Question

For all complex numbers $z$ of the form $1+i\alpha ,\alpha ϵR,$ if $z^2=x+iy$, then :

• A. $y^2-4x+2=0$
• B. $y^2+4x-4=0$
• C. $y^2-4x+4=0$
• D. $y^2+4x+2=0$

Answer 1

The correct option is B $y^2+4x-4=0$

It is given that

$z=1+i\alpha$

Hence, $z^2=(1+i\alpha {)}^2$

$=(1-{\alpha }^2)+i\left(2\alpha \right)$ $=x+iy$

Hence, $x=1-{\alpha }^2$ And $y=2\alpha$

Thus $\frac{y}{2}=\alpha$

Substituting in the equation of $x$ gives us

$x=1-{\left(\frac{y}{2}\right)}^2$

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

$4x=4-y^2$

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