Mathematics - Complex Number Question with Solution | TestHub

Given $z=1+\mathit{i\alpha }, z^2=x+\mathit{iy}$

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

$\Rightarrow 1^2+i^2{\alpha }^2+2i\alpha =x+iy$

Using $i^2=-1,$

$\Rightarrow 1-{\alpha }^2+2i\alpha =x+\mathit{iy}$

Equating the real and imaginary parts, we get

$x=1-{\alpha }^2$ and $y=2\alpha$

$\Rightarrow {\alpha }^2=1-x$ and $y^2=4{\alpha }^2$

$\Rightarrow y^2=4\left(1-x\right)$

$\Rightarrow y^2=4-4x$

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