Prove $(z)^{1/2}$ sends horizontal and vertical lines to hyperbolae. I need to prove that  $(z)^{1/2}=(r)^{1/2}e^{i\frac{\theta}{2}}$ defined by branch of logarithm $[0,2\pi)$ sends horizontal and vertical lines in $C-{R^+ \cup  \{0\}}$ to hyperbolae.
How can I show this? I see it geometrically, but I don't know how to parameterize it. Any help will be apprecciated.
 A: Let $z=x+yi$ and $w=u+vi$ such that $w=\sqrt{z}$, then
\begin{align*}
  u+vi &= \sqrt{x+yi} \\
 (u^2-v^2)+2uvi &= x+yi \\
\end{align*}
For vertical line $x=a$,

$$u^2-v^2 = a$$

which is a rectangular hyperbola in $uv$-plane where $a\ne 0$.
For horizontal line $y=b$,

$$2uv = b$$

which is a rectangular hyperbola in $uv$-plane where $b\ne 0$.
The two families of hyperbolae are orthogonal.
A: The horizontal line at height $b\in\Bbb R$ is given by $L_b:=\{z\in\Bbb C\;:\;z=x+ib,\;x\in\Bbb R\}$.
You have to get the above set $A_b$ in the polar form: the modulus is given by
$$
r=\sqrt{x^2+b^2}
$$
and the (principal) argument $\theta$ (suppose $b\neq0$) is given by (once you wrote $x=r\cos\theta$ and $b=r\sin\theta$)
$$
\theta=\arctan(b/x).
$$
Let's see the image of $A_b$ thru the function $f(z)=z^{1/2}=r^{1/2}e^{i\theta/2}$ (intended with the principal branch):
$$
f(x+ib)=f(re^{i\theta})=(x^2+b^2)^{1/4}e^{\frac i2\arctan(b/x)}.
$$
From this, you should write explicitly the real and the imaginary part of these last numbers; in such a way, you'd have written the image $f(A_b)$; then you should parametrize this set, and the parametrization should give you the desired form.
