A) \[z\]is purely real
B) \[z\]is purely imaginary
C) Either \[z\]is purely real or purely imaginary
D) None of these
Correct Answer: C
Solution :
Let\[z=x+iy\], then its conjugate \[\overline{z}=x-iy\] Given that \[{{z}^{2}}={{(\overline{z})}^{2}}\] Þ \[{{x}^{2}}-{{y}^{2}}+2ixy={{x}^{2}}-{{y}^{2}}-2ixy\]Þ \[4ixy=0\] If \[x\ne 0\] then \[y=0\]and if \[y\ne 0\]then \[x=0\]
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