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, \(\left| \frac{z_1 + \bar {z_2}}{\bar {z_1} + z_2}\right| = 1\)



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