The question is:
Consider the transformation
$w=z^2+3z$ from the $z$-plane where $z=x+iy$ and $w=u+iv$.
Determine the image of the line $y=1$ and $x=3$ in the $w$-plane
My attempt at a solution: $\begin{align} u+iv=w&=z^2+3z\\ &=(x+iy)^2+3(x+iy)\\ &=x^2+ixy+ixy+(iy)^2+3x+3iy\\ &=x^2+2ixy-y+3x+3iy\\ &=x^2+3x-y+i(2xy+3y) \end{align}$ So $\begin{align} &u=x^2+3x-y\\ &v=2xy+3y \end{align}$
$\begin{align} w&=z^2+3z\\ z^2+3z-w&=0\\\text(Which is a Quadratic euation)\\\\ x+iy=z&=\frac{-3\pm\sqrt{9-4w}}{2}\\ &=\frac{-3\pm\sqrt{9-4(u+iv)}}{2} \end{align}$ So how do I now break $u$ and $iv$ appart?