{z}}_{1}-2 \overline{\mathrm{z}}_{2}}{\frac{1}{2}-\overline{\mathrm{z}}_{1} \mathrm{z}_{2}}=4$

$\left|z_{1}\right|^{2} 2 z_{1} \bar{z}_{2}-2 \bar{z}_{1} z_{2}+4\left|z_{2}\right|^{2}$

$=4\left(\frac{1}{4}-\frac{\bar{z}_{1} \mathrm{z}_{2}}{2}-\frac{\mathrm{z}_{1} \overline{\mathrm{z}}_{2}}{2}+\left|\mathrm{z}_{1}\right|^{2}\left|\mathrm{z}_{2}\right|^{2}\right)$

$\mathrm{z}_{1} \overline{\mathrm{z}}_{1}+2 \mathrm{z}_{2} \cdot 2 \overline{\mathrm{z}}_{2}-\mathrm{z}_{1} \overline{\mathrm{z}}_{1} 2 \mathrm{z}_{2} 2 \overline{\mathrm{z}}_{2}-1=0$ $\qquad$

$\left(\mathrm{z}, \overline{\mathrm{z}}_{1}-1\right)\left(1-2 \mathrm{z}_{2} \cdot 2 \overline{\mathrm{z}}_{2}\right)=0$

$\left(\left|z_{1}\right|^{2}-1\right)\left(\left|2 z_{2}\right|^{2}-1\right)=0$