Final remark: conformal mappings

Imagine two smooth curves in the complex plane: $z_1(t), z_2(t)$ where $z_1(t_0) = z_2(t_0) = z_0$. The angle between the curves is determined by the angle between the tangent vectors $z'_1(t_0), z'_2(t_0)$ at $z_0$. The angle between the vectors can be thought of as $arg(z'_1(t_0)) -arg(z'_2(t_0))$ since the curves meet at $t_0$

Now let $f$ be analytic at $z_0$ and let us focus on just a small part of the two cures in question: let $w_1(t) = f(z_1(t)), w_2(t) = f(z_2(t))$. These will be smooth curves as well, albeit in the domain space.

We have the chain rule and so $w'_1(t) = f'(z_1(t))z'_1(t)$ and $w'_2(t) = f'(z_2(t))z'_2(t)$

Now look at the arguments: $arg(w'_1(t)) = arg(f'(z_1(t)) + arg(z'_1(t))$ and $arg(w'_2(t)) = arg(f'(z_2(t)) + arg(z'_2(t))$ PROVIDED $f'(z_1(t_0))= f'(z_2(t_0)) \neq 0$ Now calculate $arg(w'_1(t)) -arg(w'_2(t)) = arg(z'_1(t)) - arg(z'_2(t))$ which is the angle between the original curves.

Any function that preserves angles in this manner is called conformal. So we showed that, if $f$ is analytic on an open disk and $f'$ is not zero on the disk, then $f$ is conformal. Furthermore, our previous work shows that if $f$ is analytic and one-to-one on a disk, $f'$ never zero on the disk. So functions which are analytic and one-to-one (at least locally) are conformal (at least locally).

Examples: find where $e^z, sin(z), cos(z), \sum_{k=0}^n a_k z^k$ are conformal. What about $f(z) = \frac{a + bz}{c+dz}$ ?

Related exercise: where is $f(z) = \frac{a + bz}{c+dz}$ analytic and one to one?