Why are square roots of lines in the complex plane hyperbolas? I was recently asked in a homework problem to sketch the image of
the half plane $\Re(z) \ge 1$ under the square root map (defined in such a way as to take 1 to itself and be continuous).
It turns out that the square root of the line $1 + it$ (for $t \in \mathbb{R}$) is the the right branch of the hyperbola $x^2 - y^2 = 1$ (I've included the way I arrived at this answer below). Using this it is clear that the image of any other line (or ray, if we wish to remain in this half-plane) under the map is a rotated hyperbola.
I was wondering whether there is some way to understand 'why' these hyperbolas appear. For example, in the case of the map $z \mapsto \frac{1}{z}$ it is easy to verify that lines that don't cross the origin become circles touching the origin, and I can rationalize this by observing that inversion corresponds to a rotation of the Riemann sphere.
However, as far as I am aware, the square root map can't even be defined properly on the Riemann sphere. Moreover, the inverse image of a hyperbola under the stereographic projection seems to be half of a lemniscate-like figure that I can't even recognize, much less visualise as the square root of a circle (on the Riemann sphere).
Which brings me to the question: is there some `natural way' to see that this is a hyperbola?

Computing the square root (@Nate gives a quicker way below)
We can parametrize the vertical line $1 + it$ as
$$\sec \theta e^{i \theta} \text{ for } -\frac{\pi}{2} < \theta < \frac{\pi}{2}.$$
Taking the square root and setting $\phi := \frac{\theta}{2}$
the resulting curve is
$$\sqrt{\sec{2\phi}} e^{i \phi} \text{ for } |\phi| < \frac{\pi}{4}$$
which is, in Cartesian coordinates,
\begin{align*}
 x &= \sqrt{\sec 2 \phi} \cos \phi \\
 y &= \sqrt{\sec 2 \phi} \sin \phi \\
\end{align*}
But (squaring, subtracting, and using the addition formula for cosines) this is precisely the right branch of the hyperbola $x^2 - y^2 = 1$.

Caveat
Please don't answer with a method of showing that the square root of a line is a hyperbola; if there are more elegant ways of deriving this I'd be happy to hear of them, but I'm quite satisfied with my method and am more curious about the other things I've mentioned.

Addendum:
As @Andrew D. Hwang points out below, it takes some work to establish a correspondence between clines in the plane and circles on the Riemann sphere, something (my words) I've conveniently glossed over. My reason for doing this was that once this correspondence has been established, the computation required to show that inversion is a rotation is straightforward, so the Riemann sphere is in some sense a `natural setting' for this operation. (I should note here that I am fairly new to the subject and am fully aware that I don't have the authority to call things natural.)

Second addendum: I've received some nice answers (a number of which I've upvoted), but I'm staving off the moment when I accept one as I'm still hoping that somebody may jump in with a picture of a cone slicing the complex plane or something similarly dramatic. I do realize that this is perhaps over optimistic.
 A: Don't know that square roots can be intuited in terms of inversions (or, more generally, Mobius transforms), but it can be shown that $\,z^2 = 1 + i y \implies \big||z+\sqrt{2}|-|z-\sqrt{2}|\big|=2\,$ which is the geometric definition of the same hyperbola $\,x^2-y^2=1\,$.
This follows from $\,z^2=1+iy \implies z^2+\bar z ^2=2 \implies (z+\bar z)^2=2 + 2 |z|^2\,$. With $\,z\,$ in the right half plane $\,z+\bar z = \sqrt{2(|z|^2+1)}\,$, then:
$$
\begin{align}
\big|z\pm\sqrt{2}\big|^2 &= \big(z\pm\sqrt{2}\big)\big(\bar z \pm \sqrt{2}\big)
\\ &= |z|^2 \pm \sqrt{2}(z+\bar z)+2
\\ &= |z|^2 \pm 2\sqrt{|z|^2+1}+2
\\ &= \left(\sqrt{|z|^2+1}\,\pm\,1\right)^2
\end{align}
$$
Therefore $\require{cancel}\,\big|z+\sqrt{2}\big|-\big|z-\sqrt{2}\big|= \left(\cancel{\sqrt{|z|^2+1}}\,+\,1\right) - \left(\cancel{\sqrt{|z|^2+1}}\,-\,1\right) = 2\,$.
A: If you work out a sketch in polar form, you will get an immediate visual hint of what kind of mapping the square root is giving for lines that do not pass through the origin.
You will notice that, since the argument of the line goes from $- \pi /2 $ to $\pi /2$, the argument of the root shall stay within $(- \pi /4,  \pi /4)$, which are the asymptote lines, shall pass through the point $(x,y)= (r, \theta)=(1,0)$, etc ..
In fact, converting to polar coordinates you have
$$
1 + it = 1 + i\tan \alpha  = \sqrt {1 + \tan ^2 \alpha } \;e^{i\,\alpha }  = \frac{1}
{{\cos \alpha }}\;e^{i\,\alpha } \quad \left| {\; - \frac{\pi }
{2} < \alpha  < \frac{\pi }{2}} \right.
$$
from which
$$
\eqalign{
  & \sqrt {1 + it}  = \frac{1}
{{\sqrt {\cos \alpha } }}\;e^{i\,\alpha /2}  = \frac{1}
{{\sqrt {\cos ^2 \left( {\alpha /2} \right) - \sin ^2 \left( {\alpha /2} \right)} }}\;e^{i\,\alpha /2}  =   \cr 
  &  = \frac{1}
{{\sqrt {\cos ^2 \beta  - \sin ^2 \beta } }}\;e^{i\,\beta }  = \frac{{\cos \beta  + i\sin \beta }}
{{\sqrt {\cos ^2 \beta  - \sin ^2 \beta } }} = \left( {x + iy} \right)  \cr 
  &  \Rightarrow \quad x^2  - y^2  = 1 \cr} 
$$

A: Let's identify $\mathbb{C}$ with $\mathbb{R}^2$ via the obvious map $x+iy \to (x,y)$. The squaring map is given by $(x,y) \to (x^2-y^2, 2xy)$.  Instead of thinking about the square root map let's just ask: What gets sent to the line $ax+by = c$?
The condition that a point gets send to this line via the squaring map is just $a(x^2-y^2) + 2bxy = c$.  This is a degree 2 polynomial equation so it defines a (possibly degenerate) conic.  Since the coefficients of $x^2$ and $y^2$ have opposite signs the only possibilities are a hyperbola or a pair of lines.
