A) \[\operatorname{Re}({{z}^{2}})=0\]
B) \[\operatorname{Im}({{z}^{2}})=0\]
C) \[\operatorname{Re}({{z}^{2}})=\operatorname{Im}({{z}^{2}})\]
D) \[\operatorname{Re}({{x}^{2}})=-\operatorname{Im}({{z}^{2}})\]
E) \[{{z}^{2}}=0\]
Correct Answer: A
Solution :
Let\[z=x+iy\Rightarrow zz=(x+iy)(x+iy)\] \[={{x}^{2}}-{{y}^{2}}+2ixy\] \[=0+2ixy\] \[\because \] \[\operatorname{Re}(z)=\operatorname{Im}(z)\] \[\Rightarrow \] \[\operatorname{Re}({{z}^{2}})=0\]
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