A) \[\frac{1}{|z-1|}\]
B) \[\frac{1}{|z+1|}\]
C) \[\left| \frac{z}{z-1} \right|\]
D) \[0\]
Correct Answer: D
Solution :
Let \[z=x+iy\] \[\therefore \] \[w=\frac{x+iy+1}{x+iy-1}=\frac{(x+1)+iy}{(x-1)+iy}\times \frac{(x-1)-iy}{(x-1)-iy}\] \[=\frac{{{x}^{2}}-{{1}^{2}}+{{y}^{2}}+i(xy-y-xy-y)}{{{(x-1)}^{2}}+{{y}^{2}}}\] \[=\frac{{{x}^{2}}+{{y}^{2}}-1-2yi}{{{(x-1)}^{2}}+{{y}^{2}}}\] \[(\because \,\,|z|=1\,\,\,\Rightarrow \,\,\,{{x}^{2}}+{{y}^{2}}=1)\] \[=\frac{-2yi}{{{(x-1)}^{2}}+{{y}^{2}}}\] \[\therefore \] \[\operatorname{Re}\,(w)=0\]
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