A) 1
B) \[\sqrt{2}\]
C) \[\frac{1}{2}\]
D) 2
Correct Answer: D
Solution :
Let \[z=x+iy\] \[\therefore \]\[|z|=\sqrt{{{x}^{2}}+{{y}^{2}}}=2\Rightarrow {{x}^{2}}+{{y}^{2}}=4\] Now,\[\frac{z-\alpha }{z+\alpha }=\frac{x+iy-\alpha }{x+iy+\alpha }=\frac{(x-\alpha )+iy}{(x+\alpha )+iy}\times \frac{(x+\alpha )-iy}{(x+\alpha )-iy}\] \[=\frac{({{x}^{2}}+{{y}^{2}}-{{\alpha }^{2}})}{{{(x+\alpha )}^{2}}+{{y}^{2}}}\times \frac{i2\alpha y}{{{(x+\alpha )}^{2}}+{{y}^{2}}}\] \[\Rightarrow \]\[\frac{{{x}^{2}}+{{y}^{2}}-{{\alpha }^{2}}}{{{(x+\alpha )}^{2}}+{{y}^{2}}}=0\] \[\left[ \because \frac{z-\alpha }{z+\alpha }is\text{ }purely\text{ }imaginary \right]\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-{{\alpha }^{2}}=0\]\[\Rightarrow \]\[{{\alpha }^{2}}=4\]\[\Rightarrow \]\[\alpha =\pm 2\]
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