A) \[|z|\,=0\]
B) \[|z|\,=1\]
C) \[|z|\,>1\]
D) \[|z|\,<1\]
Correct Answer: B
Solution :
Let \[\frac{z-1}{z+1}=iy\]where \[y\in R\] This gives \[z=\frac{1+iy}{1-iy}=\frac{1+iy}{1-iy}\times \frac{1+iy}{1+iy}=\frac{(1-{{y}^{2}})+2iy}{1+{{y}^{2}}}\] \[\therefore \] \[|z|=\frac{1}{1+{{y}^{2}}}\sqrt{{{(1-{{y}^{2}})}^{2}}+4{{y}^{2}}}=\frac{1+{{y}^{2}}}{1+{{y}^{2}}}=1\].
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