If $ z $ and $ \omega $ are two complex numbers such that $ \left| z\omega \right|=1 $ and \[\arg \left( z \right)-\arg \left( \omega \right)=\dfrac{\pi }{2}\], then
(a) $ \overline{z}\omega =i $
(b) $ z\overline{\omega }=i $
(c) $ z\overline{\omega }=\dfrac{1-i}{\sqrt{2}} $
(d) $ z\overline{\omega }=\dfrac{-1+i}{\sqrt{2}} $ .

Hint: We start solving the problem by assuming the arguments for the complex numbers $ z $ and $ \omega $ . We then write these complex numbers in polar form by recalling the fact that the polar form of a complex number $ {{z}_{1}} $ is defined as $ {{z}_{1}}=\left| {{z}_{1}} \right|\left( \cos \alpha +i\sin \alpha \right) $ , where $ \arg \left( {{z}_{1}} \right)=\alpha $ and the complex conjugate of complex number $ {{z}_{1}} $ is $ \overline{{{z}_{1}}}=\left| {{z}_{1}} \right|\left( \cos \alpha -i\sin \alpha \right) $ . We then consider $ \overline{z}\omega $ and substitute the polar forms in it. We then make use of the facts $ \sin A\cos B-\sin B\cos A=\sin \left( A-B \right) $ , $ \cos A\cos B+\sin A\sin B=\cos \left( A-B \right) $ and make the necessary calculations to get the value of $ \overline{z}\omega $ . We then apply conjugate on both sides to get the value of $ z\overline{\omega } $ which will give us the required answer.

Complete step by step answer:
According to the problem, we are given that $ z $ and $ \omega $ are two complex numbers such that $ \left| z\omega \right|=1 $ and \[\arg \left( z \right)-\arg \left( \omega \right)=\dfrac{\pi }{2}\]. We need to find which of the given options are true.
Let us assume the \[\arg \left( z \right)={{\theta }_{1}}\] and \[\arg \left( \omega \right)={{\theta }_{2}}\].
We know that the polar form of a complex number $ {{z}_{1}} $ is defined as $ {{z}_{1}}=\left| {{z}_{1}} \right|\left( \cos \alpha +i\sin \alpha \right) $ , where $ \arg \left( {{z}_{1}} \right)=\alpha $ . We know that complex conjugate of complex number $ {{z}_{1}} $ is $ \overline{{{z}_{1}}}=\left| {{z}_{1}} \right|\left( \cos \alpha -i\sin \alpha \right) $ .
So, we get polar forms of $ z $ and $ \omega $ as \[z=\left| z \right|\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)\], \[\omega =\left| \omega \right|\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)\].
Now, we have given \[\arg \left( z \right)-\arg \left( \omega \right)=\dfrac{\pi }{2}\]. So, we get $ {{\theta }_{1}}-{{\theta }_{2}}=\dfrac{\pi }{2} $ ---(1).
Now, let us consider $ \overline{z}\omega $ .
 $ \Rightarrow \overline{z}\omega =z=\left| z \right|\left( \cos {{\theta }_{1}}-i\sin {{\theta }_{1}} \right)\left| \omega \right|\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right) $ .
 $ \Rightarrow \overline{z}\omega =z=\left| z \right|\left| \omega \right|\left( \cos {{\theta }_{1}}\cos {{\theta }_{2}}+i\cos {{\theta }_{1}}\sin {{\theta }_{2}}-i\sin {{\theta }_{1}}\cos {{\theta }_{2}}-{{i}^{2}}\sin {{\theta }_{1}}\sin {{\theta }_{2}} \right) $ .
We know that in a complex number $ \left| {{z}_{1}} \right|\left| {{z}_{2}} \right|=\left| {{z}_{1}}{{z}_{2}} \right| $ and $ {{i}^{2}}=-1 $ .
 $ \Rightarrow \overline{z}\omega =z=\left| z\omega \right|\left( \cos {{\theta }_{1}}\cos {{\theta }_{2}}+i\left( \cos {{\theta }_{1}}\sin {{\theta }_{2}}-\sin {{\theta }_{1}}\cos {{\theta }_{2}} \right)-\left( -1 \right)\sin {{\theta }_{1}}\sin {{\theta }_{2}} \right) $ .
From the problem, we have $ \left| z\omega \right|=1 $ .
 $ \Rightarrow \overline{z}\omega =z=\left( 1 \right)\left( \left( \cos {{\theta }_{1}}\cos {{\theta }_{2}}+\sin {{\theta }_{1}}\sin {{\theta }_{2}} \right)+i\left( \cos {{\theta }_{1}}\sin {{\theta }_{2}}-\sin {{\theta }_{1}}\cos {{\theta }_{2}} \right) \right) $ .
We know that $ \sin A\cos B-\sin B\cos A=\sin \left( A-B \right) $ and $ \cos A\cos B+\sin A\sin B=\cos \left( A-B \right) $ .
 $ \Rightarrow \overline{z}\omega =z=\left( \left( \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right)+i\left( \sin \left( {{\theta }_{2}}-{{\theta }_{1}} \right) \right) \right) $ .
From equation (1), we get $ \overline{z}\omega =\left( \left( \cos \left( \dfrac{\pi }{2} \right) \right)+i\left( \sin \left( \dfrac{-\pi }{2} \right) \right) \right) $ .
 $ \Rightarrow \overline{z}\omega =\left( 0+i\left( -1 \right) \right) $ .
 $ \Rightarrow \overline{z}\omega =-i $ .
 $ \Rightarrow z\overline{\omega }=\overline{\left( \overline{z}\omega \right)}=\overline{\left( -i \right)} $ .
 $ \Rightarrow z\overline{\omega }=i $ .
So, we have found the complex numbers $ \overline{z}\omega =-i $ and $ z\overline{\omega }=i $ .
 $\therefore$ The correct option for the given problem is (b).

Note:
We can see that the given problem contains a huge amount of calculation, so we need to perform each step carefully in order to avoid confusion and calculation mistakes. We can also solve by taking Euler’s formula for the given complex numbers to solve this problem. Similarly, we can expect problems to find the maximum possible value of $ \arg \left( z \right)+\arg \left( \omega \right) $ .