A) 1
B) 4
C) 16
D) 25
Correct Answer: A
Solution :
[a] We have \[1+2|z{{|}^{2}}=|{{z}^{2}}+1{{|}^{2}}+2|z+1{{|}^{2}}\] \[\Rightarrow \,\,1+2z\bar{z}=({{z}^{2}}+1)({{\bar{z}}^{2}}+1)+2(z+1)(\bar{z}+1)\] \[\Rightarrow \,\,{{(z\bar{z})}^{2}}+2(z+\bar{z})+{{z}^{2}}+{{\bar{z}}^{2}}+2=0\] \[\Rightarrow \,\,{{(z\bar{z}-1)}^{2}}+{{(z+\bar{z}+1)}^{2}}=0\] \[\Rightarrow \,\,\,z\bar{z}=1\] and \[\Rightarrow \,\,\,z+\bar{z}+1=0\] \[\Rightarrow \,\,\,{{z}^{2}}+z=-1\Rightarrow |{{z}^{2}}+z|=1\]You need to login to perform this action.
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