Analysis Some notes for Chapter 6

Some notes for Chapter 6 Theorem 1: (Cauchy’s Residue Theorem) If Γ is a simple closed positively oriented contour and f is analytic inside and on Γ except at points z1 , z2 , . . . , zn inside Γ, then I f (z) dz = 2π i n X Res (f, zk ). k=1 Γ We have seen various ways throughout the textbook to find the residue of a function f at a singularity. See Example 2 in Section 6.1 for a method that can be useful in some cases. Here are a few examples to illustrate Cauchy’s Residue Theorem. For the record, simple closed curves are assumed to be traversed once in the positive direction unless explicitly stated otherwise. I Example 2: Evaluate eiz dz, where Γ is the unit circle. z 2 (z − 2)(z + 5i) Γ Solution: Let f be the function that appears as the integrand of the given integral. It is clear that f is analytic on and inside Γ except for a singularity when z = 0. Using Theorem 1 in Section 6.1, we find that Res (f, 0) =  d eiz dz (z − 2)(z + 5i) = z=0 (−10i)(i) − (1)(−2 + 5i) 12 − 5i = . 2 (−10i) −100 (We have combined the differentiate and evaluate steps, but carefully included the appropriate values.) It follows that I z 2 (z  12 − 5i   12i + 5  eiz π 6π dz = 2π i =π =− − i. − 2)(z + 5i) −100 −50 10 25 Γ I Example 3: Evaluate 3z + 2 dz, where Γ is the rectangle with vertices −2, 4, 4 + 3i, and −2 + 3i. z4 + 1 Γ Solution: Let f be the function that appears as the integrand of the given integral. The singularities of f occur at the four fourth roots of −1. These are 1 z1 = eiπ/4 = √ (1 + i); 2 1 z2 = e3iπ/4 = √ (−1 + i); 2 1 z3 = e5π/4 = √ (−1 − i); 2 1 z4 = e7iπ/4 = √ (1 − i); 2 z12 = i = z32 , z22 = −i = z42 , which satisfy z3 = −z1 ; z4 = −z2 ; z1 z2 = −1 = z3 z4 ; √ z1 + z2 = i 2 = −(z3 + z4 ); √ z1 − z2 = 2 = −(z3 − z4 ). We may not use all of these facts, but we record them for the sake of completeness. In addition, note that only z1 and z2 lie within Γ so we do not need the residues at the other two points. However, we determine all of the residues for further practice. Here are several options for finding the value of the integral. 1 Using the result in Example 2 of Section 6.1, we find that 3z1 + 2 1 2 1 3z1 + 2 = = 3+ = (3 − 2z2 ); 3 2 4z1 4z1 · z1 4i z1 4i   3z2 + 2 −1 2 3z2 + 2 1 = = 3+ Res (f, z2 ) = = (−3 + 2z1 ); 4z23 4z2 · z22 4i z2 4i 3z3 + 2 1 2 1 3z3 + 2 = = 3 + = (3 + 2z2 ); Res (f, z3 ) = 4z33 4z3 · z32 4i z3 4i   2 3z4 + 2 3z4 + 2 −1 1 3+ Res (f, z4 ) = = = = (−3 − 2z1 ). 3 2 4z4 4z4 · z4 4i z4 4i Res (f, z1 ) = Notice that the sum of all four residues is 0. Hence, the value of the integral in Exercise 6.1.3d is 0. For our problem, we find that I √
2πi 3z + 2 dz = 2πi Res (f, z1 ) + Res (f, z2 ) = (−2z2 + 2z1 ) = π(z1 − z2 ) = π 2. 4 z +1 4i Γ The answer is the same for any contour that contains z1 and z2 and does not contain z3 and z4 . Suppose a contour contains just z1 and z3 . (You should sketch a few appropriate possibilities for such contours.) In that case, the value of the integral is 3π. To illustrate another way to find these residues, note that f (z) = 3z + 2 . (z − z1 )(z − z2 )(z − z3 )(z − z4 ) It then follows that 3z + 2 (z − z2 )(z − z3 )(z − z4 ) z=z1 3z1 + 2 = (z1 − z2 )(z1 − z3 )(z1 − z4 ) 3 √ (1 + i) + 2 2 √ =√ √ 2 · 2(1 + i) · 2 i √ 3 + 2 2 + 3i = 4(−1 + i) √ 1 3 + 2 2 + 3i 1 + i =− · · 4 1−i 1+i √ √ 1 2 2 + (6 + 2 2 )i =− · 4 2 √
1 √ 2 + (3 + 2 )i =− 4 √
1 = 3 + 2(1 − i) 4i 1 = (3 − 2z2 ). 4i Res (f, z1 ) = The last two steps are not necessary, but they do show that the two calculations give the same answer. You can certainly see how the first method has some clear advantages in terms of reducing the number of computations. 2 Suppose that a and b are real numbers that satisfy 0 < a < b and consider the four integrals Z 2π 0 2π Z 1 dθ, b + a sin θ 0 2π Z 1 dθ, b − a sin θ 0 Z 1 dθ, b + a cos θ 2π 0 1 dθ. b − a cos θ We claim that all four of these integrals are equal. Given that the functions are periodic and the integral is over one full period, this is not all that surprising. To prove these facts more carefully is a bit tedious, but here goes. We first note that the substitutions φ = θ + π and φ = θ − π yield π Z 0 Z 2π π 1 dθ = b + a sin θ Z 1 dθ = b + a sin θ Z 2π π π 0 Z 2π 1 1 dφ = dφ; b + a sin(φ − π) b − a sin φ π Z π 1 1 dφ = dφ; b + a sin(φ + π) b − a sin φ 0 respectively. Adding these two integral equalities shows that the first two integrals listed above are equal. We can prove that the third and fourth integrals are equal in the same way. To prove that the first and fourth integrals are equal, we first make the substitution φ = θ + 12 π to obtain Z 0 2π 1 dθ = b + a sin θ Z 5π/2 π/2 1 dφ = b + a sin(φ − 12 π) Z 5π/2 π/2 1 dφ b − a cos φ and then note that Z 5π/2 2π 1 dφ = b − a cos φ Z π/2 0 1 dφ b − a cos φ due to the period of cosine. The equality of the second and third integrals is verified in the same way. As a result of the equality of these four integrals, the definite integral that appears as Exercise 6.2.5 gives the value of many related integrals. The key result for Section 6.3 can be put in the following form. It is rather intriguing that integrals involving real-valued functions can be evaluated using complex numbers. Theorem 4: Let P and Q be polynomials with real coefficients such that Q(x) 6= 0 for all x and the degree of Q is at least two more than the degree of P . If the zeros of Q in the upper half plane are z1 , z2 , . . . , zn , then Z ∞ −∞ n X P (x) dx = 2πi Res (P/Q, zk ). Q(x) k=1 Here are a few examples of integrals that could be evaluated using Theorem 4: Z ∞ −∞ x2 dx, x4 + 1 Z ∞ −∞ x2 dx, x6 + 1 Z ∞ −∞ x4 dx, x6 + 1 Z 0 ∞ x4 − 3x2 + 5 dx. x8 + 12 Note that the last integral has a different lower bound. However, since the function is even, we can integrate over the entire real line and then divide by two. You might note that the second integral can be evaluated using simple calculus techniques. 3 Referring to functions f : R → R, recall that f is an even function if f (−x) = f (x) for all x and that f is an odd function if f (−x) = −f (x) for all x. It is then easy to show that Z a Z a f (x) dx = 2 −a Z f (x) dx a and f (x) dx = 0 −a 0 for even and odd functions, respectively. For example, we see that Z ∞ −∞ x2 dx = 2 4 x +1 Z ∞ 0 x2 dx, 4 x +1 Z ∞ −∞ x3 dx = 0, 8 x + 12 Z ∞ −∞ x cos x dx = 0. (x2 + 1)2 It is helpful be keep these facts in mind when doing integrals of these types. Z ∞ Example 5: Evaluate −∞ 3 dx. (x2 + 9)2 Solution: Consider the rational function R(z) = (z 2 3 . The only zero of the denominator in the upper half + 9)2 plane is 3i. Noting that Res (R, 3i) = d 3 dz (z + 3i)2 = z=3i −6 1 = , 3 (6i) 36i we find that (using Theorem 4) Z ∞ −∞ 1 π 3 dx = 2πi · = . (x2 + 9)2 36i 18 You should use this result to verify the value of the integral in Exercise 6.3.5 by making an appropriate substitution and WITHOUT doing any actual integration. Z ∞ Example 6: Evaluate −∞ 1 dx. x6 + 1 1 . The zeros of the denominator in the upper half plane are z6 + 1 z1 = eiπ/6 , z2 = eiπ/2 , and z3 = ei5π/6 . Using Theorem 4 and our quick way to find the residues of simple poles Solution: Consider the rational function R(z) = (Example 2 in Section 6.1), we find that Z ∞ −∞ x6  1 1 1  1 dx = 2πi + 5+ 5 5 +1 6z1 6z2 6z3  πi 1 1 1  = + + 3 z13 z12 z23 z22 z33 z32  πi 1 1 1  = − + 3 iz12 iz22 iz32
π −iπ/3 = e − e−iπ + e−i5π/3 3 √ √  π1 3 1 3 = − i − (−1) + + i 3 2 2 2 2 2π = . 3 The steps involving the cubes of the roots are not necessary (you could go straight to the exponentials); they are included to point out simplifications/observations that are helpful in some cases. 4 The key result for Section 6.4 can be put in the following form. Theorem 7: Let P and Q be polynomials with real coefficients such that Q(x) 6= 0 for all x and the degree of Q is at least one more than the degree of P . If the zeros of Q in the upper half plane are z1 , z2 , . . . , zn , then Z n ∞ −∞ X P (x)eimx dx = 2πi Res (f, zk ). Q(x) k=1 where m is a positive number and f is the function defined by f (z) = eimz P (z)/Q(z). ∞ Z Example 8: Evaluate −∞ cos(3x) dx. x2 + 4 Solution: This is Example 1 in Section 6.4, but we will solve the problem in a simpler way. Note that ∞ Z cos(3x) dx = x2 + 4 −∞ Z ∞ −∞ cos(3x) dx + i x2 + 4 Z ∞ −∞ sin(3x) dx = x2 + 4 Z ∞ −∞ e3ix dx x2 + 4 since sin(3x) is an odd function. Let f (z) = e3iz /(z 2 + 4) and note that the only 0 of the denominator that lies in the upper half plane is 2i. Applying Theorem 7, we find that Z ∞ e3iz e3ix dx = 2πi Res (f, 2i) = 2πi · 2 x +4 z + 2i −∞ = 2πi · z=2i e−6 π = 6. 4i 2e Hence, the value of the requested integral is π/(2e6 ). Z ∞ Example 9: Evaluate −∞ x sin x dx, where a is a positive number. (x2 + a2 )2 Solution: This time, we note that Z ∞ −∞ x sin x dx = (x2 + a2 )2 Z ∞ −∞ x sin x dx − i (x2 + a2 )2 Z ∞ −∞ x cos x dx = (x2 + a2 )2 Z ∞ −∞ −ixeix dx (x2 + a2 )2 since x cos x is an odd function. Let f (z) = −izeiz /(z 2 + a2 )2 and note that the only 0 of the denominator that lies in the upper half plane is ai. Applying Theorem 7, we find that Z ∞ −∞ x sin x dx = (x2 + a2 )2 Z ∞ −∞ −ixeix dx (x2 + a2 )2 = 2πi Res (f, ai) = 2π · d zeiz dz (z + ai)2 2 = 2π · (2ai) ai(ie z=ai −a
) + e−a − aie−a 2(2ai) (2ai)4 2π · (−a + 1 − 1) (2ai)2 ea π = . 2aea = Finding integral results involving a parameter such as this one can be helpful. 5 Theorem 10: (Rouché’s Theorem) Suppose that f and g are analytic on and inside a simple closed contour C. If |g(z)| < |f (z)| for all z ∈ C, then f and f + g have the same number of zeros inside C. Proof: For each real number t ∈ [0, 1], let N (t) = 1 2πi I C f 0 (z) + tg 0 (z) dz. f (z) + tg(z) Since |f (z) + tg(z)| ≥ |f (z)| − |tg(z)| ≥ |f (z)| − |g(z)| > 0 for all t ∈ [0, 1] and all z ∈ C, the Argument Principle (see Corollary 1 in Section 6.7) tells us that N (t) is the number of zeros of f (z) + tg(z) inside C. (Note that the hypotheses reveal that f (z) + tg(z) does not have any poles inside C.) Since the function N (t) is integer-valued and continuous, we must have N (0) = N (1). It follows that f and f + g have the same number of zeros inside C. Example 11: Show that all of the roots of p(z) = z 7 + 4z 5 − 3z 4 + 6z 2 − z + i lie within the circle |z| = 3. Solution: Let f (z) = z 7 and g(z) = 4z 5 − 3z 4 + 6z 2 − z + i. Both f and g are entire functions and f has seven roots (counting multiplicities) within the circle |z| = 3. On the circle |z| = 3, we find that |g(z)| ≤ 4|z|5 + 3|z|4 + 6|z|2 + |z| + |i| = 4 · 35 + 35 + 6 · 32 + 3 + 1 < 5 · 35 + 3 5 < 37 = |f (z)|. By Rouché’s Theorem, the functions f and p = f + g have the same number of zeros inside the circle |z| = 3. Hence, all seven roots of the polynomial p lie inside the circle |z| = 3. Example 12: Show that q(z) = z 4 − (2 + i)z + 25 has no roots inside the circle |z| = 2. Solution: Let f (z) = 25 and g(z) = z 4 − (2 + i)z. Both f and g are entire functions and f has no roots inside the circle |z| = 2. On the circle |z| = 2, we find that √ |g(z)| ≤ |z|4 + |2 + i| |z| = 16 + 2 5 < 25 = |f (z)|. By Rouché’s Theorem, the functions f and q = f + g have the same number of zeros inside the circle |z| = 2. Hence, the polynomial q has no roots inside the circle |z| = 2. 6 Recall that a domain is an open connected set. (You should review the terms ‘open’ and ‘connected’ if you do not remember them.) If f is analytic on a domain D, then the set f (D) = {f (z) : z ∈ D} is the range of f . We sometimes refer to f (D) as the image of D under the mapping f . For functions f : C → C, both D and f (D) are subsets of the complex plane. On the real line, the only connected sets are intervals. However, for functions f : R → R the image of an open interval may not be an open interval. For example, the function f (x) = 1 − x2 maps the open interval (−1, 1) to the half-open interval (0, 1], and the function g(x) = sin x maps the open interval (0, 8) to the closed interval [−1, 1]. This situation does not occur for complex valued functions Theorem 13: (Open Mapping Theorem) If f is analytic and nonconstant in a domain D, then f (D) is a domain. Proof: To show that f (D) is connected, let w1 and w2 be two points in f (D). Since w1 and w2 are in the range of f , we can select points z1 and z2 in D for which f (z1 ) = w1 and f (z2 ) = w2 . Since D is a connected set, there exists a path γ in D that joins the points z1 and z2 . It then follows that f ◦ γ is a path in f (D) that joins the points w1 and w2 . (We are using the fact that the composition of two continuous functions is continuous and that f (γ(t)) ∈ f (D) for all t values.) Hence, the set f (D) is connected. To show that f (D) is open, we must show that all of its points are interior points. Let w0 ∈ f (D) and choose z0 ∈ D so that f (z0 ) = w0 . Since the zeros of analytic functions are isolated (apply Corollary 3 in Section 5.6 to the function f (z) − w0 ), there exists ρ > 0 such that f (z) 6= w0 for all z that satisfy 0 < |z − z0 | ≤ ρ. Since D is an open set, we may also assume that ρ has been chosen small enough so that every z that satisfies |z − z0 | ≤ ρ belongs to D. Now let δ = min{|f (z) − w0 | : |z − z0 | = ρ}. Since each of the numbers |f (z) − w0 | in this set is positive, it can be shown that the number δ is positive. Suppose that w1 is a complex number that satisfies |w1 − w0 | < δ. Noting that |w0 − w1 | < δ ≤ |f (z) − w0 | for all z that satisfy |z − z0 | = ρ, Rouché’s Theorem tells us that the functions f (z) − w0 and (f (z) − w0 ) + (w0 − w1 ) = f (z) − w1 have the same number of zeros inside the circle |z − z0 | = ρ. In particular, there exists at least one point z1 ∈ D such that f (z1 ) = w1 . This shows that w1 ∈ f (D). Since w1 was an arbitrary point in the set |w − w0 | < δ, we see that there is a disk around w0 that lies inside f (D). This shows that w0 is an interior point of f (D). We conclude that f (D) is an open set. 7