For all complex numbers z of the form 1 + i(\alpha), (\alpha... | Mathematics | SolveeIt

Question

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

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

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

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

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

Answer

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

Explanation

Solution

$(1+i \alpha)^{2}=x+i y$
$1-\alpha^{2}+2 i \alpha= x + i y$
so $x=1-\alpha^{2}, y=2 \alpha$
putting $\alpha= y / 2$
$x=1-\left(\frac{y}{2}\right)^{2}$
$\Rightarrow y^{2}+4 x-4=0$

Mathematics Question on Complex Numbers and Quadratic Equations

Solution