A) An ellipse
B) The imaginary axis
C) A circle
D) The real axis
Correct Answer: B
Solution :
We have \[|{{z}^{2}}-1|\ =\ |z{{|}^{2}}+1\] Þ \[|{{(x+iy)}^{2}}-1|\ =\ |x+iy{{|}^{2}}+1\] Þ \[|({{x}^{2}}-{{y}^{2}}-1)+2xyi|\ ={{\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)}^{2}}+1\] Þ \[\sqrt{{{({{x}^{2}}-{{y}^{2}}-1)}^{2}}+{{(2xy)}^{2}}}={{x}^{2}}+{{y}^{2}}+1\] Þ \[{{x}^{4}}+{{y}^{4}}+1-2{{x}^{2}}{{y}^{2}}+2{{y}^{2}}-2{{x}^{2}}+4{{x}^{2}}{{y}^{2}}\] \[={{x}^{4}}+{{y}^{4}}+1+2{{x}^{2}}{{y}^{2}}+2{{y}^{2}}+2{{x}^{2}}\] Þ \[2{{x}^{2}}{{y}^{2}}=2{{x}^{2}}{{y}^{2}}+4{{x}^{2}}\]Þ \[x=0\] then, \[z=x+iy=0+iy=iy\] Hence \[z\] lies on imaginary axis.
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