A) \[x-\]axis
B) \[y-\]axis
C) circle with unity radius
D) none of the above
Correct Answer: A
Solution :
We have\[\left| \frac{1-iz}{z-i} \right|=1\] \[\Rightarrow \] \[|1-iz|=|z-i|\] \[\Rightarrow \] \[|1-i(x+iy)|=|x+iy-i|\] \[\Rightarrow \] \[|1-ix+y|=|x+i(y-1)|\] \[\Rightarrow \] \[|{{(1+y)}^{2}}+{{x}^{2}}={{x}^{2}}+{{(y-1)}^{2}}\] \[\Rightarrow \] \[1+{{y}^{2}}+2y+{{x}^{2}}={{x}^{2}}+{{y}^{2}}+1-2y\] \[\Rightarrow \] \[4y=0\] \[\Rightarrow \] \[y=0\] \[\Rightarrow \] \[x-\]axis.
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