A) \[\pm \,\,1\]
B) \[\pm \,\,i\]
C) \[1\pm \,\,i\]
D) \[\pm \,\,1\pm i\]
Correct Answer: D
Solution :
Given equation is \[{{z}^{2}}+|z|=0\] ?.(i) Let \[z=x+iy\] So that, \[{{(x+iy)}^{2}}+|x+iy|=0\] [from Eq. (i)] \[\Rightarrow \] \[{{x}^{2}}-{{y}^{2}}+2ixy+\sqrt{{{x}^{2}}+{{y}^{2}}}=0\] \[\Rightarrow \] \[({{x}^{2}}-{{y}^{2}}+\sqrt{{{x}^{2}}+{{y}^{2}}})+i(2xy)=0\] On equating real and imaginary parts, we get \[{{x}^{2}}-{{y}^{2}}+\sqrt{{{x}^{2}}+{{y}^{2}}}=0\] ?.(ii) and \[2xy=0\] \[\Rightarrow \] \[x=0\] or \[y=0\] ?..(iii) if \[(x=0),\] then from Eq. (ii), we get \[-{{y}^{2}}\pm \,y=0\] \[\Rightarrow \] s\[y(\pm \,1-y)=0\] \[\Rightarrow \] \[y=0\] or \[y=\pm 1\] If \[(y=0),\] then from Eq. (ii), we get \[{{x}^{2}}\pm x=0\] \[\Rightarrow \] \[x(x\pm 1)=0\] \[\Rightarrow \] \[x=0\] or \[x=\pm 1\] Hence, the non-zero solutions are \[+\,\,i\] and \[\bar{+}\,\,1.\]
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