A) \[x=0\]
B) \[y=0\]
C) \[x=1\]
D) \[y=1\]
E) none of these
Correct Answer: B
Solution :
\[\because \]\[\left| \frac{z-i}{z+i} \right|=1\] Let \[z=x+iy\] \[\therefore \] \[\left| \frac{x+iy-i}{x+iy+i} \right|=1\] \[\Rightarrow \] \[\left| \frac{x+i(y-1)}{x+i(y+1)} \right|=1\] \[\Rightarrow \] \[{{x}^{2}}+{{(y-1)}^{2}}={{x}^{2}}+{{(y+1)}^{2}}\] \[\Rightarrow \] \[-2y=2y\]\[\Rightarrow \]\[4y=0\] Thus, the locus of z is\[y=0\].
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