A) 0
B) any non-zero real number other than 1.
C) any non-zero real number.
D) a purely imaginary number.
Correct Answer: C
Solution :
Let \[z=x+iy\] \[\frac{z-i}{z+i}\]is purely imaginary means its real part is zero. \[\frac{x+iy-i}{z+iy+i}=\frac{x+i(y-1)}{x+i(y+1)}\times \frac{x-i(y+1)}{x-i(y+1)}\] \[=\frac{{{x}^{2}}-2ix(y+1)+xi(y-1)+{{y}^{2}}-1}{{{x}^{2}}+{{(y+1)}^{2}}}\] \[=\frac{{{x}^{2}}+{{y}^{2}}-1)}{{{x}^{2}}+{{(y+1)}^{2}}}-\frac{2xi}{{{x}^{2}}+{{(y+1)}^{2}}}\] for pure imaginary, we have \[=\frac{{{x}^{2}}+{{y}^{2}}-1)}{{{x}^{2}}+{{(y+1)}^{2}}}=0\]\[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}=1\] \[\Rightarrow \]\[(x+iy)(x-iy)=1\] \[\Rightarrow \]\[x+iy=\frac{1}{x-iy}=z\]and\[\frac{1}{z}=x-iy\] \[z+\frac{1}{z}=(x+iy)+(x-iy)=2x\] \[\left( z+\frac{1}{z} \right)\]is any non-zero real number
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