A) 1
B) 2
C) 3
D) 4
Correct Answer: D
Solution :
Let \[z=x+iy,\] so that \[\overline{z}=x-iy,\] therefore \[{{z}^{2}}+\overline{z}=0\,\Leftrightarrow ({{x}^{2}}-{{y}^{2}}+x)+i\,(2xy-y)=0\] Equating real and imaginary parts, we get \[{{x}^{2}}-{{y}^{2}}+x=0\] .....(i) and \[2xy-y=0\] Þ \[y=0\]or \[x=\frac{1}{2}\] If \[y=0\], then (i) gives \[{{x}^{2}}+x=0\,\,\Rightarrow x=0\]or \[x=-1\] If \[x=\frac{1}{2},\] Then \[{{x}^{2}}-{{y}^{2}}+x=0\]Þ\[{{y}^{2}}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\]Þ\[y=\pm \frac{\sqrt{3}}{2}\] Hence, there are four solutions in all.
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