Complex image of $\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)z^2$ over a line I need to take the image of 
$$f(z) = \left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)z^2$$
over the set $S=\{z=(x,y)\in \mathbb{C}; y=3x+1\}$.
I suspect that $z^2$ might be a parabola but I don't see how. When I have such parabola, I just multiply by the constant $\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)$, that is, I rotate the parabola by the angle of this number, and stretch by its absolute value.
I must see that $(x+i(3x+1))^2$ is a parabola:
$$x^2+2i(3x+1)+(3x+1)^2 = x^2+6ix+2i+9x^2+6x+1 = 10x^2+6x+1+i(6x+2)$$
$$u = 10x^2+6x+1$$
$$v = 6x+2$$
Then we can see that $u = 10x^2+v-1\implies \frac{u-v+1}{10} = x^2\implies$
$$x = \sqrt{\frac{u-v+1}{10}}$$
but $x$ gets ugly, I don't know how to make $u$ dependent on $v$ only. Can somebody help me?
Is there a generalization for $az^2$?

$(x+(3x+1)i)^2=x^2 +2xi(3x+1)-(3x+1)^2=-8x^2 -6x-1+i(6x^2+2x)$ Now $$-u=8x^2+6x+1$$ and $$v=6x^2+2x$$ completing the square by adding and subtracting $\frac{1}{6}$   you get  $v+\frac{1}{6}=(\sqrt{6}x+\frac{1}{\sqrt{6}})^2$ 
for $v \geq -1/6$ $$6x=\sqrt{6v+1}-1=$$ now going back to the first you get $$-u=8/6(\sqrt{6v+1}-1)^2 +\sqrt{6v+1}$$ which is a parabola can you see why?
