A) \[0\]
B) \[1\]
C) \[\sqrt{2}\]
D) None of these
Correct Answer: B
Solution :
Key Idea If \[z\] is purely imaginary, then real coefficient will be zero. Let\[z=x+iy\] \[\therefore \] \[\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}\] \[=\frac{(x-1)+iy}{(x+1)+iy}\times \frac{(x+1)+iy}{(x+1)-iy}\] \[=\frac{(x-1)(x+1)-iy(x-1)+iy(x+1)-{{i}^{2}}{{y}^{2}}}{{{(x+1)}^{2}}-{{i}^{2}}{{y}^{2}}}\] \[=\frac{{{x}^{2}}-1-iyx+iy+iyx+iy+{{y}^{2}}}{{{(x+1)}^{2}}+{{y}^{2}}}\] \[=\frac{{{x}^{2}}-1+{{y}^{2}}+2iy}{{{(x+1)}^{2}}+{{y}^{2}}}\] \[=\frac{{{x}^{2}}-1+{{y}^{2}}}{{{(x+1)}^{2}}+{{y}^{2}}}+\frac{2y}{{{(x+1)}^{2}}+{{y}^{2}}}i\] Since, it is purely imaginary, then \[\frac{{{x}^{2}}-1+{{y}^{2}}}{{{(x+1)}^{2}}+{{y}^{2}}}=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-1=0\] \[\Rightarrow \] \[|z|=1\] \[(\because |z|={{x}^{2}}+{{y}^{2}})\]
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