Step 1: Write down the limit
Step 2: Use the arc-cosine formula
A complex function, f(z), would assign to each complex number z (z in C) a new complex number, f(z). It is a map from C to C. It sort of deforms C. It is like how f(x) = x^2 deforms the real number line R^1 to look like a parabola.
So in the complex plane C, imagine I have a point A. Suppose there are curves C1 and C2 that intersect at A. When they intersect, both the curves have their own direction. Let's call these directions U and V. We say that a transformation by a complex function f(z) is conformal if the angle between U and V remain unchanged. f(z) is called a conformal map.
Backtrack. Let’s look at point A again. At that point, C1 is passing through. Suppose that the “next point” is point B1. C1, at point A, points from A, toward B1. That direction it is pointing in is the direction U. U is a vector defined as position vector B1 minus position vector A.
Now, C1 is just a bunch of points on the complex plane that, together, trace out a line. C2 is also just a bunch of points on the complex plane that, together, trace out a line.
If you apply f(z) to C, what happens to the points that make up C1 and C2. They also transform. So C1 becomes f(C1) and C2 becomes f(C2). f(C1) is the “image” of C1 under the f(z) transformation.
Now, what does f(B1) - f(A) become? B1 - A use to be the vector U. Now it is the vector f(U). The question is, has the angle between f(U) and f(V) changed?
Suppose f is holomorphic, so that the limit definition of a derivative holds. f’(A) = limit as B1 approaches A (f(B1) - f(A))/(B1 - A). Remember from calculus that the derivative at A is the direction from A to an infinitesimal neighbor A + epsilon. Here we have B1 instead of A + epsilon.
As B1 gets closer to A, we can approximate f(B1) by f(A) + f’(A)(B1 - A). So, f(B1) - f(A) is f’(A)(B1 - A), which is also f’(A)*U.
So we have proved that f(U) = f(B1) - f(A) = f’(A)*(B1 - A) = f’(A)*U. The question is, is the angle between f(U) and f(V), that is to say f’(A)*U and f’(A)*V, the same as the angle between U and V.
It suffices to show that multiplication by a complex number preserves angles.
f’(A) is a complex number. U, remember, is a 2D vector. It pointed from a point A in the complex plane (isomorphic to R^2) to another point B. We can say that U = (u1,u2) = u1 + i*u2. A complex number, here f’(A), can be written as r*e^(iT) (T is theta), or r(cosT + i*sinT).
f’(A)U = r(cosT + i*sinT) * (u1,u2) = r(cosT + i*sinT) * (u1 + i*u2) = r(cosT * u1 - sinT * u2 + i * (sinT*u1 + cosT*u2)).
So the new vector f’(A)*U looks like:
- r * [cosT * u1 - sinT * u2]
- r * [sinT * u1 + cosT * u2]
If you look very closely, you’ll notice that we can rewrite this as a matrix with entries cosT in the top left, -sinT in the top right, sinT in the bottom left and cosT in the bottom right. Let’s call this matrix K. So U becomes f’(A)*U which can be written as r*K*U.
The angle between two vectors U and V is arccos[(U dot V) / (mag(U)*mag(V))]. The angle between f’(A)*U and f’(A)*V is arccos[(f’(A)*U dot f’(A)*V)) / (mag(f’(A)*U)*mag(f’(A)*V))]. With a lot of effort you can prove that the top is r*r*U dot V. Both K’s go away. Similarly, you can prove that the bottom is r*mag(U)*r*mag(V). Dividing leads to r*r canceling on the top and bottom, leaving us with U dot V over mag(U)*mag(V). So we have proven that the angle between U and V and the angle between f(U) and f(V) are the same.
Note that a conformal map f(z) will preserve the angle between any two vectors at any point in the complex plane.
