Why is $z^2$ a conformal mapping? It's not a one-to-one mapping, by the Fundamental Theorem of Algebra.
$e^z$ is one-to-one, when restricted to a horizontal strip of width = $2\pi i$.
Is it a similar argument for $z^2$?  
Thanks,
Edit: $z^2$ doubles angles (and squares magnitudes of numbers in complex exponential form.), so it seems one-to-one - I guess I am possibly confusing myself with the conformal mapping $z^2$ with the polynomial equation, $z^2$ + az = 0.
 A: It is conformal on $\mathbb{C} \mathop{\backslash} \{0\}$. At $0$ the derivative is $0$ and it is not conformal (as you say, it doubles angles subtended at $0$).
My mental model of $z \mapsto z^2$ is two turns of a spiral staircase projected onto the ground floor: that mapping is two-to-one but preserves (plan view) angles except at the centre of the staircase.
A: $z^2$ is a conformal mapping because it is complex differentiable and it's derivative is non-zero everywhere except at $0$ (where it is not conformal).
You can see why analytic functions are conformal as follows.  Let $z_0$ be a point where $f$ is analytic with non-zero derivative.  Then for two points $z_i$ close to $z_0$ we have by Taylor that
$$f(z_i) \approx f(z_0) + f'(z_0)(z_i - z_0)$$
This mapping is conformal if the angle between the vector $z_1 - z_0$ and $z_2 - z_0$ is preserved.  Using the above then
$$\frac{f(z_2) - f(z_0)}{f(z_1) - f(z_0)} \approx \frac{f'(z_0)}{f'(z_0)}\frac{z_2 - z_0}{z_1 - z_0}$$
In general, the ratio of two complex numbers $w_2$ and $w_1$ is given by
$$\frac{w_2}{w_1} = \frac{r_2}{r_1}e^{i(\phi_2 - \phi_1)}$$
where $r_i, \phi_i$ are the polar coordinates of $w_i$. From this we see that the angles are preserved through the mapping.
