A) 1 or i
B) i or \[-i\]
C) 1 or \[-i\]
D) i or \[-1\]
Correct Answer: C
Solution :
We have \[2=\left| z+i\omega \right| \le \left| \,z\, \right|+\left| \omega \right|\] ... (i) \[\therefore \,\,\,~\,\,\,\left| \,z\, \right|+\,\,\left| \omega \right|\ge 2\] But given that \[\left| z \right| \le 1 and \left| \omega \right| \le 1\] ... (ii) \[\Rightarrow \,\,\,\,\left| \,z\, \right|+\left| \omega \right|\le 2\] From (1) and (2) \[\left| \,z\, \right|=\left| \omega \right|=1\] Also \[|z+i\omega |=|z-i\,\overline{\omega }|\,\,\Rightarrow \,\,\,{{\left| z+i\,\omega \right|}^{2}}{{\left| z-\,\,i\overline{\omega } \right|}^{2}}\] \[\Rightarrow \,\,\,\left( z+i\,\omega \right) \left( \bar{z}\,-i\,\bar{\omega } \right)=\,\,\left( \overline{z} +\,\,i \omega \right)\,\left( z\,-i\,\overline{\omega } \right)\] \[\Rightarrow \,\,\,\operatorname{z}\bar{z}+i\omega \bar{z}\,\,-\,\,iz\bar{\omega }+\omega \bar{\omega }=z\bar{z}-i\,\bar{z}\bar{\omega }+i\omega z+\omega \bar{\omega }\] \[\Rightarrow \,\,\omega \bar{z}-\bar{\omega }z\,+\bar{\omega }\bar{z}\,-\omega z=0\] \[=\,\,(\omega +\bar{\omega })\,(\bar{z}\,-z)=0\] \[\Rightarrow \,\,z=\bar{z}\,\,or\,\,\omega =-\bar{\omega }\,\,\Rightarrow \,\,{{\operatorname{I}}_{m}}(z)=0\Rightarrow Re\left( \omega \right)=0\] Also \[\left| z \right|=1,\,\,\left| \omega \right|=1\,\,\Rightarrow \,\,z=1\,\,or\,\,-1\,\,and\,\,\omega =i\] or \[-\,i\]
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