A) \[{{y}^{2}}+4x+2=0\]
B) \[{{y}^{2}}-4x+4=0\]
C) \[{{y}^{2}}+4x-4=0\]
D) \[{{y}^{2}}-4x+2=0\]
Correct Answer: C
Solution :
\[z=1+i\alpha ,\,\,\alpha \in R\] Now, \[{{z}^{2}}=x+iy\] \[\Rightarrow \,\,{{(1+i\alpha )}^{2}}+x+iy\] \[\Rightarrow \,\,x=1-{{\alpha }^{2}},\,\,y=2\alpha \] \[\therefore \] On eliminating \[\alpha ,\]we get \[{{y}^{2}}+4x-4=0\]
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