A) 0
B) 1
C) 2
D) None of these
Correct Answer: B
Solution :
Here \[\frac{z-i}{z+i}=\frac{x+i(y-1)}{x+i(y+1)}.\frac{x-i(y+1)}{x-i(y+1)}\] \[=\frac{({{x}^{2}}+{{y}^{2}}-1)+i(-2x)}{{{x}^{2}}+{{(y+1)}^{2}}}\] As \[\frac{z-i}{z+i}\] is purely imaginary, we get \[{{x}^{2}}+{{y}^{2}}-1=0\]Þ \[{{x}^{2}}+{{y}^{2}}=1\]Þ\[z\overline{z}=1\].You need to login to perform this action.
You will be redirected in
3 sec