Prove algebraically that for any complex number
$w?0$ the equation
$z^2=w$ has exactly two solutions.
When I sub in
$w$, I get
$z^2 = x + iy$. Knowing
$z$ is complex, then I sub in
$x+iy$ for that as well to get
$x^2+2xiy+(iy)^2 = x + iy$. Subtracting over I get
$x^2-x+2xiy+(iy)^2-iy$. From here I am not sure what to do or if I am doing it right. Any help would be greatly appreciated! Thanks!