COMPLEX DIFFERENTIAL FORMS. 1. Complex 1-forms, the -operator and the Winding Number

Transcription

1 CHAPTER 3. COMPLEX DIFFERENTIAL FORMS 1. Complex 1-forms, the -operator and the Winding Number Now we consider differentials for complex functions. Take a complex-valued function defined on an open region in the complex plane C, say f. The variable of f is designated by the symbol z so that we may write f f(z) if we wish. The real and the imaginary parts of z are, as usual, denoted by x and y respectively: z x+iy. Since f(z) f(x+iy), we may regard f as a function of two real variables x and y. Thus the partial derivatives f x and f y make sense and the differential df can be expressed in terms of them: df f f dx + dy. (1.1) x y However, often we do not need this identity in actual computation. Consider: Example 1.1. Find df for f(z) z. Solution: To compute df via (3.1), we put f(x + iy) (x + iy) (x y ) + ixy. Hence df d(x y )+id(xy) (xdx ydy)+i(ydx+xdy). This answer is certainly correct but does not look very nice. Why don t we take differential directly from f z to get df z dz? After all, substituting z x + iy and dz dx + idy in zdz, we can check that z dz is the same answer. Indeed, we should carry out the computation in a straightforward manner: d(z ) z dz. This example shows that (1.1) is not a good way for studying complex functionals or differentials. In whay follows we look for a better way. Let us take a general complex function f with z x + iy as its variable (which runs through an open domain). Let us write z x iy, the complex conjugate of z. Now we have two pairs of variables: {x, y} and {z, z}, which are related by z x + iy, z x iy, and x z + z, y z z. i These identities can be formally carried over to differentials: dz dx + idy, d z dx idy, dx dz + d z, dy dz d z. (1.) i As we know, the real variables x and y are independent of each other and hence so are their differentials. But the complex variables z and z are related. They cannot 1

2 be regarded as independent variables. However, their differentials dz and d z are linearly independent in the sense of the following: Rule CD1. If g dz + h d z 0, then g 0 and h 0. Here g and h are arbitrary (complex-valued) functions in z. We have to justify this rule. Indeed, substituting dz dx + idy and d z dx idy in gdz + hd z 0, we have g(dx + idy) + h(dx idy) 0, which is equivalent to (g + h)dx + (ig ih)dy 0. Since dx and dy are independent, we have g + h 0 and ig ih 0 (by Rule DF6 in 1.), from which it follows that g 0 and h 0. In (1.1), the differential df of f is expressed in terms of dx and dy. Hence, by using (3.), df can also be expressed in terms of dz and d z. For a complex function f, expressing df in terms of dz and d z turns out to be more natural and desirable, as shown in the following quick examples: d(z ) zdz, d( z ) zd z, d(z 3 ) 3z dz, d( z 3 ) 3 z d z, d(z 1 ) z dz, etc. Example 1.. Express d z in terms of dz and d z. Solution: Recall the identity z z z. Taking differentials on both sides, we have d z d(z z), or z d z zdz + zd z and hence d z ( zdz + zd z)/ z. Alternatively, we have d z d(z z) 1/ 1 (z z) 1/ d(z z) d(z z) zdz + zd z. z z Using the last two identities in (1.), we may rewrite (1.1) as df f f f dx + dy x y x 1 f (dz + d z) + y 1 ( ) f x if dz + 1 ( ) f y x + if dz. y ( i ) (dz d z) (1.3) Now we define f/z and f/z in such a way that the identity df f f dz + d z (1.4) z z holds. Comparing (1.4) with the previous identity (1.3), naturally we define: f z 1 ( ) f x if, y f z 1 ( ) f x + if. (1.5) y The second identity gives the definition of the so-called - operator pronounced as dee bar ). Keep it in mind! / z; (here is

3 Example 1. gives d z ( z/ z )dz + (z/ z )d z. In view of (1.4), we obtain z z z z z z, z z z z. We will see more computations involving operators /z and / z later. Next we interpret these operators as rates of change. Consider the change f f(z + z) f(z) in f caused by an increment z x+i y in z x+iy. Under what condition does the limit lim z 0 f/ z exist, which gives the complex derivative f (z) of f at z? To answer this question, let us begin with identity (1.1). This identity can be viewed as a polished version of the following precise one: f f f { } x + x y y + o ( x) + ( y). ( ) (The little o notation o{ } will be explained soon.) Substitute x by 1 z + z, ( ) and y by 1 i z z. Replace ( x) + ( y) by z. Following a computation similar to (1.3), it is easy to get f f f z + z + o{ z }. z z Here, o{ z } is a quantity less than ɛ z if z is small enough, where ɛ is an arbitrarily small positive number given beforehand. Dividing both sides by z, we see that the difference quotient f/ z has a limit as z tends to zero, provided that f satisfies the following so-called Cauchy-Riemann equation: f z 0, and, in that case, the complex derivative f (z) is given by f/z. Notice that, when z 0, the quotient z/ z has a limits +1 if we let z move along the real axis and a different limit 1 if z moves along the imaginary axis. Hence, if the Cauchy-Riemann equation fails at a certain point, then the difference quotient f/ z does not have a limit as z tends to zero. We have shown, Rule CD (CR-equation). A (smooth) function f defined in C is complex differentiable if and only if it satisfies the Cauchy-Riemann equation f/ z 0, and in that case the derivative of f is given by f f/z. A (smooth) function f defined on an open set in C is called an analytic function or a holomorphic function if it satisfies the Cauchy-Riemann equation, or, equivalently, it has a complex derivative at each point in its domain. 3

4 The operator 4 z z ( x i ) ( y x + i ) y is the Laplacian. Indeed, an easy computation shows ( ) ( ) + x y x + y. The usual definition of the Laplacian is given by the last expression. Thus 4 z z x + y. (1.6) We call a (smooth) function f defined on an open set in C a harmonic function if it satisfies the so-called Laplace equation f 0. (We apologize for using the symbol f to denote two completely different notions: increment in f and the Laplacian of f. The context will make the meaning of this symbol clear whenever it appears.) Analytic functions are harmonic since a function satisfying the Cauchy - Riemann equation clearly satisfies the Laplace equation. The Laplacian, unlike the -operator /z, is a real operator: if f is a real-valued function, then its Laplacian f is also real-valued. Thus, the real part (or the imaginary part) of a harmonic function is also harmonic. But the real part of an analytic function in general is not analytic. In fact, a real-valued analytic function must be a constant. Example 1.3. Prove the last statement. Solution: Let f be a real-valued analytic function. Then f z f x + if y 0. As f is real, f/x and f/y are also real and hence they must vanish. Thus we have df (f/x)dx + (f/y)dy 0 and hence f is a constant. The operators /z and / z behave like usual partial derivatives, if we consider a function of x, y as a function of z, z instead. The following computations give you some rough idea how it works: z y z z z i 1 i i, (x + y ) 4 z Next we give an important but more complicated example. z z z 4 z z 4. Example 1.4. Show that the so-called Poisson kernel P (z, e it ), given below, which is regarded as a function of z for z < 1, is a harmonic function; (the Poisson kernel is used for describing solutions to Dirichlet s problem over the unit disk). P (z, e it ) 1 z e it ; t R, z < 1. (1.7) z 4

5 (Here e it cos t + i sin t.) Solution: First we apply the -operator to P (z, e it ). For simplicity, write w for e it. z P (z, w) 1 z z w z 1 z (w z)( w z) z z z (w z)( w z) 1 1 w z z w z z z w z z w z 1 1 w z ( w z) z w w z ( w z) 1 z w (w z)( w z) (w z) w (w z)( w z) w (1 zw). Since the last expression does not involve z, we have (/z) w/(1 zw) 0. Hence z P (z, e it ) 4 z z P (z, w) 4 z w (1 zw) 0. (Here z means that the Laplacian is taken with respect to z.) Therefore the Poisson kernel P (z, e it ), as a function of z varying within z < 1, is a harmonic function. Just like the way Rule CD1 is derived from other rules from our previous sections, we can also derive some rules for the -operator / z and its conjugate operator /z. Here we only show how to derive the product rule for these operators. Let us begin with the well known d(fg) fdg + gdf. For simplicity, let us write f z for f/z and f z for f/ z. By means of (3.3), we have d(fg) (fg) z dz + (fg) z d z. On the other hand, we have fdg + gdf f(g z dz + g z d z) + g(f z dz + f z d z) (fg z + f z g)dz + (fg z + f z g)d z. By Rule CD1, we get (fg) z f z g + fg z and (fg) z f z g + fg z. Rule CD3. z fg f g z + g f z, z fg f g z + g f z. A consequence of the product rule is, if f and g are analytic, then so is their product. Indeed, from f z 0 and g z 0 we obtain (fg) z fg z + f z g 0. Of course this fact can also be seen in the usual way: the existence of the complex derivatives f and g for f and g implies the existence of (fg), which is given by fg + f g. Let f be an analytic function. Then f/ z 0 and df (f/z) dz f (z)dz, from which it follows f (z)dz df f(z(b)) f(z(a)), where is a parametric curve given by z z(t) (a t b). Thus, to find the value of a complex line integral g(z)dz, in principle one may proceed like the first year calculus by finding a primitive function for g first, that is, a function f satisfying g f, or, written in differential forms, gdz df. 5

6 Example 1.5. Compute dz/z, where is any path running from to 1 + i. Solution: z dz 1+i d( z 1 ) z 1 ((1 + i) 1 1 ) i. But the method does not work when we fails to find a primitive function. In that case we go back to the first principle by using f(z)dz b a (f(z)dz) b f(z(t)) dz(t), where a is described by the parametric equation z z(t); a t b. Consider the differential form dz/z, which is important not just because of its simplicity in appearance. The complex logarithm log z, if properly defined, should satisfy d log z dz/z, i.e. log z is a primitive of dz/z. A proper definition of the complex logarithm log z is not easy to come by. A careful study of dz/z certainly helps us to understand log z. Putting z x + iy and dz dx + idy, we have dz z dx + idy x + iy d(x + y )/ x + y (x iy)(dx + idy) (x iy)(x + iy) xdy ydx + i x + y (xdx + ydy) + i( ydx + xdy) x + y 1 d log(x + y ) + iω d log r + iω, (1.8) where r x + y z and ω is the angular form. Restricting to the slit plane D {z x + iy : y 0 or x > 0}, the polar angle θ is unambiguously defined, with π < θ < π and ω dθ. It follows from (1.8) above that, for z in D, dz/z d log r+idθ d(log r + iθ); (remember, both r and θ here are functions of z which varies in D). Since we would like to have d log z dz/z, naturally we define the complex logarithm on D by putting log z log r + iθ. Let us repeat that z x + iy is in D if and only if the polar coordinates r, θ of the point (x, y) satisfy r > 0 and π < θ < π. Rule CD4 (Complex Logarithm). If the polar coordinates r, θ of (x, y) satisfy r > 0 and π < θ < π, then the logarithm of z x + iy is log z log r + iθ. Relation w log z can be rewritten as z e w. Formally, from w log z log r + iθ we deduce e w e log r+iθ e log r e iθ re iθ. Since r, θ are polar coordinates, we have x r cos θ, y r sin θ and consequently z x + iy r cos θ + ir sin θ r(cos θ + i sin θ). Thus e w z gives re iθ r(cos θ + i sin θ). Canceling r, we have e iθ cos θ + i sin θ, which is called Euler s formula when the restriction π < θ < π is dropped. Rule CD5 (Complex Exponential). e z e x+iy e x e iy e x (cos y + i sin y). 6

7 As one may expect, de z e z dz. Indeed, from e z e x (cos y + i sin y), we have de z (cos y+i sin y)de x +e x d(cos y+i sin y) (cos y+i sin y)e x dx+ e x ( sin y+i cos y)dy e x (cos y + i sin y)dx + e x i(i sin y + cos y) e x (cos y + i sin y)(dx + idy) e z dz. Consider line integrals of the form dz/z, where is a loop in the punctured complex plane C\{0}. Let z(t) x(t) + iy(t) (a t b) be the parametric equation for, with z(a) z(b). Write r r(t) z(t) {x(t) + y(t) } 1/. Since does not run through the origin, we have r(t) 0 for all t, which allows us to define w(t) r(t) 1 z(t). Intuitively, w(t) is the parametric equation of a loop going along the unit circled, obtained by tying down to the circle. Since w(t) is a point on the unit circle changing smoothly as t varies, the argument of w(t) also changes smoothly. So there is a smooth angular function θ θ(t) such that w(t) cos θ(t) + i sin θ(t) e iθ(t), a t b. (If you believe such a function exists, fine! Otherwise, please do Exercise 5. In the language of topology, this says that a path in T can be lifted to a path in its universal cover R.) Thus z(t) r(t)w(t) r(t)e iθ(t). To compute dz/z, we have to find the pull-back: dz z dz(t) z(t) z (t)dt {r (t)e iθ(t) + r(t)e iθ(t). iθ (t)} dt z(t) r(t)e { iθ(t) r } (t) r(t) + iθ (t) dt d{log r(t) + iθ(t)}. So dz/z b a (dz/z) log r(t)+iθ(t) b i(θ(b) θ(a)); (notice that from z(a) z(b) a we have r(a) z(a) z(b) r(b) and hence log r(t) b 0). When winds around the a origen N times, then θ(b) θ(a) is πn and hence dz/z πin, or (1/πi) dz/z N, 1 the winding number of about 0. Thus the line integral πi number of the curve about the origin, which is an integer. dz/z is the winding Recall from (1.8) that dz/z d log r + iω where ω is the angular form (xdy ydx)/(x + y ). Since d log r is an exact form, we have d log r 0 and hence dz/z i ω. In summary, letting W(, 0) be the winding number of about 0, we have W(, 0) 1 dz πi z 1 xdy ydx π x + y. (1.9) Example 1.5 shows that the winding number of the path going around the unit circle once in the anticlockwise direction is 1. In general, the winding number about z 0 of a curve not going through a point z 0 x 0 + iy 0 is given by W(, z 0 ) 1 dz 1 (x x 0 )dy (y y 0 )dx πi z z 0 π (x x 0 ) + (y y 0 ), (1.10) 7

8 which is an integer. The winding number W(, z 0 ) remains the same as z 0 or varies continuously, as long as they stay away from each other. Example 1.6. Find the winding number of the circular path 1 : z e it (0 t π) about any point a not on the path, that is, a 1. Solution: Recipe (1.10) gives W(, a) (πi) 1 πi 0 {ie it /(e it a)}dt, which is very difficult to compute. However, we can get the right answer by the following argument. Suppose first that a is inside the unit circle, that is a < 1. We can gradually moves a to the origin without crossing the circle. The integral cannot change abruptly in this process. Since it is an integer, it has to remain constant. We know that the winding number is 1 when a 0 and hence the answer is 1 when a < 1. More precisely, for 0 s 1, consider W(, sa) (πi) 1 π 0 {ie it /(e it sa)}dt, which is a continuous function of s. Since W (, sa) is always an integer, this continuous function must be a constant. In particular, its values at 1 and at 0 are the same, giving W(, a) W(, 0) 1. When a > 1, we can argue in the same fashion but this time we let a run to infinity. Since ie it /(e it a) tends to zero as a, we have W(, a) 0 for a > 1. (This example illustrates that sometimes an exact formula is useless in practical situations.) Take an analytic function w f(z) of a complex variable z and a point z 0 in its domain D. Draw a circular path around z 0 inside the domain, say : z(t) z 0 + re it (0 π). Then f is the path in the w-plane given by w f(z(t)) (0 t π). Assume that this path does not go through the origin of the w-plane. The winding number of this path around 0 is, according to Rule PB, W(f, 0) 1 dw πi w 1 πi f f dw w 1 πi df f 1 πi f f dz. As we shall see, the last expression gives the number of solutions to f(z) 0 inside the circle, counting multiplicity. Here the expression f /f is called the logarithmic derivative of f, for an obvious reason. Exercises 1. Convert the following complex differential forms into the standard form f dz + gd z, where f and g are functions expressed in terms of z and z, (without using the substitution x (z + z)/, y (z z)/i, dx (dz + d z)/ and dy (dz d z)/i, if possible): (a) xdx + ydy (b) xdy + ydx (c) xdy ydx (d) xdx + ydy + i(ydx xdy). 8

9 . Calculate (a) 3. Verify the identity f f(x, y) z x ( ) f 4. Verify the identities f z z x z, (b) y z, (c) x z, (d) z x z + f y y z. ( ) f and z f z. 5. Show that the -operator / z in polar coordinates is eiθ 1 z 1 + z. ( r + i ). r θ 6. (a) Show that if f u + iv is a complex function, where u and v are the real part and the imaginary part of f respectively, then the Cauchy-Riemann equation f/ z 0 is equivalent to u x v y, u y v x. (1.11) (b) Use (1.11) to check that the function f(x, y) e x cos y + ie x sin y is analytic. (c) Use the previous exercise or other methods to show that in polar coordinates (1.11) is u r 1 r v θ, v r 1 r u θ. 7. Show that if f(z) is analytic and if f(z) constant, then f(z) constant. Hint: Apply / z to f f f. 8. Solve the -equation u/ z x. Hint: Write x z + z. Treat z as a constant. 9. (a) Show that if a (smooth) complex function f satisfies f/ z 0 if and only if f g z + h for some analytic functions g and h. (b) Let f u + iv be an analytic function of a complex variable z x + iy. Show that, at each point in the domain of f, the gradients u (u/x, u/y) and v (v/x, v/y) of u and v are perpendicular to each other. Also check this for f(z) z. 10. Let p(x, y) be a polynomial in two variables x and y with complex coefficients. Show that p is analytic if and only if it can be written in the form p(x, y) n k0 c kz k, where c k are some constants and z x + iy. Hint: By the substitution x (z + z)/ and y (z z)/i, any polynomial in x, y can be converted into a polynomial of z and z. Apply CR-equation p/ z For a loop in the complex plane, we have 1 xdy 1 zdz. Why? π πi 9

10 1 1. Compute the line integral πi (z z 0) 1 dz, where is the semi-circular path of radius r given by z z 0 + re it (0 t π). 13. In this exercise we prove the statement a path in T can be lifted to a path in R, which is used for defining winding numbers. Let : w(t) u(t) + iv(t) (a t b) be a smooth path around the unit circle, starting from 1: w(t) u(t) + v(t) 1 for all t and w(a) 1. Verify that (dz/z) (uv vu )dt. This suggests the function θ on [a, b] defined by putting θ(t) t a (u(τ)v (τ) v(τ)u (τ))dτ. Show that w(t) e iθ(t), that is, u(t) cos θ(t) and v(t) sin θ(t). (Hint: Consider the function g(t) (u(t) cos θ(t)) + (v(t) sin θ(t)). Check that g 0 by using uu + vv 0, which is derived from u + v 1.) 14. Use the identity 1 + z + z + + z n (1 z n+1 )/(1 z) and Euler s identity (z ) e iθ cos θ + i sin θ to derive n 1 + cos kθ k1 sin n+1 θ sin θ and n sin kθ k1 sin nθ n+1 sin θ sin θ. π 15. Compute I e iθ dθ and I 1 1 e iθ dθ. (The integral I 1 π 0 π 0 represents the average distance from the unit circle to 1.) Answers: I 1 4/π, I. 16. Suppose that w (z z 0 )/(1 z 0 z), where z 0 re iθ is a constant with r z 0 < 1. π Show that dw/w (1 z 0 )dz/(z z 0 )(1 z 0 z), from which derive that e is eit re iθ 1 re i(t θ) implies ds 1 r 1 r cos(t θ) + r dt. 17. For two loops j (j 1, ) given by z z j (t) (a t b, j 1, ) in the complex plane, the product 1 is the closed path defined by the parametric equation z z 1 (t)z (t) (a t b). Show that we have W( 1, 0) W( 1, 0) + W(, 0), assuming that none of 1 and goes through the origin. Use this identity to find the winding number W (, 0) for described parametrically as z(t) p(e it ) (0 t π), where p is the polynomial given by p(z) z 4 3z 3 z. 18. Use the result of previous exercise to prove Rouchè s theorem, or the dog walking theorem described as follows. Let µ and δ be two closed paths described by the 10

11 parametric equations z z µ (t) and z z δ (t) (a t b) respectively. Then µ + δ is the path described by z z µ (t) + z δ (t) (a t β).this theorem says, assuming z µ (t) < z δ (t) for all t, we have W(δ + µ, 0) W(δ, 0). (This theorem can be interpreted as follows. The origin 0 is a tree. A dog walks around it along the path δ. A man walking the dog follows the path µ relative to the dog. His actual path (relative to the tree) is δ + µ. To avoid the leash entangling the tree, or worse, hurling his head upon the tree, the man has to make sure that his distance to the dog is less than the distance between the dog and the tree. Then, if the dog goes around the tree 99 times, the man also has to walk around it 99 times.) 19. Use Rouchè s theorem to prove the fundamental theorem of algebra, which says that every nonconstant polynomial of a complex variable has at least one root, by following the suggestion given below. Take any polynomial with leading coefficient 1, say p(z) z n + a n 1 z n a 1 z + a 0. Let q(z) z n. Take a circular path R of radius R : z Re it (0 t π). Show that, when R is large enough, Rouchè s theorem gives W(p R, 0) W(q R, 0) n. If p had no roots, it would be possible to let p R shrink to the origin by decreasing R so that W(p R, 0) 0. 11