• # question_answer If ${{z}_{1}}=a+ib$ and ${{z}_{2}}=c+id$ are complex numbers such that  $|{{z}_{1}}|\,=\,|{{z}_{2}}|=1$ and $R({{z}_{1}}\overline{{{z}_{2}}})=0,$ then the pair of complex numbers ${{w}_{1}}=a+ic$ and ${{w}_{2}}=b+id$ satisfies [IIT 1985] A) $|{{w}_{1}}|=1$ B) $|{{w}_{2}}|=1$ C) $R({{w}_{1}}\overline{{{w}_{2}}})=0,$ D) All the above

Since$|{{z}_{1}}|\,=\,|{{z}_{2}}|\,=1$, we have ${{z}_{1}}=\cos {{\theta }_{1}}+i\sin {{\theta }_{1}},{{z}_{2}}=\cos {{\theta }_{2}}+i\sin {{\theta }_{2}}$ where ${{\theta }_{1}}=arg({{z}_{1}})$ and ${{\theta }_{2}}=arg({{z}_{2}})$ Also, ${{z}_{1}}=a+ib$ and ${{z}_{2}}=c+id.$ Therefore$a=\cos {{\theta }_{1}}$,$b=\sin {{\theta }_{1}},c=\cos {{\theta }_{2}}$ and $d=\sin {{\theta }_{2}}$ Also, $R({{z}_{1}}{{\overline{z}}_{2}})=0$ Þ $R[(\cos {{\theta }_{1}}+i\sin {{\theta }_{1}})(\cos {{\theta }_{2}}-i\sin {{\theta }_{2}})]=0$ Þ $R[(\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})]=0$ Þ $\cos ({{\theta }_{1}}-{{\theta }_{2}})=0$Þ${{\theta }_{1}}-{{\theta }_{2}}=\frac{\pi }{2}$Þ ${{\theta }_{1}}={{\theta }_{2}}+\frac{\pi }{2}$ Now, ${{w}_{1}}=a+ic=\cos {{\theta }_{1}}+i\cos {{\theta }_{2}}$$=\cos {{\theta }_{1}}+i\sin {{\theta }_{1}}$ Þ $|{{w}_{1}}|\,=1$ Similarly, $|{{w}_{2}}|\,=1$ Next ${{w}_{1}}{{\overline{w}}_{2}}=(\cos {{\theta }_{1}}+i\sin {{\theta }_{1}})(\cos {{\theta }_{2}}-i\sin {{\theta }_{2}})$                   $=\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})$Þ $|{{w}_{1}}{{\overline{w}}_{2}}|\,=1$ Finally, $R({{\overline{w}}_{1}}{{w}_{2}})=R({{w}_{2}}{{\overline{w}}_{1}})$ $=R[(\cos {{\theta }_{2}}+i\sin {{\theta }_{2}})(\cos {{\theta }_{1}}-i\sin {{\theta }_{1}})]$ $=R[\cos ({{\theta }_{2}}-{{\theta }_{1}})+i\sin ({{\theta }_{2}}-{{\theta }_{1}})]$ $=\cos ({{\theta }_{2}}-{{\theta }_{1}})=\cos \left( \frac{-\pi }{2} \right)=0$