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?


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