Chapter 4 Applications

Chapter 4 Applications 4.1 Singularities Let f be analytic on U \ {z0 }, where U is an open subset of C and z0 ∈ U . In this case z0 is said to be an isolated singularity of f . The purpose of this section is to determine the behavior of f near z0 . Theorem 4.1.1 (Cauchy’s Formula for an Annulus). Let f be analytic on an open set U containing the annulus r1 ≤ |z − z0 | ≤ r2 , where 0 < r1 < r2 < ∞. Let Γi = |z − z0 | = ri , i = 1, 2, oriented counterclockwise. Then for r1 < |z − z0 | < r2 , Z 1 f (w) f (z) = dw. 2πi Γ2 −Γ1 w − z Proof. By Cauchy’s Integral Formula, 1 f (z) ind(Γ2 − Γ1 ; z0 ) = 2πi Z Γ2 −Γ1 f (w) dw. w−z By the properties of the index, if r1 < |z − z0 | < r2 , ind(Γ2 − Γ1 ; z) = ind(Γ2 ; z) − ind(Γ1 ; z) = 1 − 0 = 1. Theorem 4.1.2 (Laurent Series). Let f be analytic on U = {s1 < |z − z0 | < s2 }, where 0 ≤ s1 < s2 ≤ ∞. Then f is given by f (z) = ∞ X n=−∞ where an = 1 2πi an (z − z0 )n , Z Γ z ∈ U, f (w) dw (w − z0 )n+1 and Γ is any circle of radius r, s1 < r < s2 , and center z0 . The series converges absolutely on U , and converges uniformly on compact subsets. Proof. Choose s1 < r1 < r2 < s2 , and let Γ1 and Γ2 be the circles |w − z0 | = r1 and |w − z0 | = r2 , respectively. By Cauchy’s Integral Formula for an Annulus (Theorem 4.1.1) applied to a point z such that r1 < |z − z0 | < r2 , Z Z 1 f (w) 1 f (w) f (z) = dw − dw. 2πi Γ2 w − z 2πi Γ1 w − z If w ∈ Γ2 , then ¯ ¯ ¯ z − z0 ¯ |z − z0 | ¯ ¯ < 1, ¯ w − z0 ¯ = r2 so the series 1 = w−z ∞ X (z − z0 )n 1 µ ¶= z − z0 (w − z0 )n+1 n=0 (w − z0 ) 1 − w − z0 converges absolutely and uniformly on compact subsets in |w − z0 | < r2 , hence uniformly on 1 Γ2 . By multiplying by the bounded function f (w) (which preserves uniform convergence) 2πi and integrating terms by term, we obtain 1 2πi where 1 an = 2πi Z Z ∞ Γ2 Γ2 X f (w) dw = an (z − z0 )n w−z n=0 f (w) dw, (w − z0 )n−1 n = 0, 1, 2, · · · The second integral is treated similarly. If w ∈ Γ1 , then ¯ ¯ ¯ w − z0 ¯ r1 ¯ ¯ ¯ z − z0 ¯ ≤ |z − z0 | < 1 and the series ∞ X (w − z0 )n−1 −1 = w−z (z − z0 )n n=1 converges uniformly on Γ2 . Multiplying by − where 1 bn = 2πi 1 f (w) and integrating term by term, 2πi ∞ 1 2πi Z Z f (w)(w − z0 )n−1 dw, Γ2 Γ1 X f (w) bn dw = w−z (z − z0 )n n=0 n = 1, 2, · · · Replacing the index n = 1, 2, · above by −n = −1, −2, · · · and writing Z 1 an = b−n = f (w)(w − z0 )−n−1 dw 2πi Γ1 for n = −1, −2, · · · we obtain the expansion − 1 2πi Z Γ1 −∞ X f (w) an (z − z0 )n . dw = w−z n=−1 Example 4.1.3. Replacing z by 1/z in the power series ez = 1 + 1 1 z + z2 + · · · 1! 2! |z| < ∞ we have the Laurent series e1/z = 1 + 11 1 1 + + ··· 1! z 2! z 2 0 < |z| < ∞ Note that the series contains no positive powers of z, and that it has an infinite number of negative powers. Note that the coefficient a−1 = 1; and according to Laurent’s Series theorem, that coefficient is given by Z 1 a−1 = e1/z dz 2πi Γ where Γ is any positive oriented circle at the origin. Therefore, Z e1/z dz = 2πi. Γ This method of evaluating certain integrals will be developed further in subsequent sections. Example 4.1.4. The function f (z) = 1/(z − i)2 is already in the form of a Laurent series where z0 = i. That is f (z) = ∞ X −∞ an (z − i)n , 0 < |z − i| < ∞, where a−2 = 1 and all other coefficients are zero. From Laurent series theorem we conclude that if Γ is a positively oriented circle |z − i| = r, then ½ Z 1 0, n 6= −2 dz = n+2 2πi, n = −2 (z − i) Γ Example 4.1.5. The function f (z) = −1 1 1 = − (z − 1)(z − 2) z−1 z−2 is analytic on C \ {1, 2}. It is analytic on the annuli A1 = {|z| < 1}, A2 = {1 < |z| < 2} and A3 = {2 < |z| < ∞}, and it has a Laurent series on each of them. Their representation can be fund with the help of the representation ∞ X 1 = zn 1−z n=0 |z| < 1. To find the representation on A1 , we write f (z) = − µ ¶ 1 1 1 + 1 − z 2 1 − (z/2) and note that |z| < 1 and |z/2| < 1 on A1 , f (z) = − ∞ X zn + ∞ X n=0 n=0 1 2 zn = n+1 ∞ X n=0 (2−1−n − 1)z n for |z| < 1. As for the representation on A2 , we write µ ¶ µ ¶ 1 1 1 1 f (z) = + . z 1 − (1/z) 2 1 − (z/2) Since |1/z| < 1 and |z/2| < 1 if 1 < |z| < 2, it follows that f (z) = ∞ X n=0 1 z n+1 + ∞ X 1 2n+1 n=0 zn 1 < |z| < 2. By replacing n by n − 1 in the first series, and then interchange them, we obtain f (z) = ∞ X n=0 1 2n+1 n z + −1 X zn n=−∞ 1 < |z| < 2. This must be the Laurent series for f on the annulus A2 because there is at most one such series. The representation in A2 is obtained in a similar way. First write µ ¶ µ ¶ 1 1 1 1 f (z) = + , z 1 − (1/z) z 1 − (2/z) and note that if 2 < |z| < ∞, then |1/z| < 1 and |2/z| < 1. Therefore, f (z) = ∞ X n=0 1 z n+1 − ∞ X 2n , z n+1 n=0 2 < |z| < ∞ That is f (z) = 1 X n=−∞ (1 − 2n−1 )z n , 2 < |z| < ∞. f (n) (z0 ) for n! n ≥ 0, because f is only analytic on s1 < |z − z0 | < s2 and may not have an analytic extension to |z − z0 | < s2 . Remark. The coefficient an of the Laurent expansion of f is not necessarily Theorem 4.1.6. Suppose that f (z) = 0 ≤ s1 < s2 ≤ ∞. Then bn is given by P∞ n=−∞ bn (z 1 bn = 2πi Z Γ − z0 )n on U = {s1 < |z − z0 | < s2 }, f (w) dw. (w − z0 )n+1 That is, the Laurent expansion of f on U is unique. Proof. The power series representing f converges uniformly on compact subsets of U . Mul1 and integrate over Γ to obtain tiply both sides by (z − z0 )k+1 Z Z ∞ X f (z) dz = bn (z − z0 )n−k−1 dz k+1 (z − z ) 0 Γ Γ n=−∞ = bk by the argument in ??. Definition 4.1.7. Let f have an isolated singularity at z0 , so that f has a Laurent expansion about z0 of the form ∞ X f (z) = an (z − z0 )n n=−∞ valid for 0 < |z − z0 | < r, for some r > 0. The sum of the negative powers of the Laurent series, that is, −1 X n=−∞ an (z − z0 )n is called the principal part of the Laurent expansion of f about z0 . If the Laurent expansion contains no negative powers of (z − z0 ), then f is said to have a removable singularity at z0 . in this case, f can be extended to z0 by setting f (z0 ) = a0 . If the principal part has finitely many non-zero term, that is, if ∞ f (z) = X a−m a−1 + ··· + + an (z − z0 )n m (z − z0 ) (z − z0 ) n=1 and a−m 6= 0, then f is said to have a pole of order m at z0 ( a simple pole if m = 1). In this case (z−z0 )m f (z) has a removable singularity at z0 , and lim (z − z0 )m f (z) = a−m 6= 0. z→z0 In this case, setting f (z0 ) = ∞ we obtain an analytic mapping of U into the Riemann sphere. Finally, if the principal part of the Laurent series of f about z0 contains infinitely many non-zero terms, then f is said to have an isolated essential singularity at z0 . Lemma 4.1.8. Let f have an isolated singularity at z0 , and let M (f, z0 , r) = max{|f (z)| | |z − z0 | = r}. If there are constants k > 0 and α ≥ 0 such that M (f, z0 , r) ≤ kr −α for all sufficiently small r > 0, then f has either a removable singularity at z 0 or a pole of order ≤ α. Proof. The coefficient a−n of the Laurent series can be estimated by using the integral representation |a−n | ≤ krn−α which converges to 0 as r → 0 if n > α; thus a−n = 0 in this situation. Theorem 4.1.9 (Casorati-Weierstrass Theorem). Let f have an essential singularity at z0 . Then for any r > 0, the image of the punctured disk 0 < |z − z0 | < r is dense in C. Proof. Justifying the thesis is equivalent to proving that for any complex number w, the 1 function g(z) = is unbounded in any deleted neighborhood of z0 . f (z) − w Assume that g is bounded on V = {0 < |z − z0 | < b}. In particular, f (z) 6= w for all z ∈ V , hence g is analytic on V . Now M (g, z0 , r) ≤ K, for some constant K and all 0 < r < b, by assumption. It follows from Lemma 4.1.8 (with α = 0) that g has a removable singularity 1 at z0 . But on V , f (z) = w + ; therefore, if m is the order of the zero of g at z0 (setting g(z) m = 0 if g(z0 ) 6= 0), then (z − z0 )m f (z) has a removable singularity at z0 . Consequently, m (z − z0 ) f (z) = ∞ X n=0 bn (z − z0 )n , z∈V and it follows, after dividing by (z − z0 )m , that f has either a removable singularity at z0 or a pole of order m at z0 , contradicting the hypothesis. Theorem 4.1.10 (Classification of Singularities). Let f have an isolated singularity at z0 . (a) There is a removable singularity at z0 if and only if f (z) approaches a finite limit as z → z0 . (b) There is a pole of order m at z0 (m = 1, 2 · · · ) if and only if (z − z0 )m approaches a finite non-zero limit as z → z0 , and in this case f (z) → ∞ as z → z0 . (c) There is an essential singularity at z0 if and only if f (z) does not approach a finite or infinite limit as z → z0 . Proof. (a) The “only if” part follows from Definition 4.1.7; the “if”part follows from Theorem 2.2.11. (b) The “only if” part follows from Definition 4.1.7; for the “if” part, note that if (z − z0 )m f (z) approaches a finite non-zero limit, then (z − z0 )m f (z) has a removable singularity ∞ X at z0 by (a), hence is given by bn (z − z0 )n with b0 6= 0. Thus, by dividing by (z − z0 )m , n=0 it follows that f has a pole of order m at z0 . (c) The “only if” part follows from Casorati-Weierstrass Theorem (Theorem 4.1.9); the “if” part from (a) and (b). The behavior of a complex function f at ∞ may be studied by considering the function g(z) = f (1/z) at z = 0. 4.2 Meromorphic Functions Definition 4.2.1. The function f has an isolated singularity at ∞ if and only if f is analytic on a set |z| > r and the function g(z) = f (1/z) has an isolated singularity at z = 0. Removable singularities, poles, and essential singularities at ∞ are defined similarly. Definition 4.2.2. A function f on an open subset of the Riemann sphere P is meromorphic on U if it is analytic on U except for poles and removable singularities. Example 4.2.3. Let R(z) = P (z)/Q(z) be a rational function, where P and Q are polynomials. The R is a meromorphic function on P. Theorem 4.2.4. If f is meromorphic on P, then f is a rational function. 4.3 Calculus of Residues In this section we present a technique which allows for rapid evaluation of integrals γ is a cosed path in U and f is analytic on U except for isolated singularities. R γ f where Definition 4.3.1. If f has an isolated singularity at z0 , the coefficient a−1 of the Laurent expansion of f about z0 is called the residue of f at z0 , and denoted by res(f ; z0 ). Lemma 4.3.2. Let f be analytic on U \ {z0 }. Let R be a rectangle whose closure is contained in U . If z0 ∈ R, then Z 1 f (z) dz. res(f ; z0 ) = 2πi ∂R Proof. We may replace ∂R by a circle Γ with center z0 , by the First Cauchy Theorem (Theorem 3.3.6. Then, Theorem 4.1.2, 1 2πi Z Z ∞ X an f (z) dz = (z − z0 )n dz 2πi Γ Γ n=−∞ = a−1 by the calculation in the proof of the Cauchy Integral Formula for a Circle (Theorem 2.2.7.) Lemma 4.3.3. Let γ be a closed path in C, S a subset of C whose closure is disjoint from γ ∗ . Assume that whenever w is a limit point of S, then ind(γ; w) = 0. Then ind(γ; z) = 0 for all but finitely many z ∈ S. Proof. The set A consisting of those z satisfying ind(γ; z) = 0 is an open subset of C \ γ ∗ which contains |z| > r, for r sufficiently large. Therefore C \ A is compact. If infinitely many points of S belong to C \ A, then S has a limit point in C \ A, contradicting the hypothesis. Theorem 4.3.4 (Residue Theorem). Let f be analytic on U except for isolated singularities at the points w1 , w2 , · · · . Let γ be a closed path (or cycle) in U such that ind(γ; z) = 0 for all z not in U , and such that none of the wj belong to γ ∗ . Then 1 2πi Z f= γ X j ind(γ; wj ) res(f ; wj ). Proof. First note that ind(γ; wj ) = 0 for all but finitely many wj , hence the sum in the statement is finite. Indeed, let S = {wj }. If w is a limit point of S, then w ∈ / U since all the singularities are isolated; thus ind(γ; w) = 0. Moreover, the closure of S does not meet γ ∗ because neither S meets γ ∗ nor the limit point of S meet γ ∗ . Let w1 , · · · , wn be the singularities for which ind(γ; wj ) 6= 0. It may be assumed that γ is a polygonal path with edges parallel to the coordinate axes and contained in U \{w 1 , w2 , · · · }. By a slight modification of γ, if necessary, it may also be assumed that none of the w 1 , · · · , wn lie on the rectangular grid induced by the polygonal path. m X By Lemma 3.3.4, the path γ is equivalent to a cycle of the form ind(γ; zj )∂Rj , and j=1 for each k = 1, · · · , n, the singularity wk lies in some Rj . The grid may be taken so fine that no two singularities among {w1 , · · · , wn } lie in the same Rj . Let V = U \ {wn+1 , · · · }. If the closure of Rj is not contained in V , let z0 ∈ Rj \ V . Since z0 ∈ / V , it cannot be in γ ∗ ⊂ V , and so the segment [z0 , zj ] does not meet γ ∗ . It follows that z0 and zj lie in the same component of C \ γ ∗ , and thus the indexes ind(γ; z0 ) = ind(γ; zj ). If z0 6 inU , then ind(γ; z0 ) = 0 by hypothesis; if z0 = wk for some k > n, then ind(γ; z0 ) = 0 by the present construction. In any case, γ is equivalent to a cycle of the r X form σ = ind(γ; zj )∂Rj , where each rectangle Rj has closure contained in V and none of j=1 w1 , · · · , wn is in the boundary of Rj . Therefore f is defined and continuous on σ ∗ , and the equivalence of σ and γ yields Z r Z X f= f. γ j=1 ∂Rj Each of the rectangles appearing in this sum contains at most one of the singularities R w1 , · · · , wn . If wk ∈ Rj , for some k = 1, · · · , n, then ind(γ; wk ) = ind(γ; zk ) and ∂Rk f = R 2πi res(f ; wj ). If Rj contains no wk , then ∂Rj f = 0. It follows that 1 2πi Z f= Example 4.3.5. Evaluate the integral n X ind(γ; wj ) res(f ; wj ). j=1 R Log z γ 1 + ez dz, where γ is the curve: The advantage of expressing an integral in terms of residues is that it is often possible to compute the residues. There is in fact an explicit formula for the residue at a pole. Theorem 4.3.6 (Residue at a Pole). If f has a pole of order m at z0 , then µ m−1 ¶ 1 d res(f ; z0 ) = lim (z − z0 )f (z) . (m − 1)! z→z0 dz m−1 In particular, if z0 is a simple pole, res(f ; z0 ) = lim (z − z0 )f (z). z→z0 Proof. Multiply the Laurent expansion ∞ f (z) = X a−m a−1 an (z − z0 )n + ··· + + m (z − z0 ) z − z0 n=0 by (z − z0 )m , differentiate m − 1 times, and take the limit as z → z0 . Lemma 4.3.7. Let f be analytic at z0 and have a zero of order m there. Then f 0 /f has a simple pole at z0 with res(f 0 /f ; z0 ) = m. Proof. Write f (z) = (z − z0 )m g(z), where g is analytic at z0 and z0 6= 0. Then f 0 (z) m g 0 (z) = , + f (z) z − z0 g(z) Since g 0 /g is analytic at z0 , the residue res(f 0 /f ; z0 ) = m. Example 4.3.8. If f has a pole of order m at z0 , then f 0 /f has a simple pole at z0 with residue res(f 0 /f ; z0 ) = −m. Theorem 4.3.9 (Argument Principle). Let f be analytic on the connected open set U ⊂ C, and let γ be a closed path in U such that f is never 0 on γ ∗ and such that ind(γ; z) = 0 for every z ∈ / U . If z1 , · · · are the distinct zeros of f with multiplicities m1 , · · · , then X ind(f ◦ γ; 0) = mj ind(γ; zj ). j Proof. This follows directly form Theorem 3.2.5. Indeed, Z 0 1 f (z) ind(f ◦ γ; 0) = dz by Theorem 3.2.5 2πi γ f (z) X = res(f 0 /f ; zj ) ind(γ; zj ) by Residue Theorem j = X mj ind(γ; zj ) by Lemma 4.3.7 j Remark. Intuitively, the number of times that f (z) winds about the origin as z traverses the path γ is the number of zeros of f inside γ ∗ , each zero counted according to its multiplicity and its index with respect to γ. If U is not assumed to be connected, then we have to add the hypothesis that f is not identically zero on each component of U . Example 4.3.10. Let f (z) = (z − a)m and let γ(t) = eikt , 0 ≤ t ≤ 2π and k an integer. Then f has a zero at a of multiplicity m. The index of a with respect to γ is 0 if |a| > 1 and k if |a| < 1. Therefore, ind(f ◦ γ; 0) = mk if |a| < 1 and 0 otherwise. Theorem 4.3.11 (Generalized Argument Principle). Let f and g be analytic on the open set U , with neither f nor g identically zero on a component of U . Let γ be a closed path on U such that f and g are never 0 on γ ∗ , and such that ind(γ; z) = 0 for all z ∈ / U . If z1 , z2 , · · · are the zeros of f with multiplicities n1 , · · · , and w1 , w2 , · · · are the zeros of g with multiplicities m1 , m2 , · · · , then µ ¶ X X f ind nj ind(γ; zj ) − mk ind(γ; wk ). ◦ γ; 0 = g j k Proof. Theorem 4.3.9 implies that ¶ µ Z 1 f (f /g)0 ◦ γ; 0 = . ind g 2πi γ f /g But (f /g)0 f 0 g0 = − f /g f g and the result follows from the previous theorem. Example 4.3.12. Let f (z) = (z − 1)(z − 3 + 4i) , and let γ be the circle of center 0 and (z + 2i)2 radius 3. Find ind(f ◦ γ; 0). Solution. The numerator has simple zeros at z = 1 and z = 3 − 4i. The denominator has a zero of order 2 at −2i. The index of 1 and −2i with respect to γ is 1, and the index of 3 − 4i is 0. The Generalized Argument Principle yields ind(f ◦ γ; 0) = 1 · ind(γ; 1) + 1 · ind(γ; 3 − 4i) − 2 · ind(γ; −2i) = 1 + 0 − 2 = −1 Theorem 4.3.13 (Rouche’s Theorem). Let f and g be analytic on the connected open set U . Suppose that f has zeros z1 , · · · with multiplicities n1 , · · · and g has zeros w1 , · · · with multiplicities m1 , · · · . Let γ be a path in U such that ind(γ; z) = 0 for all z ∈ / U . Assume also that |f (z) − g(z)| < |f (z)| for all z in γ ∗ . Then ind(f ◦ γ; 0) = ind(g ◦ γ; 0); hence X j nj ind(γ; zj ) = X mk ind(γ; wk ) k Thus f and g have the same number of zeros inside γ ∗ , counting index and multiplicity. Proof. The hypothesis that |f − g| < |f | on γ ∗ implies that f and g are never 0 on γ ∗ . g−f Therefore, f is never 0 in some neighborhood of γ ∗ . If h = 1 + , then g = hf and h f is never 0 on γ ∗ . In fact, |1 − h| < δ on γ ∗ , so that the curve h ◦ γ is contained in the disk D(1; 1). Now g0 f 0 h0 = + g f h so that by Theorem 3.2.5 Z 0 1 g ind(g ◦ γ; 0) = 2πi γ g Z 0 Z 0 1 1 f h + = 2πi γ f 2πi γ h = ind(f ◦ γ; 0) + ind(h ◦ γ; 0). But h ◦ γ is a closed path lying in the disk D(1, 1), and thus ind(h ◦ γ; 0) = 0 because that disk does no contain 0. Example 4.3.14. A polynomial of degree n ≥ 0 has exactly n zeros, counting multiplicity. Example 4.3.15. Rouche’s theorem gives an easy proof of the following version of the maximum principle: if f is analytic at z0 , then there is z near z0 such that |f (z)| ≥ |f (z0 ). For if |f (z)| < |f (z0 )| in a disk |z − z0 | ≤ r, then f − f (z0 ) and the constant f (z0 ) would have the same number of zeros inside |z − z0 | = r. Now f − f (z0 ) has a zero at z0 , and the only way that f (z0 ) can have a zero is if f (z0 ) = 0. This gives |f (z)| < 0, a contradiction. Example 4.3.16. Show that all the zeros of p(z) = z 4 + 6z + 1 are inside the circle |z| = 2. Moreover, three of the roots are in 1 < |z| < 2. √ Solution. The polynomial q(z) = z 4 + 6z has roots at 0 and at z = 3 −6. For these z, √ 1 < |z| = 3 6 < 2 On |z| = 2, |z 4 + 6z + 1| ≥ 16 − 13 = 3 > |p(z) − q(z)| = 1. The zeros of q(z) all have index 1 with respect to |z| = 2. The zeros of p(z) have index either 0 or 1, and the sum of theirs multiplicities is 4. On |z| = 1, |p(z) − q(z)| = 1, and |q(z)| ≥ |6z| − 1 = 5. Therefore q has exactly one root inside |z| = 1. To show that there are three roots in 1 < |z| < 2, look at |z| = 1 + ε for small ε. ∞ X Example 4.3.17. If f (z) = an z n is analytic on |z| ≤ R, and if a0 6= 0, then f cannot n=0 |a0 |R vanish on |z| < , where M (r) = sup |f (z)|. M (R) + |a0 | |z|=r Theorem 4.3.18. Let U ⊂ C be a connected open set whose boundary is a finite collection Γ of simple closed paths. Let f be analytic on U = U ∪ Γ, except for a finite number of poles in U , and moreover, f is not zero on Γ. Then Z 0 1 f = N − P, 2πi Γ f where N is the number of zeros of f in U and P is the number of poles of f in U , counted according to their multiplicity. Here Γ is the cycle obtained by giving each component γ of Γ the orientation that makes U to be locally on the left of γ. 4.4 Integrals The Residue Theorem can be used for evaluating integrals. Rational functions of sin and cos To evaluate Z 2π R(cos θ, sin θ) dθ 0 where R is a rational function, we substitute z = eiθ and note that µ ¶ µ ¶ 1 1 1 1 cos θ = z+ and sin θ = z− 2 z 2i z so that the integral becomes · µ ¶ µ ¶¸ 1 1 1 1 −i R z+ , z− dz. 2 z 2i z |z|=1 Z Example 4.4.1. If a > b > 0, then show that Z 2π 1 2π dθ = √ . 2 a + b cos θ a − b2 0 Improper Integrals Cauchy’s theorem can be applied to compute improper integrals, like Z ∞ sin x dx (1) x 0 One defines the improper integrals of a real-valued function f (x) of a real variable x > 0 which is bounded near 0 by Z R Z ∞ f (x) dx. f (x) dx = lim R→∞ 0 0 In general, f need not be absolutely integrable on (0, ∞), as the example f (x) = sin x/x above. To compute (1), consider the function f (z) = eiz /z, which is analytic on z 6= 0. Let R > r > 0 be positive real numbers, and let γ be the closed path depicted below, oriented counterclockwise. −R −r r R Then, since the index ind(γ; 0) = 0, Cauchy’s Theorem implies that Z f (z) = 0 γ Parameterizing γ in the obvious way, this integral can be written as (2) Z −r −R eix dx + x Z R eix dx x r Z π Z −r sin θ+ir cos θ −i e dθ + i 0 Now −r eix dx + x R π e−R sin θ+iR cos θ dθ. 0 eix dx = 2i x R sin x dx x r 0 −R Rπ and letting r → 0, the third integral in (2) has the limit −i 0 dθ = −πi. Thus Z 2 Z R 0 Z sin x dx − π + x Z π Z e−R sin θ+iR cos θ dθ = 0. 0 Since this holds true for all R > 0, it also holds in the limit as R → ∞. We now show that the last integral in (2) above approaches 0 as R → ∞. First note that ¯ −R sin θ+iR cos θ ¯ ¯e ¯ = e−R sin θ , so that ¯Z ¯ ¯ ¯ π e −R sin θ+iR cos θ 0 ¯ Z ¯ dθ¯¯ ≤ π e−R sin θ dθ = 2 0 Z π/2 e−R sin θ dθ. 0 If 0 ≤ θ ≤ π/2, then 2θ/π ≤ sin θ. Thus 2 Z π/2 e −R sin θ 0 dθ ≤ 2 Z π/2 e−2Rθ/π dθ = 0 π (1 − e−R ). R This converges to 0 as R → ∞, and therefore lim 2 R→∞ or Z Z R 0 ∞ 0 Example 4.4.2. Evaluate the integral Z sin x π dx = . x 2 ∞ 0 sin x dx − π = 0, x 1 dx, x4 + a 4 where a > 0. Solution. Since 1/(x4 + a4 ) is even, Z ∞ 0 1 1 dx = x4 + a 4 2 Z ∞ −∞ 1 dx. x4 + a 4 1 has poles at aekπ/4 (k = 0, 1, 2, 3) of which aeπ/4 and aei3π/4 z 4 + a4 are in the upper half plane. Integrating along the boundary of a semicircle of radius R and center 0 in the upper half plane, noticing that |f (x)| ≤ 1/R 4 on |z| = R, and letting R → ∞ we obtain µ ¶ √ Z 1 ∞ 1 1 2π πi/4 3πi/4 ) + res(f ; ae ) = dx = 2πi res(f ; ae . 4 4 2 −∞ x + a 2 4a3 The function f (z) = 4.5 The Open Mapping Theorem The purpose of this section is to show that a non-constant analytic function maps open sets to open sets, and that a one-one analytic function has an analytic inverse. The next theorem examines the number of solutions of the equation f (z) = w, where w is fixed and z ranges over a neighborhood of a zero of f . Theorem 4.5.1. Let f be analytic and not identically constant on the disk D(z 0 , r), and assume that f has a zero of order k at z0 . Choose r1 < r so small that neither f nor f 0 is zero on 0 < |z − z0 | < r1 , and let m = min{|f (z)| | |z − z0 | = r1 }. If 0 < |w| < m, then the equation f (z) − w has exactly k solutions z in D(z0 , r1 ). Proof. Note that such r1 must exists, for otherwise either f or f 0 has a limit point of zeros, hence f is identically constant on D(z0 , r). Let γ(t) = z0 + r1 eit , 0 ≤ t ≤ 2π; Then |f (z)| ≥ m > |w| on γ ∗ . Apply Rouche’s Theorem to f and g = f − w to obtain ind(f ◦ γ; 0) = ind((f − w) ◦ γ; 0). By hypothesis, f has a single 0 inside γ ∗ (that is, in |z − z0 | < r1 ) of multiplicity k. By the Argument Principle, X ind((f − w) ◦ γ; 0) = kj ind(γ; zj ) j where the zj are the zeros of f − w inside γ ∗ , and kj is theirs respective multiplicity. Because ind(γ; zj ) = 1, if there are fewer than k such points, then kj > 1 for at least one index j, and therefor f − w has a zero of order greater than 1 at z j . This inn turn implies that f 0 (zj ) = 0. This leads to a contradiction because the only point inside γ ∗ where f 0 = 0 is at z0 , and f (z0 ) − w = w 6= 0. Theorem 4.5.2. Let f be analytic and not identically constant on a disk D(z 0 , r) and suppose that f has a zero of order k at z0 . Then there is an open set U ⊂ D(z0 , r), with z0 ∈ U , such that for each z ∈ U , z 6= z0 , (a) f (z) 6= 0, and (b) there are exactly k points z 0 in U \ {z0 } such that f (z 0 ) = f (z). Proof. Let r1 and m be as in previous Theorem 4.5.1. Let U = D(z0 , r1 ) ∩ f −1 (D(0, m)). If z ∈ U \ {z0 }, then f (z) 6= 0, by the choice of r1 , and |f (z)| < m by the choice of U . Theorem 4.5.1 implies that there are exactly k points z 0 in D(z0 , r1 ) with f (z 0 ) = f (z). Since |f (z)| < m, all such z 0 belong to U \ {z0 }. Corollary 4.5.3. Let f be analytic at z0 . If f 0 (z0 ) 6= 0, then there exists a neighborhood of z0 on which f is one-one. If f 0 (z0 ) = 0, then f cannot be one-one in any neighborhood of z0 . Proof. Apply Theorem 4.5.2 to f (z) − f (z0 ); note that f (z) − f (z0 ) has a zero of order 1 at z = z0 if f 0 (z0 ) 6= 0, and a zero of order > 1 at z = z0 if f 0 (z0 ) = 0. Theorem 4.5.4 (Open Mapping Theorem). Let f be analytic on the open set U ⊂ C, and not identically constant on any component of U . Then f : U → C is an open mapping. Proof. It must be proved that if V is an open subset of U , then f (V ) is an open subset of C. Let z0 ∈ V and define g(z) = f (z) − f (z0 ). Let D(z0 ; r) ⊂ V and construct r1 and m as in Theorem 4.5.1. (Note that g is not identically constant on D(z0 ; r).) It follows from Theorem 4.5.1 that D(0, m) ⊂ g(V ). But then D(f (z0 ), m) ⊂ f (V ), for if |w − f (z0 )| < m, then there is a point z in V such that g(z) = w − f (z0 ), and therefore f (z) = w. Thus f (V ) is open. Lemma 4.5.5. Let U and V be open subset s of C, f a one-one mapping of U onto V , with inverse g. Assume that (1) f is continuous, (2) g is differentiable, and (3) g 0 is never 0 on 1 . U . Then f is differentiable and f 0 = 0 g ◦f Theorem 4.5.6. Let f be analytic and one-one on an open set U . Then f −1 is analytic on the open set f (U ). Proof. Since f is one-one, it is not identically constant on any component of U , hence f (U ) is open by the Open Mapping Theorem. Also by the Open Mapping Theorem, g = f −1 is continuous on V = f (U ). Because of Theorem 4.5.2, f 0 is never 0 on U . The result follows from the previous lemma. The next theorem gives a more explicit description of the nature of an analytic function near a zero of order k. Theorem 4.5.7. Let f be analytic and non-constant on U . Let z0 ∈ U and set f (z0 ) = w0 . Let k be the order of the zero of f − w0 at z0 . Then there exists a neighborhood V of z0 in U , and analytic function ϕ on V such that (a) f (z) = w0 + (ϕ(z))k for all z in V . (b) ϕ0 has no zero on V , and ϕ is a one-one mapping of V onto D(0; r). Proof. It may be assumed that U is a convex neighborhood of z0 , so small that f (z) 6= w0 for all z in U \ {z0 }. Then there is an analytic function g on U such that f (z) − w0 = (z − z0 )k g(z) for all z in U and such that g is never zero on U . Therefore g 0 /g is analytic on U , and since U is convex, it has an analytic logarithm h on U . If ϕ(z) = (z − z0 ) exp h(z)/k, then (a) holds for all z in U . Also ϕ(z0 ) = 0 and, because this is a simple zero, ϕ0 (z0 ) 6= 0. The rest now follows at once from the Open mapping Theorem. Exercise 4.5.8. Let f be analytic at z0 , with f (z0 ) = 0 and f 0 (z0 ) 6= 0. Let Γ = {|z − z0 | = r}, where r is chosen small enough so that f is one-one on the disk D(z0 ; s) for some s > r. If h is analytic on D(z0 ; s) and |w| < min{|f (z)| | z ∈ Γ}, then prove that Z 1 f 0 (z) −1 h(f (w)) = dz h(z) 2πi Γ f (z) − w In particular, f −1 (w) = 1 2πi Z z Γ f 0 (z) dz f (z) − w