For all complex numbers z of the form 1 + ial | vedclass

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(B) Given $z = 1 + i\alpha$,where $\alpha \in R$.Squaring both sides,we get $z^2 = (1 + i\alpha)^2 = 1^2 + (i\alpha)^2 + 2(1)(i\alpha)$.$z^2 = 1 - \al

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

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(B) Given $z = 1 + i\alpha$,where $\alpha \in R$.Squaring both sides,we get $z^2 = (1 + i\alpha)^2 = 1^2 + (i\alpha)^2 + 2(1)(i\alpha)$.$z^2 = 1 - \alpha^2 + 2i\alpha$.Given $z^2 = x + iy$,we equate the real and imaginary parts:$x = 1 - \alpha^2$ and $y = 2\alpha$.From $y = 2\alpha$,we have $\alpha = \frac{y}{2}$.Substituting this into the expression for $x$:$x = 1 - (\frac{y}{2})^2$$x = 1 - \frac{y^2}{4}$Multiplying by $4$:$4x = 4 - y^2$$y^2 + 4x - 4 = 0$.

For all complex numbers $z$ of the form $1 + i\alpha$,where $\alpha \in R$,if $z^2 = x + iy$,then which of the following relations holds?

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