Critical points of holomorphic functions

Transcription

1 Critical points of holomorphic functions Alberto Garcı a-raboso October 27, 2014 Let w = f (z) a complex-valued function of one complex variable. One possible way of representing graphically such a function consists of assigning a color to each complex number in the w-plane, and to paint each point in the z-plane with the color that corresponds to its image. The HSL model describes colors using three coordinates. Hue describes the perceived tone of the color (red, green, blue, etc.). It is naturally an angular variable and thus usually measured in degrees. Saturation codifies the degree of purity of the color. It takes values between 0 and 1, with the latter representing a pure color and the former a shade of grey. Lightness describes how bright the color is, with 0 corresponding to black and 1 to white. A picture is worth a thousand words, or so they say, so here s one: As you can see, the HSL color solid is usually depicted as a cylinder. A better representation in my mind is that of a solid sphere: since the top and bottom ends of the cylinder yield white and black, respectively, we can squish them to a point, like in the following picture (which, by the way, is said to have been inscribed on Archimedes tombstone in accordance with his own wish). 1

2 Here s another picture of the same idea of representing colors using a sphere; this one is due to German painter Phillip Otto Runge ( ), and is known as Runge s Farbenkugel (literally, color ball). Okay, enough with the trivia... How do we map complex numbers to colors? Well, the boundary of the solid sphere above is a (hollow) sphere... and so is the Riemann sphere! We can thus fix the saturation to 1, and map argument to hue and modulus to lightness. But there are a couple of subtleties here. 2

3 The first issue is that lightness takes value between 0 and 1, while w is unbounded. To compress the whole positive real line into the interval (0, 1), we can use the following formula: L(w) = x x + 9 There are, of course, many functions that map R >0 to (0, 1); I chose this one because the pictures looked best (not too dark, not too light). The second modification is a bit more perceptual in origin. If you look at the L = 1/2 slice of the HSL cylinder (it s right there on the first page), you ll see that red corresponds to a hue of 0 o, and cyan to 180 o. The naïve assignment 1 H(w) = (60/2π) arg w would then make the real line easy to identify in pictures of complex-valued functions (see examples below): red is positive real, cyan is negative real. However, purely imaginary numbers would be mapped to some rather forgettable tones: greenish yellow for positive imaginary, purplish blue for negative imaginary. A better choice would be to assign yellow and blue to these (look at how much more identifiable they look in that L = 1/2 slice). This is precisely what the DLMF does, using the piecewise linear function 120(arg w/π) if 0 arg w < π/2 240(arg w/π) 60 if π/2 arg w < π H(w) = 120(arg w/π) + 60 if π arg w < π/2 240(arg w/π) 120 if π/2 arg w < 2π Here s how the identity function f(z) = z looks like with these choices. 1 Here we let the argument take values between 0 and 2π.

4 Notice how clear it is that f(0) = 0 we colored it black, and that the modulus grows when we get away from the origin we get lighter and lighter colors. The red and blue lines superimposed to the color representation we just discussed are the level curves of u = Re f and v = Im f. For the function f(z) = z, they are just vertical and horizontal lines, respectively, for u = x and y = z. At each point in the z-plane, only one red line and one blue line cross. This is indicative that the transformation is conformal everywhere (indeed, f (z) = 1 0). Here s the function f(z) = z 2. We see again the zero at the origin, and the modulus growing as we get far from the origin. As we traverse the unit circle in the z-plane, we go from red to yellow to cyan to blue and back to red again twice: the transformation f(z) = z 2 covers the w-plane twice. We know that the origin is also a critical point of order one of f(z) = z 2 : f (0) = 2z = 0, f (0) = 2 = 2 0 This is reflected in the fact that there are two red lines and two blue lines crossing at the origin. At all other points in the z-plane, the derivative of f is nonzero, and hence f is conformal at those points (so only one red and one blue line pass through each of them). Going one step up the power ladder, we now look at f(z) = z. 4

5 The only noteworthy difference with the cases above is that the origin is now a critical point of order 2: f 0 (0) = z 2 = 0, f 00 (0) = 6z = 0, f () (0) = 6 = 6 6= 0. That s why there are three red lines and three blue lines crossing there. I m sure you know by now how f (z) = z m looks like for m >, so let s move on. Here s a not-so-trivial example: f (z) = z / z. 5

6 This is a cubic polynomial, so it has three zeroes, located at z = 0 and z = ±i and clearly visible in the picture they are colored in black. But f (z) = w0 is also a cubic polynomial, and so each value w0 comes from three (different, most of the times) values of z. This is also apparent in the picture: e.g., there are three regions in red. Another way of phrasing this is saying that f (z) covers the w-plane three times. As for critical points, f (z) has two of them: f 0 (z) = z 2 1 = 0 z = ±i. They are both of order 1, since f 00 (±i) = 2z z=±i = ±2i 6= 0. But notice that we could have identified them and their order directly from the picture: at z = ±i we see two red curves crossing! Now, locally around these points, f (z) is given by 2i f (z) = i(z i)2 + O (z i) and 2i = i(z + i)2 + O (z + i), f (z) + respectively. That is, it looks locally like z 7 z 2 followed by a rotation by π/2 or π/2, respectively. This is made clearer if we plot the shifted functions g± (z) = f (z) so that g± (±i) = i

7 Wait! These don t look like z z 2 rotated by ±π/2, do they? Well, the issue is where the rotation is happening: it is not applied to z z 0, but to (z z 0 ) 2. In other words, it is a rotation in color. Let us concentrate on the left-hand picture (i.e., on g + ). Numbers to the right of z 0 get sent to positive real numbers by z (z z 0 ) 2, i.e., to red; then we rotate the color from red to yellow (positive imaginary). We can transport the rotation to the z-coordinate by pulling the factor of i in front of (z z 0 ) 2 inside of the parentheses: g + (z) = [ ] 2 e iπ/4 ( (z i) + O (z i) ) According to this, if we want to get red near i, we need e iπ/4 (z i) to be real. That ll happen if arg(z i) = π/4 (or arg(z i) = π/4). All in all, a rotation by π/2 around 2i/ in the w-plane is the same as a rotation by π/4 (or π/4 that gives the same thing) about i in the z-plane. 7