(a) हम जानते हैं कि $|{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2}$

$⇒ |{z_1}{|^2} + |{z_2}{|^2} + 2|{z_1}||{z_2}|\cos ({\theta _1} – {\theta _2}) = |{z_1}{|^2} + |{z_2}{|^2}$ 

जहाँ ${\theta _1} = arg({z_1}),{\theta _2} = arg({z_2})$

$ ⇒ \cos ({\theta _1} – {\theta _2}) = 0\,\,\,\, \Rightarrow {\theta _1} – {\theta _2} = \frac{\pi }{2}$

$⇒ arg\left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \frac{\pi }{2} \Rightarrow {\mathop{\rm Re}\nolimits} \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \frac{{|{z_1}|}}{{|{z_2}|}}\cos \left( {\frac{\pi }{2}} \right) = 0$

नोट : हमें ज्ञात है कि ${\mathop{\rm Re}\nolimits} \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = 0 \Rightarrow {\mathop{\rm Re}\nolimits} ({z_1}\overline {{z_2}} ) = 0$

$⇒ {z_1}\overline {{z_2}} $ पूर्णत: काल्पनिक हैं।