A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
Correct Answer: D
Solution :
Given, \[{{z}^{2}}+\bar{z}=0\] Let \[z=x+iy\] \[\therefore \] \[{{(x+iy)}^{2}}+(\overline{x+iy})=0\] \[\Rightarrow \] \[{{x}^{2}}-yh2+2ixy+x-iy=0\] On equating real and imaginary parts, we get \[{{x}^{2}}+x-{{y}^{2}}=0\] and \[2xy-y=0\] \[\Rightarrow \] \[y=0,\,x=\frac{1}{2}\] when \[y=0,\,{{x}^{2}}+x=0\] \[\Rightarrow \] \[x=0,-1\] when \[x=\frac{1}{2}\] \[{{\left( \frac{1}{2} \right)}^{2}}+\frac{1}{2}-{{y}^{2}}=0\] \[\Rightarrow \] \[\frac{3}{4}-{{y}^{2}}=0\] \[\Rightarrow \] \[{{y}^{2}}=\frac{3}{4}\] \[\Rightarrow \] \[y=\pm \frac{\sqrt{3}}{2}\] i.e, \[(x,y)=(0,0),(-1,0),\] \[\left( \frac{1}{2},\frac{\sqrt{3}}{2} \right),\,\left( \frac{1}{2},-\frac{\sqrt{3}}{2} \right)\] Hence, the number of solutions are 4.
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