Can anyone please let me know if I did something wrong?
Let $L$ be a line on the complex plane. Find its image under $f(z) =\sqrt{z}$ when:
- $L$ is horizontal.
- $L$ is vertical.
- $L$ is given by the equation $y=x\sqrt{3}$.
I started by writing $w^2=z$. Then replacing $z=x+yi$ and $w=u+vi$. Thus I get
$$u^2-v^2+2uvi=x+yi. $$
In the first case, $z\in L$ means that $z=x+0i$. It follows that
$$u^2-v^2=x$$
which represents a hyperbola.
If $L$ is vertical, then $z\in L$ is the same of $z=yi$. Consequently,
$$2uv=y, $$
That it represents a hyperbola too.
Finally, an element in 3rd case is just like $z=x+x\sqrt{3} i$. So, $$u^2-v^2=x,\ 2uv=x\sqrt{3} . $$ And I can't go on from this point.
Thanks for your attention.
Write $w=u\,\mathrm{e}^{\mathrm{i}\alpha}$ and $z=v\,\mathrm{e}^{\mathrm{i}\beta}$. Then $w^2=z$ implies that either $$ w=\sqrt{z}\quad\text{and}\quad \alpha=\beta/2$$ or $$ w=-\sqrt{z}\quad\text{and}\quad \alpha=\beta/2+\pi/2\;.$$ So the slopes of the image line is half of the slope of the original line modulo $\pi/2$. In the first case, the image is either a horizontal or vertical line, in the second case, these are the 45 and 135 degree lines and in the last case the slope is either $\sqrt{3}/2$ or the line perpendicular to that one.