| 1. |
Find \( \left|\frac{z_{1}+\overline{z_{2}}}{\overline{z_{1}}+z_{2}}\right| \) assuming \( \left|\overline{z_{1}}+z_{2}\right| \neq 0 \) |
|
Answer» \(\left| \frac{z_1 + \bar {z_2}}{\bar {z_1} + z_2}\right| = \left|\frac{(z_1 + \bar{z_2})}{(\bar{z_1} + z_2)} \times \frac{z_1 + \bar{z_2}}{z_1 + \bar{z_2}} \right|\) \(= \left|\frac{(z_1 + \bar{z_2})^2}{(\bar{z_1}+z_2)(\overline{\bar{z_1} + z_2})}\right|\) \(\)\(= \left|\frac{(z_1 + \bar{z_2})^2}{(\bar{z_1}+z_2)^2}\right|\) \((\because z\;\bar z = |z|^2)\) \(= \frac{|(z_1 + \bar{z_2})^2|}{|\bar {z_1} + z_2|^2}\) \(= \frac{|z_1 + \bar{z_2}|^2}{|\overline{z_1 + \bar{z_2}}|^2}\) \(= \frac{|z_1 + \bar{z_2}|^2}{|{z_1 + \bar{z_2}}|^2}\) \((\because |\bar z| = |z|)\) \(= 1\) Hence, \(\left| \frac{z_1 + \bar {z_2}}{\bar {z_1} + z_2}\right| = 1\) |
|