For all complex numbers z of the form 1+iα, α∈R if z2=x+iy, then

z=1+iα⇒z2=(1+iα)21−α2+i(2α)=x+iy Hence, x=1−α2 And y=2α y2=αx=1−y22⇒x=1−y244x=4−y2⇒y2+4x−4=0

For all complex numbers z of the form 1+iα, α∈R if z2=x+iy, then

z=1+iα⇒z2=(1+iα)21−α2+i(2α)=x+iy Hence, x=1−α2 And y=2α y2=αx=1−y22⇒x=1−y244x=4−y2⇒y2+4x−4=0

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For all complex numbers z of the form $1 + iα, α \in R$ if $z^{2} = x + iy,$ then

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$y^{2} - 4 x + 2 = 0$

$y^{2} + 4 x - 4 = 0$

$y^{2} + 4 x + 2 = 0$

$y^{2} - 4 x + 4 = 0$

answer is B.

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Detailed Solution

$z = 1 + i α$ $\Rightarrow z^{2} = (1 + i α)^{2}$

$(1 - α^{2}) + i (2 α) = x + i y$

$Hence, x = 1 - α^{2} And y = 2 α$

$\frac{y}{2} = α$

$x = 1 - {(\frac{y}{2})}^{2} \Rightarrow x = 1 - \frac{y^{2}}{4}$

$4 x = 4 - y^{2} \Rightarrow y^{2} + 4 x - 4 = 0$

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