Question

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) $\operatorname{Im}\left( {{w}_{1}},{{w}_{2}} \right)=0$

Answer 1

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, $\operatorname{Re}\left( {{z}_{1}}.{{z}_{2}} \right)=2\cos A\cos B-2\sin A\sin B$
$0=2\left[ \cos \left( A+B \right) \right]$

& \Rightarrow \cos \left( A+B \right)=0 \\\ & \Rightarrow A+B=\dfrac{\pi }{2} \\\ & \Rightarrow B=\dfrac{\pi }{2}-A \\\ \end{aligned}$$ And $${{z}_{2}}=2\left( \cos \left( \dfrac{\pi }{2}-A \right) \right)+i\sin \left( \dfrac{\pi }{2}-A \right)$$ $$\begin{aligned} & {{z}_{2}}=2\left( \sin A+i\cos A \right) \\\ & {{z}_{1}}=\cos A+i\sin A \\\ \end{aligned}$$ So, we can write $${{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{aligned} & \left| {{w}_{1}} \right|=\sqrt{{{\cos }^{2}}A+{{\sin }^{2}}A} \\\ & \left| {{w}_{1}} \right|=1 \\\ \end{aligned}$$ And similarly, we can write $$\begin{aligned} & {{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{aligned}$$ Therefore, we can write $$\begin{aligned} & {{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{aligned}$$ $$\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}$$.