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: B
Solution :
Let \[z=1+i\alpha ,\alpha \in R\] \[{{z}^{2}}=(1+i\alpha )(1+i\alpha )\] \[x+iy=(1+2i\alpha -{{\alpha }^{2}})\] On comparing real and imaginary parts, we get \[x=1-{{\alpha }^{2}},y=2\alpha \] Now, consider option (b), which is \[{{y}^{2}}+4x-4=0\] LHS : \[{{y}^{2}}+4x-4={{(2\alpha )}^{2}}+4(1-{{\alpha }^{2}})-4\] \[=4{{\alpha }^{2}}+4-4{{\alpha }^{2}}-4\] = 0 = R.H.S. Hence, \[{{y}^{2}}+4x-4=0\]
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