If \[{{z}_{1}}={{a}_{1}}+i{{b}_{1}}\] and \[{{z}_{2}}={{a}_{2}}+i{{b}_{2}}\] are complex numbers such that \[\left| {{z}_{1}} \right|=1,\left| {{z}_{2}} \right|=2\] and \[\operatorname{Re}\left( {{z}_{1}},{{z}_{2}} \right)=0\], then the pair of complex numbers \[{{w}_{1}}={{a}_{1}}+\dfrac{i{{a}_{2}}}{2}\] and \[{{w}_{2}}=2{{b}_{1}}+i{{b}_{2}}\] satisfy
(a) \[\left| {{w}_{1}} \right|=1\]
(b) \[\left| {{w}_{2}} \right|=2\]
(c) \[\operatorname{Re}\left( {{w}_{1}},{{w}_{2}} \right)=0\]
(d) \[\operatorname{Im}\left( {{w}_{1}},{{w}_{2}} \right)=0\]

Hint: We will first assume \[{{z}_{1}}=\cos {{\theta }_{1}}+\sin {{\theta }_{1}}\] and \[{{z}_{2}}=\cos {{\theta }_{2}}+\sin {{\theta }_{2}}\] where \[{{\theta }_{1}}\] is the argument of \[{{z}_{1}}\] and \[{{\theta }_{2}}\] is the argument of \[{{z}_{2}}\]. After that put these values in given condition and find the modulus and then finally find the relation between \[{{\theta }_{1}}\] and \[{{\theta }_{2}}\]. The modulus of a complex number \[z=x+iy\] is given by –
\[\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\]

Complete step by step answer:
Let the complex number \[{{z}_{1}}\] be \[{{z}_{1}}={{a}_{1}}+i{{b}_{1}}\] and complex number \[{{z}_{2}}\] be \[{{z}_{2}}={{a}_{2}}+i{{b}_{2}}\].
Since, it is given in question that \[\left| {{z}_{1}} \right|=1,\left| {{z}_{2}} \right|=2\] and \[\operatorname{Re}\left( {{z}_{1}},{{z}_{2}} \right)=0\].
So, let us consider \[{{z}_{1}}={{a}_{1}}+i{{b}_{1}}\]
\[\left| {{z}_{1}} \right|=1\Rightarrow {{a}_{1}}=r\cos A,{{b}_{1}}=r\sin A\]
\[\Rightarrow r=\left| {{z}_{1}} \right|=1\]
So, we can write it as,
\[{{z}_{1}}=\cos A+i\sin A\]
And \[{{z}_{2}}={{a}_{2}}+i{{b}_{2}}\]
\[\left| {{z}_{2}} \right|=2\Rightarrow {{a}_{2}}=r\cos B,{{b}_{2}}=r\sin B\]
\[\Rightarrow r=\left| {{z}_{2}} \right|=2\]
So, we can write it as
\[{{z}_{2}}=2\cos B+2i\sin B\]
\[=2\left( \cos B+i\sin B \right)\]
Then, \[{{z}_{1}}.{{z}_{2}}=\left( \cos A+i\sin A \right)\left( 2\cos B+2i\sin B \right)\]
So, \[\operatorname{Re}\left( {{z}_{1}}.{{z}_{2}} \right)=2\cos A\cos B-2\sin A\sin B\]
\[0=2\left[ \cos \left( A+B \right) \right]\]
\[\begin{align}
  & \Rightarrow \cos \left( A+B \right)=0 \\
 & \Rightarrow A+B=\dfrac{\pi }{2} \\
 & \Rightarrow B=\dfrac{\pi }{2}-A \\
\end{align}\]
And \[{{z}_{2}}=2\left( \cos \left( \dfrac{\pi }{2}-A \right) \right)+i\sin \left( \dfrac{\pi }{2}-A \right)\]
\[\begin{align}
  & {{z}_{2}}=2\left( \sin A+i\cos A \right) \\
 & {{z}_{1}}=\cos A+i\sin A \\
\end{align}\]
So, we can write \[{{w}_{1}}={{a}_{1}}+i\dfrac{{{a}_{1}}}{2}\]
\[=\cos A+\dfrac{i\left( 2\sin A \right)}{2}\]
\[{{w}_{1}}=\cos A+i\sin A\]
\[\begin{align}
  & \left| {{w}_{1}} \right|=\sqrt{{{\cos }^{2}}A+{{\sin }^{2}}A} \\
 & \left| {{w}_{1}} \right|=1 \\
\end{align}\]
And similarly, we can write
\[\begin{align}
  & {{w}_{2}}=2b+i{{b}_{2}} \\
 & {{w}_{2}}=2\sin A+i\left( 2\cos A \right) \\
 & {{w}_{2}}=2\sin A+2i\cos A \\
 & \left| {{w}_{2}} \right|=2\left| {{w}_{1}} \right| \\
 & \left| {{w}_{2}} \right|=2.1=2 \\
\end{align}\]
Therefore, we can write
\[\begin{align}
  & {{w}_{1}}.{{w}_{2}}=\left( \cos A.\left( 2\sin A \right)-2\sin A\cos A \right)+i\left( 2{{\cos }^{2}}A+2{{\sin }^{2}}A \right) \\
 & {{w}_{1}}.{{w}_{2}}=2i\left( 1 \right) \\
 & {{w}_{1}}.{{w}_{2}}=0+2i \\
 & \operatorname{Re}\left( {{w}_{1}}.{{w}_{2}} \right)=0 \\
\end{align}\]
\[\operatorname{Im}\left( {{w}_{1}}.{{w}_{2}} \right)=2\]

So, the correct answer is “Option d”.

Note: Students should not take note that the argument of the complex numbers is the angled measured from the positive real axis to the line segment. For a complex number, \[z=x+iy\]. The argument is given by \[\arg \left( z \right)={{\tan }^{-1}}\dfrac{y}{x}\].