A) (-1,1)
B) [-1,1]
C) [0, 1)
D) (-1,0]
Correct Answer: A
Solution :
Let\[z=x+iy\Rightarrow {{z}^{2}}={{x}^{2}}-{{y}^{2}}+2ixy\]Now, \[\frac{1+{{z}^{2}}}{2iz}=\frac{1+{{x}^{2}}-{{y}^{2}}}{2i(x+iy)}=\frac{({{x}^{2}}-{{y}^{2}}+1)+2ixy}{2ix-2y}\] \[\frac{({{x}^{2}}-{{y}^{2}}+1)+2ixy}{-2y+2ix}\times \frac{-2y-2ix}{-2y-2ix}\] \[=\frac{y({{x}^{2}}-{{y}^{2}}-1)+x({{x}^{2}}+{{y}^{2}}+1)i}{2({{x}^{2}}+{{y}^{2}})}\] \[a=\frac{x({{x}^{2}}-{{y}^{2}}+1)}{2({{x}^{2}}+{{y}^{2}})}\] Since, \[|z|=1\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}}=1\]\[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}=1\] \[\therefore \]\[a=\frac{x(1+1)}{2\times 1}=x\] Also\[z\ne 1\Rightarrow x+iy\ne 1\] \[\therefore \]\[A=(-1,1)\]You need to login to perform this action.
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