Why is $z^2$ a conformal mapping?

Question

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.

Answer 1

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.

Answer 2

$z^2$ 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.