A) z lies on the imaginary axis
B) z lies on the real axis
C) z lies on the unit circle
D) None of these
Correct Answer: B
Solution :
lf \[z=x+iy\]and \[\omega =\frac{1-iz}{z-i}\] i.e. \[\omega =\frac{1-(ix-y)}{x+iy-i}\] \[\because \,\,|\omega |=1\Rightarrow \frac{|1-iz|}{|z-i|}=1\] \[\Rightarrow \,\,\left| 1-iz \right|=\left| z-i \right|\] \[\Rightarrow \,\,\left| 1-i(x+iy) \right|=\left| x+iy-i \right|\] \[\Rightarrow \,\,\left| 1+y-ix \right|=\left| x+i(y-1) \right|\] \[\Rightarrow \,\,\sqrt{{{(1+y)}^{2}}+{{x}^{2}}}=\sqrt{{{x}^{2}}+{{(y-1)}^{2}}}\] \[\Rightarrow \,\,1+{{y}^{2}}+2y+{{x}^{2}}={{x}^{2}}+{{y}^{2}}+1-2y\] \[\Rightarrow \,\,4y=0\Rightarrow y=0\] \[\Rightarrow \,\,z=x+i.0=x\] \[\therefore \] z lies on real axis.
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