A) Purely real
B) Purely imaginary
C) Zero
D) Undefined
Correct Answer: B
Solution :
\[z=x+iy\Rightarrow |z{{|}^{2}}={{x}^{2}}+{{y}^{2}}=1\] .....(i) Now, \[\left( \frac{z-1}{z+1} \right)=\frac{(x-1)+iy}{(x+1)+iy}\times \frac{(x+1)-iy}{(x+1)-iy}\] \[=\frac{({{x}^{2}}+{{y}^{2}}-1)+2iy}{{{(x+1)}^{2}}+{{y}^{2}}}\]\[=\frac{2iy}{{{(x+1)}^{2}}+{{y}^{2}}}\] [by equation (i)] Hence, \[\left( \frac{z-1}{z+1} \right)\]is purely imaginary.
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