11.1 Cauchy’s Integral Formula Result 11.1.1 Cauchy’s Integral Formula If f (ζ) is analytic in a compact, closed, connected domain D and z is a point in the interior of D then f (z) = 1 ı2π ∂D f (ζ) 1 dζ = ζ −z ı2π Ck k f (ζ) dζ ζ −z (11.1) Here the set of contours {Ck } make up the positively oriented boundary ∂D of the domain D More generally, we have f (n) (z) = n! ı2π ∂D n! f (ζ) dζ = (ζ − z)n+1 ı2π k Ck f (ζ) dζ (ζ − z)n+1 (11.2) Cauchy’s Formula shows that the value of f (z) and all its derivatives in a domain are determined by the value of f (z) on the boundary of the domain Consider the first formula of the result, Equation 11.1 We deform the contour to a circle of radius δ about the point ζ = z C f (ζ) dζ = ζ −z Cδ = Cδ f (ζ) dζ ζ −z f (z) dζ + ζ −z Cδ f (ζ) − f (z) dζ ζ −z We use the result of Example 10.8.1 to evaluate the first integral C f (ζ) dζ = ı2πf (z) + ζ −z 494 Cδ f (ζ) − f (z) dζ ζ −z The remaining integral along Cδ vanishes as δ → 0 because f (ζ) is continuous We demonstrate this with the maximum modulus integral bound The length of the path of integration is 2πδ lim δ→0 Cδ f (ζ) − f (z) 1 dζ ≤ lim (2πδ) max |f (ζ) − f (z)| δ→0 ζ −z δ |ζ−z|=δ ≤ lim 2π max |f (ζ) − f (z)| |ζ−z|=δ δ→0 =0 This gives us the desired result f (z) = 1 ı2π C f (ζ) dζ ζ −z We derive the second formula, Equation 11.2, from the first by differentiating with respect to z Note that the integral converges uniformly for z in any closed subset of the interior of C Thus we can differentiate with respect to z and interchange the order of differentiation and integration f (ζ) 1 dn dζ n ı2π dz C ζ − z dn f (ζ) 1 dζ = ı2π C dz n ζ − z n! f (ζ) = dζ ı2π C (ζ − z)n+1 f (n) (z) = Example 11.1.1 Consider the following integrals where C is the positive contour on the unit circle For the third integral, the point z = −1 is removed from the contour sin cos z 5 1 dz C 2 C 1 dz (z − 3)(3z − 1) 495 √ 3 z dz C 1 Since sin (cos (z 5 )) is an analytic function inside the unit circle, sin cos z 5 dz = 0 C 2 1 (z−3)(3z−1) has singularities at z = 3 and z = 1/3 Since z = 3 is outside the contour, only the singularity at z = 1/3 will contribute to the value of the integral We will evaluate this integral using the Cauchy integral formula 1 1 ıπ dz = ı2π =− (1/3 − 3)3 4 C (z − 3)(3z − 1) √ 3 Since the curve is not closed, we cannot apply the Cauchy integral formula Note that z is single-valued and analytic in the complex plane with a branch cut on the negative real axis Thus we use the Fundamental Theorem of Calculus √ C eıπ 2√ 3 z dz = z 3 e−ıπ 2 ı3π/2 e = − e−ı3π/2 3 2 = (−ı − ı) 3 4 = −ı 3 Cauchy’s Inequality Suppose the f (ζ) is analytic in the closed disk |ζ − z| ≤ r By Cauchy’s integral formula, f (n) (z) = n! ı2π C f (ζ) dζ, (ζ − z)n+1 496 where C is the circle of radius r centered about the point z We use this to obtain an upper bound on the modulus of f (n) (z) n! f (ζ) dζ 2π C (ζ − z)n+1 n! f (ζ) ≤ 2πr max |ζ−z|=r (ζ − z)n+1 2π n! = n max |f (ζ)| r |ζ−z|=r f (n) (z) = Result 11.1.2 Cauchy’s Inequality If f (ζ) is analytic in |ζ − z| ≤ r then f (n) (z) ≤ n!M rn where |f (ζ)| ≤ M for all |ζ − z| = r Liouville’s Theorem Consider a function f (z) that is analytic and bounded, (|f (z)| ≤ M ), in the complex plane From Cauchy’s inequality, M |f (z)| ≤ r for any positive r By taking r → ∞, we see that f (z) is identically zero for all z Thus f (z) is a constant Result 11.1.3 Liouville’s Theorem If f (z) is analytic and |f (z)| is bounded in the complex plane then f (z) is a constant The Fundamental Theorem of Algebra We will prove that every polynomial of degree n ≥ 1 has exactly n roots, counting multiplicities First we demonstrate that each such polynomial has at least one root Suppose that an 497 nth degree polynomial p(z) has no roots Let the lower bound on the modulus of p(z) be 0 < m ≤ |p(z)| The function f (z) = 1/p(z) is analytic, (f (z) = p (z)/p2 (z)), and bounded, (|f (z)| ≤ 1/m), in the extended complex plane Using Liouville’s theorem we conclude that f (z) and hence p(z) are constants, which yields a contradiction Therefore every such polynomial p(z) must have at least one root Now we show that we can factor the root out of the polynomial Let n pk z k p(z) = k=0 We note that n−1 n n cn−1−k z k (z − c ) = (z − c) k=0 Suppose that the n th degree polynomial p(z) has a root at z = c p(z) = p(z) − p(c) n n pk z k − = k=0 n pk ck k=0 pk z k − ck = k=0 n k−1 ck−1−j z j pk (z − c) = j=0 k=0 = (z − c)q(z) Here q(z) is a polynomial of degree n − 1 By induction, we see that p(z) has exactly n roots Result 11.1.4 Fundamental Theorem of Algebra Every polynomial of degree n ≥ 1 has exactly n roots, counting multiplicities 498 Gauss’ Mean Value Theorem Let f (ζ) be analytic in |ζ − z| ≤ r By Cauchy’s integral formula, 1 ı2π f (z) = C f (ζ) dζ, ζ −z where C is the circle |ζ − z| = r We parameterize the contour with ζ = z + r eıθ f (z) = 2π 1 ı2π 0 f (z + r eıθ ) ıθ ır e dθ r eıθ Writing this in the form, 1 f (z) = 2πr 2π f (z + r eıθ )r dθ, 0 we see that f (z) is the average value of f (ζ) on the circle of radius r about the point z Result 11.1.5 Gauss’ Average Value Theorem If f (ζ) is analytic in |ζ − z| ≤ r then 1 f (z) = 2π 2π f (z + r eıθ ) dθ 0 That is, f (z) is equal to its average value on a circle of radius r about the point z Extremum Modulus Theorem Let f (z) be analytic in closed, connected domain, D The extreme values of the modulus of the function must occur on the boundary If |f (z)| has an interior extrema, then the function is a constant We will show this with proof by contradiction Assume that |f (z)| has an interior maxima at the point z = c This means that there exists an neighborhood of the point z = c for which |f (z)| ≤ |f (c)| Choose an so that the set |z − c| ≤ lies inside this neighborhood First we use Gauss’ mean value theorem f (c) = 1 2π 2π f c + eıθ dθ 0 499 We get an upper bound on |f (c)| with the maximum modulus integral bound 1 |f (c)| ≤ 2π 2π f c + eıθ dθ 0 Since z = c is a maxima of |f (z)| we can get a lower bound on |f (c)| |f (c)| ≥ 1 2π 2π f c + eıθ dθ 0 If |f (z)| < |f (c)| for any point on |z −c| = , then the continuity of f (z) implies that |f (z)| < |f (c)| in a neighborhood of that point which would make the value of the integral of |f (z)| strictly less than |f (c)| Thus we conclude that |f (z)| = |f (c)| for all |z − c| = Since we can repeat the above procedure for any circle of radius smaller than , |f (z)| = |f (c)| for all |z − c| ≤ , i.e all the points in the disk of radius about z = c are also maxima By recursively repeating this procedure points in this disk, we see that |f (z)| = |f (c)| for all z ∈ D This implies that f (z) is a constant in the domain By reversing the inequalities in the above method we see that the minimum modulus of f (z) must also occur on the boundary Result 11.1.6 Extremum Modulus Theorem Let f (z) be analytic in a closed, connected domain, D The extreme values of the modulus of the function must occur on the boundary If |f (z)| has an interior extrema, then the function is a constant 500 11.2 The Argument Theorem Result 11.2.1 The Argument Theorem Let f (z) be analytic inside and on C except for isolated poles inside the contour Let f (z) be nonzero on C 1 ı2π C f (z) dz = N − P f (z) Here N is the number of zeros and P the number of poles, counting multiplicities, of f (z) inside C First we will simplify the problem and consider a function f (z) that has one zero or one pole Let f (z) be analytic and nonzero inside and on A except for a zero of order n at z = a Then we can write f (z) = (z − a)n g(z) where g(z) (z) is analytic and nonzero inside and on A The integral of f (z) along A is f 1 ı2π A f (z) 1 dz = f (z) ı2π 1 = ı2π 1 = ı2π 1 = ı2π =n A A A A d (log(f (z))) dz dz d (log((z − a)n ) + log(g(z))) dz dz d (log((z − a)n )) dz dz n dz z−a 501 Now let f (z) be analytic and nonzero inside and on B except for a pole of order p at z = b Then we can write g(z) (z) f (z) = (z−b)p where g(z) is analytic and nonzero inside and on B The integral of f (z) along B is f 1 ı2π B f (z) 1 dz = f (z) ı2π 1 = ı2π 1 = ı2π 1 = ı2π = −p B B B B d (log(f (z))) dz dz d log((z − b)−p ) + log(g(z)) dz dz d log((z − b)−p )+ dz dz −p dz z−b Now consider a function f (z) that is analytic inside an on the contour C except for isolated poles at the points b1 , , bp Let f (z) be nonzero except at the isolated points a1 , , an Let the contours Ak , k = 1, , n, be simple, positive contours which contain the zero at ak but no other poles or zeros of f (z) Likewise, let the contours Bk , k = 1, , p be simple, positive contours which contain the pole at bk but no other poles of zeros of f (z) (See Figure 11.1.) By deforming the contour we obtain C f (z) dz = f (z) p n j=1 Aj f (z) dz + f (z) k=1 Bj f (z) dz f (z) From this we obtain Result 11.2.1 11.3 Rouche’s Theorem Result 11.3.1 Rouche’s Theorem Let f (z) and g(z) be analytic inside and on a simple, closed contour C If |f (z)| > |g(z)| on C then f (z) and f (z) + g(z) have the same number of zeros inside C and no zeros on C 502 From this we see that the integral along C is equal to the sum of the integrals along C1 , C2 and C3 (We could also see this by deforming C onto C1 , C2 and C3 ) C z3 z dz = −9 C1 z3 z dz + −9 C2 z3 z dz + −9 C3 z3 z dz −9 We use the Cauchy Integral Formula to evaluate the integrals along C1 , C2 and C2 C z dz = z3 − 9 C1 z− + √ 3 C2 z− C3 z− + = ı2π + ı2π + ı2π z− √ 3 √ 3 √ 3 z− z− z √ dz 9 eı2π/3 z − 3 9 e−ı2π/3 z √ √ dz 3 ı2π/3 z − 9e z − 3 9 e−ı2π/3 z √ √ dz 3 z − 9 eı2π/3 z − 3 9 e−ı2π/3 z− 9 9 9 √ 3 z 9 eı2π/3 √ 3 √ 3 z− √ 3 9 e−ı2π/3 9 z √ z − 3 9 e−ı2π/3 9 z √ z − 3 9 eı2π/3 √ z= 3 9 √ z= 3 9 eı2π/3 √ z= 3 9 e−ı2π/3 = ı2π3−5/3 1 − eıπ/3 + eı2π/3 =0 2 The integrand has singularities at z = 0 and z = 4 Only the singularity at z = 0 lies inside the contour We use 518 4 2 C2 -4 -6 C C1 -2 C3 2 4 6 -2 -4 Figure 11.4: The contours for z z 3 −9 the Cauchy Integral Formula to evaluate the integral C d sin z sin z dz = ı2π − 4) dz z − 4 z=0 cos z sin z = ı2π − z − 4 (z − 4)2 ıπ =− 2 z 2 (z z=0 3 We factor the integrand to see that there are singularities at z = 0 and z = −ı C (z 3 + z + ı) sin z dz = z 4 + ız 3 C (z 3 + z + ı) sin z dz z 3 (z + ı) Let C1 and C2 be contours around z = 0 and z = −ı See Figure 11.5 Let D be the domain between C, C1 and C2 , i.e the boundary of D is the union of C, −C1 and −C2 Since the integrand is analytic in D, the integral along the boundary of D vanishes = ∂D + + −C1 C 519 =0 −C2 From this we see that the integral along C is equal to the sum of the integrals along C1 and C2 (We could also see this by deforming C onto C1 and C2 ) = C + C1 C2 We use the Cauchy Integral Formula to evaluate the integrals along C1 and C2 C (z 3 + z + ı) sin z (z 3 + z + ı) sin z dz + dz z 3 (z + ı) z 3 (z + ı) C1 C2 (z 3 + z + ı) sin z ı2π d2 (z 3 + z + ı) sin z = ı2π + z3 2! dz 2 z+ı z=−ı (z 3 + z + ı) sin z dz = z 4 + ız 3 2 z=0 3 3z + 1 z + z + ı − cos z z+ı (z + ı)2 6z 2(3z 2 + 1) 2(z 3 + z + ı) z 3 + z + ı − + − + z+ı (z + ı)2 (z + ı)3 z+ı = 2π sinh(1) = ı2π(−ı sinh(1)) + ıπ 2 sin z z=0 4 We consider the integral C ezt dz z 2 (z + 1) There are singularities at z = 0 and z = −1 Let C1 and C2 be contours around z = 0 and z = −1 See Figure 11.6 We deform C onto C1 and C2 = C + C1 520 C2 4 2 -4 C1 C2 2 -2 C 4 -2 -4 Figure 11.5: The contours for (z 3 +z+ı) sin z z 4 +ız 3 We use the Cauchy Integral Formula to evaluate the integrals along C1 and C2 C ezt dz = z 2 (z + 1) ezt ezt dz + dz 2 2 C1 z (z + 1) C1 z (z + 1) ezt d ezt = ı2π 2 + ı2π z z=−1 dz (z + 1) z=0 zt ezt te = ı2π e−t +ı2π − (z + 1) (z + 1)2 z=0 = ı2π(e−t +t − 1) Solution 11.11 Liouville’s Theorem states that if f (z) is analytic and bounded in the complex plane then f (z) is a constant 521 2 1 C2 -2 C1 -1 1 C 2 -1 -2 Figure 11.6: The contours for ezt z 2 (z+1) 1 Since f (z) is analytic, ef (z) is analytic The modulus of ef (z) is bounded ef (z) = e (f (z)) ≤ eM By Liouville’s Theorem we conclude that ef (z) is constant and hence f (z) is constant 2 We know that f (z) is entire and |f (5) (z)| is bounded in the complex plane Since f (z) is analytic, so is f (5) (z) We apply Liouville’s Theorem to f (5) (z) to conclude that it is a constant Then we integrate to determine the form of f (z) f (z) = c5 z 5 + c4 z 4 + c3 z 3 + c2 z 2 + c1 z + c0 Here c5 is the value of f (5) (z) and c4 through c0 are constants of integration We see that f (z) is a polynomial of degree at most five Solution 11.12 For this problem we will use the Extremum Modulus Theorem: Let f (z) be analytic in a closed, connected domain, D The extreme values of the modulus of the function must occur on the boundary If |f (z)| has an interior extrema, then the function is a constant 522 Since |f (z)| has an interior extrema, |f (0)| = | eı | = 1, we conclude that f (z) is a constant on D Since we know the value at z = 0, we know that f (z) = eı Solution 11.13 First we determine the radius of convergence of the series with the ratio test k 4 /4k k→∞ (k + 1)4 /4k+1 k4 = 4 lim k→∞ (k + 1)4 24 = 4 lim k→∞ 24 =4 R = lim The series converges absolutely for |z| < 4 1 Since the integrand is analytic inside and on the contour of integration, the integral vanishes by Cauchy’s Theorem 2 C f (z) dz = z3 ∞ k4 C k=0 ∞ = C k=1 ∞ = z 4 k 1 dz z3 k 4 k−3 z dz 4k (k + 3)4 k z dz 4k+3 C k=−2 = C 1 dz + 4z 2 523 C 1 dz + z ∞ C k=0 (k + 3)4 k z dz 4k+3 We can parameterize the first integral to show that it vanishes The second integral has the value ı2π by the Cauchy-Goursat Theorem The third integral vanishes by Cauchy’s Theorem as the integrand is analytic inside and on the contour f (z) dz = ı2π 3 C z 524 Chapter 12 Series and Convergence You are not thinking You are merely being logical - Neils Bohr 12.1 Series of Constants 12.1.1 Definitions Convergence of Sequences The infinite sequence {an }∞ ≡ a0 , a1 , a2 , is said to converge if n=0 lim an = a n→∞ for some constant a If the limit does not exist, then the sequence diverges Recall the definition of the limit in the above formula: For any > 0 there exists an N ∈ Z such that |a − an | < for all n > N Example 12.1.1 The sequence {sin(n)} is divergent The sequence is bounded above and below, but boundedness does not imply convergence 525 Cauchy Convergence Criterion Note that there is something a little fishy about the above definition We should be able to say if a sequence converges without first finding the constant to which it converges We fix this problem with the Cauchy convergence criterion A sequence {an } converges if and only if for any > 0 there exists an N such that |an − am | < for all n, m > N The Cauchy convergence criterion is equivalent to the definition we had before For some problems it is handier to use Now we don’t need to know the limit of a sequence to show that it converges Convergence of Series The series That is, ∞ n=1 an converges if the sequence of partial sums, SN = N −1 n=0 an , converges N −1 lim SN = lim N →∞ N →∞ an = constant n=0 If the limit does not exist, then the series diverges A necessary condition for the convergence of a series is that lim an = 0 n→∞ (See Exercise 12.1.) Otherwise the sequence of partial sums would not converge Example 12.1.2 The series ∞ (−1)n = 1 − 1 + 1 − 1 + · · · is divergent because the sequence of partial sums, n=0 {SN } = 1, 0, 1, 0, 1, 0, is divergent Tail of a Series An infinite series, ∞ an , converges or diverges with its tail That is, for fixed N , ∞ an n=0 n=0 converges if and only if ∞ an converges This is because the sum of the first N terms of a series is just a number n=N Adding or subtracting a number to a series does not change its convergence Absolute Convergence The series ∞ an converges absolutely if ∞ |an | converges Absolute convergence n=0 n=0 implies convergence If a series is convergent, but not absolutely convergent, then it is said to be conditionally convergent The terms of an absolutely convergent series can be rearranged in any order and the series will still converge to the same sum This is not true of conditionally convergent series Rearranging the terms of a conditionally convergent 526 series may change the sum In fact, the terms of a conditionally convergent series may be rearranged to obtain any desired sum Example 12.1.3 The alternating harmonic series, 1− 1 1 1 + − + ··· , 2 3 4 converges, (Exercise 12.4) Since 1 1 1 + + + ··· 2 3 4 diverges, (Exercise 12.5), the alternating harmonic series is not absolutely convergent Thus the terms can be rearranged to obtain any sum, (Exercise 12.6) 1+ Finite Series and Residuals Consider the series f (z) = terms in the series as ∞ n=0 an (z) We will denote the sum of the first N N −1 an (z) SN (z) = n=0 We will denote the residual after N terms as ∞ RN (z) ≡ f (z) − SN (z) = an (z) n=N 12.1.2 Special Series Geometric Series One of the most important series in mathematics is the geometric series, 1 ∞ zn = 1 + z + z2 + z3 + · · · n=0 1 The series is so named because the terms grow or decay geometrically Each term in the series is a constant times the previous term 527 The series clearly diverges for |z| ≥ 1 since the terms do not vanish as n → ∞ Consider the partial sum, SN (z) ≡ N −1 n n=0 z , for |z| < 1 N −1 zn (1 − z)SN (z) = (1 − z) n=0 N N −1 zn − = n=0 zn n=1 = 1 + z + · · · + z N −1 − z + z 2 + · · · + z N = 1 − zN N −1 1 − zN 1 → 1−z 1−z zn = n=0 The limit of the partial sums is 1 1−z ∞ zn = n=0 1 1−z as N → ∞ for |z| < 1 Harmonic Series Another important series is the harmonic series, ∞ n=1 1 1 1 = 1 + α + α + ··· α n 2 3 The series is absolutely convergent for (α) > 1 and absolutely divergent for Riemann zeta function ζ(α) is defined as the sum of the harmonic series ∞ ζ(α) = n=1 528 1 nα (α) ≤ 1, (see the Exercise 12.8) The The alternating harmonic series is ∞ n=1 (−1)n+1 1 1 1 = 1 − α + α − α + ··· α n 2 3 4 Again, the series is absolutely convergent for 12.1.3 (α) > 1 and absolutely divergent for (α) ≤ 1 Convergence Tests The Comparison Test Result 12.1.1 The series of positive terms an converges if there exists a convergent series bn such that an ≤ bn for all n Similarly, an diverges if there exists a divergent series bn such that an ≥ bn for all n Example 12.1.4 Consider the series ∞ n=1 We can rewrite this as 1 2n2 ∞ n=1 n a perfect square 1 2n Then by comparing this series to the geometric series, ∞ n=1 1 = 1, 2n we see that it is convergent 529 Integral Test Result 12.1.2 If the coefficients an of a series ∞ an are monotonically decreasing and n=0 can be extended to a monotonically decreasing function of the continuous variable x, for x ∈ Z0+ , a(x) = an then the series converges or diverges with the integral ∞ a(x) dx 0 Example 12.1.5 Consider the series ∞ 1 n=1 n2 Define the functions sl (x) and sr (x), (left and right), sl (x) = 1 , ( x )2 sr (x) = 1 ( x )2 Recall that x is the greatest integer function, the greatest integer which is less than or equal to x x is the least integer function, the least integer greater than or equal to x We can express the series as integrals of these functions ∞ n=1 1 = n2 ∞ ∞ sl (x) dx = 0 sr (x) dx 1 In Figure 12.1 these functions are plotted against y = 1/x2 From the graph, it is clear that we can obtain a lower and 530 upper bound for the series ∞ 1 1 dx ≤ x2 ∞ n=1 1 ≤1+ n2 ∞ 1≤ n=1 ∞ 1 1 dx x2 1 ≤2 n2 1 1 1 2 3 4 1 Figure 12.1: Upper and Lower bounds to In general, we have ∞ ∞ a(x) dx ≤ m 2 ∞ n=1 3 4 1/n2 ∞ an ≤ am + a(x) dx m n=m Thus we see that the sum converges or diverges with the integral The Ratio Test Result 12.1.3 The series an converges absolutely if lim n→∞ an+1 < 1 an If the limit is greater than unity, then the series diverges If the limit is unity, the test fails 531 If the limit is greater than unity, then the terms are eventually increasing with n Since the terms do not vanish, the sum is divergent If the limit is less than unity, then there exists some N such that an+1 ≤ r < 1 for all n ≥ N an From this we can show that ∞ n=0 an is absolutely convergent by comparing it to the geometric series ∞ ∞ rn |an | ≤ |aN | n=0 n=N = |aN | 1 1−r Example 12.1.6 Consider the series, ∞ n=1 en n! We apply the ratio test to test for absolute convergence lim n→∞ en+1 n! an+1 = lim n n→∞ e (n + 1)! an e = lim n→∞ n + 1 =0 The series is absolutely convergent Example 12.1.7 Consider the series, ∞ n=1 1 , n2 532 ... continuously deformed to C2 on the domain where the integrand is analytic Thus the integrals have the same value 5 14 -4 -2 -2 -4 Figure 11 .2: The contours and the singularities of 3z +1 -6 -4 -2 -2 -4 -6... , C2 and C2 C z dz = z3 − C1 z− + √ C2 z− C3 z− + = ? ?2? ? + ? ?2? ? + ? ?2? ? z− √ √ √ z− z− z √ dz e? ?2? ?/3 z − e−? ?2? ?/3 z √ √ dz ? ?2? ?/3 z − 9e z − e−? ?2? ?/3 z √ √ dz z − e? ?2? ?/3 z − e−? ?2? ?/3 z− 9 √ z e? ?2? ?/3... deform C onto C1 and C2 = C + C1 520 C2 -4 C1 C2 -2 C -2 -4 Figure 11.5: The contours for (z +z+ı) sin z z +ız We use the Cauchy Integral Formula to evaluate the integrals along C1 and C2