A) 1
B) -1
C) \[i\]
D) \[-i\]
Correct Answer: D
Solution :
Let \[z={{r}_{1}}{{e}^{i\theta }}\] and \[\omega ={{r}_{2}}{{e}^{i\phi }}\] \[\Rightarrow \] \[\overline{z}={{r}_{1}}\,{{e}^{-i\,\theta }}\] Given, \[\left| z\omega \right|=1\Rightarrow \left| {{r}_{1}}{{e}^{i\theta }}.\,\,{{r}_{2}}\,{{e}^{i\,\phi }} \right|=1\] \[\Rightarrow \] \[{{r}_{1}}{{r}_{2}}=1\] ... (i) and arg \[(z)=\arg \,(\omega )=\frac{\pi }{2}\] \[\Rightarrow \] \[\theta -\phi =\frac{\pi }{2}\] ... (ii) Then, \[\overline{z}\,\omega ={{r}_{1}}{{e}^{-i\,\theta }}.\,{{r}_{2}}{{e}^{i\,\theta }}={{r}_{1}}{{r}_{2}}{{e}^{-i\,(\theta -\phi )}}\] From Eqs. (i) and (ii), we get \[\overline{z}\,\omega =1.\,{{e}^{-i\pi /2}}=\cos \frac{\pi }{2}-i\sin \frac{\pi }{2}\] \[\Rightarrow \] \[\overline{z}\,\omega =-i\]
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