A) \[\{0\}\]
B) an empty set
C) \[\left\{ 0,\frac{1}{4},-\frac{1}{4} \right\}\]
D) equal to R
Correct Answer: A
Solution :
It is given that \[|z|=1\And Re\,\,z\ne \,\,1\] Lets assume \[z=x+iy\Rightarrow {{x}^{2}}+{{y}^{2}}=1\]...(1) Also, \[w=\frac{1+(1-8\alpha )z}{1-z}\] \[\Rightarrow w=\frac{1+(1-8\alpha )(x+iy)}{1-(x+iy)}\] Multiply and divide above equation by \[(1-x)+iy\], we get Solving it, we get \[\begin{align} & w=\frac{\left| (1+x(1-8\alpha ))(1-x)-(1-8x){{y}^{2}} \right|}{{{(1-x)}^{2}}+{{y}^{2}}}+ \\ & i\frac{\left[ (1+x(1-8\alpha ))y-(1-8x)y(1-x) \right]}{{{(1-x)}^{2}}+{{y}^{2}}} \\ \end{align}\] It is also given that, \[w\] is purely imaginary. Therefore, \[\operatorname{Re}w=0\] \[\operatorname{Re}w=\frac{\left[ (1+x(1-8\alpha ))(1-x)-(1-8x){{y}^{2}} \right]}{{{(1-x)}^{2}}+{{y}^{2}}}=0\] \[\Rightarrow (1-x)+x(1-8\alpha )(1-x)=(1-8x){{y}^{2}}\] \[\Rightarrow (1-x)+x(1-8\alpha )-{{x}^{2}}(1-8\alpha )=(1-8x){{y}^{2}}\] \[\Rightarrow (1-x)+x(1-8\alpha )=1-8\alpha [\because {{x}^{2}}+{{y}^{2}}=1]\] \[\Rightarrow 1-8\alpha =1\] \[\Rightarrow \alpha =0\] \[\therefore \alpha \in \{0\}\]
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