A) z will be at imaginary axis
B) z will be at real axis
C) z will be at unity circle
D) None of these
Correct Answer: B
Solution :
\[w=\frac{1-iz}{z-i}\], then \[|w|\ =1\] Þ \[\left| \ \frac{1-iz}{z-i}\ \right|\ =1\] Þ \[|1-iz|\ =\ |z-i|\] Þ \[|1-i(x+iy)|\ =\ |x+iy-i|\] Þ \[|(1+y)-ix|\ =\ |x+i(y-1)|\] Þ \[\sqrt{{{x}^{2}}+1+{{y}^{2}}+2y}=\sqrt{{{x}^{2}}+{{y}^{2}}+1-2y}\]Þ \[y=0\] Hence\[z=x+iy=x\]. So z lies on real axis.
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