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બે સંકર સંખ્યા ${z_1},{z_2}$ માટે, $|{z_1} + {z_2}{|^2} = $ $|{z_1}{|^2} + |{z_2}{|^2}$ તો
${\mathop{\rm Re}\nolimits} \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = 0$
${\mathop{\rm Im}\nolimits} \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = 0$
${\mathop{\rm Re}\nolimits} ({z_1}{z_2}) = 0$
${\mathop{\rm Im}\nolimits} ({z_1}{z_2}) = 0$
Solution
(a) We have $|{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2}$
==> $|{z_1}{|^2} + |{z_2}{|^2} + 2|{z_1}||{z_2}|\cos ({\theta _1} – {\theta _2})$
$ = |{z_1}{|^2} + |{z_2}{|^2}$
Where ${\theta _1} = arg({z_1}),{\theta _2} = arg({z_2})$
==> $\cos ({\theta _1} – {\theta _2}) = 0\,\,\,\, \Rightarrow {\theta _1} – {\theta _2} = \frac{\pi }{2}$
==> $arg\left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \frac{\pi }{2} \Rightarrow {\mathop{\rm Re}\nolimits} \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \frac{{|{z_1}|}}{{|{z_2}|}}\cos \left( {\frac{\pi }{2}} \right) = 0$
Note : Also ${\mathop{\rm Re}\nolimits} \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = 0 \Rightarrow {\mathop{\rm Re}\nolimits} ({z_1}\overline {{z_2}} ) = 0$
==> ${z_1}\overline {{z_2}} $ is purely imaginary.
