Image of line under $w=\sqrt{z}$ I have a doubt: I need to find de image of the line $y=x\sqrt{3}$, under the application $w=\sqrt{z}$. Here it is my progress so far: considering the inverse $w^2=z$ and let $z=x+yi=x+ix\sqrt{3}$ and $w=u+vi$ be, I have:
$u^2-v^2=x, \ 2uv=x\sqrt{3}$
Multiplying the first equation by $x\sqrt{3}$ and subtracting the with the second, I get
$\sqrt{3}(u^2-v^2)-2uv=0$.
I can't go on from there. What does this equation represent?
 A: Your working is correct. Solving $\sqrt{3}(u^2-v^2)-2uv=0$ for $u$, using the quadratic formula, gets $$u = \frac {\sqrt3 \pm 2\sqrt 3}{3} v$$ which are two straight lines, $u = \frac {-\sqrt3}{3} v$ and $u = 3\sqrt 3 v$ .
A different approach is as follows:
The line $y=x \sqrt{3}$ can be parametrised as $$\{te^{-\frac{\pi}{3}i}: t >0 \} \cup \{0\}\cup  \{te^{\frac{\pi}{3}i}: t>0 \}$$
Firstly suppose $z^2 \in \{te^{-\frac{\pi}{3}i}: t >0 \}$. Then $$z^2=te^{-\frac{\pi}{3}i}=te^{\frac{5\pi}{3}i}$$
$$ \implies z=\sqrt{t}e^{-\frac{\pi}{6}i} \qquad \mathrm{or} \qquad z=\sqrt{t}e^{\frac{5\pi}{6}i}$$
$$\implies z \in \{re^{\frac{-\pi}{6}i}: r>0 \}\cup  \{re^{\frac{5\pi}{6}i}: r>0 \}$$
which is the straight line $u = \frac {-\sqrt3}{3} v$ from before (minus $\{0\}$)!
Instead taking $z^2 \in \{te^{\frac{\pi}{3}i}: t >0 \}$, we would get the line $u = 3\sqrt 3 v$.
A nice extension would be to think about why the image of any straight line through the origin under $z \mapsto z^2$ is a set of two perpendicular lines.
A: $\sqrt{3} u^2 - 2uv - \sqrt 3 v^2 = 0\\
3 u^2 - 2\sqrt 3 uv - 3v^2 = 0$
Looks like a candidate for the quadratic formula.
$u = \frac {\sqrt3 \pm \sqrt 12}{3} v$
or
$ ( u + \frac{\sqrt 3}{3} v)(u - \sqrt 3 v)=0$
This is a pair of perpendicular lines.
A: Another approach is when $\sqrt{3}(u^2-v^2)=2uv$ then with $w=\rho e^{i\phi}$
$$\sqrt{3}(\rho^2\cos^2\phi-\rho^2\sin^2\phi)=2\rho^2\cos\phi\sin\phi$$
or
$$\sqrt{3}\cos2\phi=\sin2\phi$$
shows that $w$ if free of $\rho$, and $\tan2\phi=\sqrt{3}$, finally $\phi=k\dfrac{\pi}{2}+\dfrac{\pi}{6}$ is the line equation.
A: @Doug they aren't perpendicular 
