Prove algebraically that for any complex number $w?0$ the equation $z^2=w$ has exactly two solutions.
When I sub in $x+iy$ for $w$, 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!
