A) \[|{{z}^{2}}|\,=\,|z{{|}^{2}}\]
B) \[|{{z}^{2}}|\,=\,|\bar{z}{{|}^{2}}\]
C) \[z=\bar{z}\]
D) \[{{\bar{z}}^{2}}={{\bar{z}}^{2}}\]
Correct Answer: C
Solution :
L.H.S.= \[|{{z}^{2}}|\,=\,|{{(x+iy)}^{2}}|\] \[=\,\,|{{x}^{2}}-{{y}^{2}}+2ixy|=\sqrt{{{({{x}^{2}}-{{y}^{2}})}^{2}}+{{(2xy)}^{2}}}\] \[=\sqrt{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}}\] ?..(i) R.H.S. \[=|z{{|}^{2}}=|x+iy{{|}^{2}}=\sqrt{{{({{x}^{2}}+{{y}^{2}})}^{2}}}\] \[={{x}^{2}}+{{y}^{2}}\] ??(ii) Therefore \[|{{z}^{2}}|=|z{{|}^{2}}\] (b) True (c) False (since\[z\ne \overline{z}\] ).
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