Taylor and Laurent Series

Chapter 4 Taylor and Laurent Series 4.. Taylor Series 4... Taylor Series for Holomorphic Functions. In Real Analysis, the Taylor series of a given function f : R R is given by: f (x + f (x (x x + f (x 2! (x x 2 + f (x 3! (x x 3 +... We have examined some convergence issues and applications of Taylor series in MATH 233/243. We also learned that even if the function f is infinitely differentiable everywhere on R, its Taylor series may not converge to that function. In contrast, there is no such an issue in Complex Analysis: as long as the function f : C C is holomorphic on an open ball B δ (, we can show the Taylor series of f : f ( + f ( ( + f ( 2! ( 2 + f ( 3! ( 3 +... converges pointwise to f ( on B δ (, and uniformly on any smaller ball. As we shall see, it thanks to Cauchy s integral formula. Moreover, the proof of Taylor Theorem in Complex Analysis is also much easier than that in Real Analysis, again thanks to Cauchy s integral formula. In this chapter, it is more convenient to re-label the variables in the Cauchy s integral formula: f (n (α = n! γ f ( ( α n+ d f (n ( = n! γ f (ξ dξ. (ξ n+ For the re-labelled Cauchy s integral formula, we require the point to be enclosed by the simple closed curve γ. Theorem 4. (Taylor Theorem for Holomorphic Functions. Given a complex-valued function f which is holomorphic on an open ball B R (, the series: f (n ( ( n n! converges (pointwise to f ( for any B R (. 83

84 4. Taylor and Laurent Series Proof. Given any B R (, we let ε > be small enough so that < R ε. For simplicity, denote R = R ε. By Cauchy s integral formula, for any B R (, we have: ξ =R f (ξ ξ dξ. Then, the contour = R lies inside the open ball B R (. The key trick to prove the Taylor Theorem is rewriting as a geometric series. Recall that: ξ We first rewrite w = + w + w2 +... whenever w <. into this form: ξ ξ = (ξ ( = ξ ξ = ( n ξ ξ Here we have used the fact that ξ <. See the diagram below. The yellow ball is B R (, and the red circle is ξ = R. ξ ξ = R < R Then, whenever B R (, the function f ( can be expressed as: (4. f (ξ ξ =R ξ dξ = ( f (ξ n ξ =R ξ dξ ξ = f (ξ ( n ξ =R (ξ n+ dξ Next we want to see whether we can switch the integral sign and the infinite ξ =R summation. For this we need to show uniform convergence of the series below. f (ξ ( n (ξ n+.

4.. Taylor Series 85 We use Weiestrass s M-test: for any ξ on the circle { ξ = R }, we have: f (ξ ( n (ξ n+ ( n (ξ n+ sup f (ξ ξ =R }{{ } Since < R, the series = C R R C R R ( R ( R n n =:C R converges. Note that the above series does not depend on ξ (the integration variable. f (ξ ( n Hence by Weiestrass s M-test, the series (ξ n+ converges uniformly on the circle { ξ = R }, thus allowing the switch between the integral sign and the summation sign in (4.: = ξ =R f (ξ ( n (ξ n+ dξ ( f (ξ dξ ( (ξ n+ n ξ =R f = (n ( ( n. n! In the last step we have used the higher order Cauchy s integral formula. Example 4.. The function sin is an entire function. By straight-forward computations, its derivatives are given by: f ( = cos f (3 ( = cos. f ( = sin f (4 ( = sin Inductively, it is easy to deduce that f (2k+ ( = ( k, and f (2k ( = for any integer k. Hence, the Taylor series of f about is given by: f (2k+ ( (2k +! 2k+ = k= k=. ( k (2k +! 2k+ = 3 3! + 5 5! 7 7! +... This series converges to sin for any C, because sin is entire (i.e. holomorphic on every ball B R (. Example 4.2. Consider the function Log( which is holomorphic on Ω := C\{x + i : x }. Note that we can only apply Theorem 4. if the ball B R ( is contained inside Ω.

86 4. Taylor and Laurent Series Let s take = as an example. f ( = f ( = 2 f ( = f ( = f (3 ( = 2 3 f (3 ( = 2 f (4 ( = 2 3 4 f (4 ( = 2 3. Inductively, we deduce that f (n ( = ( n (n! for n. Therefore, the Taylor series for f about is given by: Log( = Log( + n= ( n (n! ( n = n!. n= = ( 2 ( 2 + 3 ( 3 4 ( 4 +... ( n ( n n Since f is holomorphic on B ( (but not on any larger ball centered at, the above Taylor series converges to Log( on B (. Example 4.3. The Taylor series for some composite functions, such as e 2, can be derived by substitution instead of deducing the general n-th derivative of the function. For example: e = + + 2 2! + 3 3! + 4 4! +... e 2 = + 2 + (2 2 2! + (2 3 3! + (2 4 4! = + 2 + 4 2! + 6 3! + 8 4! +... +... Since the series for e converges for any C, the series for e 2 converges for any C as well. Similarly, by replacing by in the Taylor series for Log(, we get: Log( = 2 2 3 3 4 4... The series for Log( about converges when <, and so the above series for Log( converges when ( <, i.e. <. Apart from using Theorem 4. to find the Taylor series of a given holomorphic function, we can also make use of the geometric series formula directly: w = + w + w2 +... where w <. This method is particularly useful for functions whose n-th derivatives are tedious to compute.

4.. Taylor Series 87 Example 4.4. Consider the function: 2 ( + 2( + 3. We are going to derive its Taylor series about. First, we do partial fractions on the function: 5 + 3 4 + 2. Then, we try to rewrite each term above in the form of 5 + 3 = 5 3 = 5 3 4 + 2 = 4 2 = 2 = 3 + = 5 3 ( 3 ( 3 a w. Note that: n = ( n 5 3 n+ n (where < 3 2 + = 2 ( 2 ( n (where < 2 2 ( n 2 n n. Hence, for < 2, we have: ( n 5 3 n+ n ( n 2 n n = ( 5 ( n 3 n+ 2 n n. 5 To derive the Taylor series of f about other center (say, we can express +3 and +2 4 into: 5 + 3 = 5 ( + 4 = 5 4 ( = = ( 5 4 4 ( n 5 4 n+ ( n 4 + 2 = 4 ( + 3 = 4 3 = = ( 4 3 3 ( n 4 3 n+ ( n. Therefore, on < 3, we have: 4 n (where < 4 ( 3 n (where < 3 ( 5 ( n 4 n+ 4 3 n+ ( n.

88 4. Taylor and Laurent Series Exercise 4.. Derive the Taylor series of each function below about the given center : (a sin 2; = 2π 3 (b cos 3; = π (c e 3 ; = (d Log(3 2; = Exercise 4.2. Find the Taylor series about of the functions below up to the 4 term: (a e cos (b Log( e Exercise 4.3. Find the Taylor series about of the function below without using Theorem 4.. State its radius of convergence. (a (b ( ( 2, = ( ( 2, = i Exercise 4.4. Let α, β and be three distinct complex numbers. Consider the function ( α( β. Find the Taylor series about of the above function, and state its radius of convergence. Exercise 4.5. Let α be a fixed non-ero complex number. Consider the principal branch of ( + α : ( + α := e αlog(+. Show that its Taylor series about is given by: ( + α = + State its radius of convergence. n= α(α (α n + n. n! 4..2. Taylor Series with Remainder Term. In Real Analysis, the Taylor Theorem with a remainder term asserts that for any smooth (C function f : R R, we have: f (x = N f (n (a n! (x a n + (N! x (x t N f (N (t dt } a {{ } =:R N (x The last integral term, commonly denoted as R N (x, measures how fast the Taylor series converges to f (x as N. If R N (x for any x in an interval I, then N the Taylor series f (n (a (x a n converges (pointwise to f (x for any x I. If n!

4.. Taylor Series 89 furthermore, we have: sup R N (x, N x I then the Taylor series converges uniformly to f on I. However, it is often not easy to show R N as the N-th derivative f (N may not be easy to find. Back to Complex Analysis, we will soon derive the remainder term for the Taylor series for holomorphic functions. One good thing about the complex version is that the remainder involves only f, but not its derivatives, making it much easier to handle the convergence issue of complex Taylor series. It again thanks to Cauchy s integral formula. Proposition 4.2. Let f be a holomorphic function defined on B R (, then for any B R (, and any simple closed curve γ in B R ( enclosing both and, we have: N f (n ( ( n + ( f (ξ N dξ n! γ ξ ξ }{{} =:R N ( Proof. We only outline the proof since it is modified from the proof of Theorem 4.. Using Cauchy s integral formula, we first have: f (ξ ξ dξ. The key step in the proof of Theorem 4. is to write: ξ = (ξ ( = ξ, ξ so that when ξ <, we have: ( n =. ξ ξ Now, to prove this proposition, we modify the above key step a bit, by considering: ( ξ N ξ = γ N ( n. ξ We leave the rest of the proof for readers (which is a good exercise to test your understanding of the proof of Theorem 4.. Exercise 4.6. Complete the proof of Proposition 4.2. Exercise 4.7. Consider the remainder term in Proposition 4.2: R N ( = ( f (ξ N dξ. γ ξ ξ Let γ be the circle ξ = R such that < R < R. Show that: R R N ( ( N R R sup f (ξ. ξ =R

9 4. Taylor and Laurent Series Exercise 4.8. Let f be a holomorphic function on B R (. Using this estimate obtained in Exercise 4.7, deduce that the Taylor series (n ( f ( n converges n! uniformly to f ( on any smaller ball B r ( where < r < R. f (n ( Remark 4.3. The uniform convergence of ( n has many remarkable n! consequences as discussed in MATH 333/343. For instance, one can integrate a Taylor series term-by-term. Exercise 4.9. Consider the Taylor series for Log( ξ: Log( ξ = ξ + ξ2 2 + ξ3 3 + + ξn + where ξ < n Show that: 2 2 + 4 3 4 + + n+ + = ( Log( + n(n + for any B (. Exercise 4.. Show that for any C, we have: e ξ2 dξ = ( n 2n+ n!(2n +.

4.2. Laurent Series 9 4.2. Laurent Series A Laurent series is a power series with negative powers of as well. The general form of a Laurent series about is: n= a n ( n + a n ( n = + a 2 ( 2 + a + a + a ( + a 2 ( 2 + which can be abbreviated as: a n ( n. n= A Laurent series is said to be convergent if both converge. n= a n ( n and a n ( n If a n = for any negative n, then the Laurent series is a Taylor series. On the other hand, if a n = for some negative n, then the Laurent series is undefined when =. As such, a Laurent series is usually defined on an annular region {r < < R} instead of a ball centered at. From now on, we denote such an annular region by: A R,r ( := { C : r < < R} where R, r [, ]. Note that: A R, ( = B R ( \{ } A,r ( = C\B r ( for r > A, ( = C\{ }. 4.2.. Examples of Laurent Series. While a Taylor series gives an analytic expression for a holomorphic function on a ball, a Laurent series gives an analytic expression for a function that has a singularity at the center of a ball. Before we discuss a general theorem about Laurent series, let s first look at some examples of writing a function as a Laurent series: Example 4.5. Consider the function f : C\{, 2} C defined by: ( ( 2. It is holomorphic on its domain C\{, 2}. Let s express the above function as a Laurent series about : 2 = ( = ( = ( n where <. Hence, on A, (, i.e. the green annulus in the figure below, we have: 2 = ( n = = ( ( 2 +. ( n

92 4. Taylor and Laurent Series y 2 x However, the green annulus A, ( is not the only annulus centered at on which f is holomorphic. There is another one A, ( = { < } centered at, i.e. the yellow annulus in the above figure, on which f is also holomorphic. It is also possible to express f as a Laurent series on this yellow annulus: 2 = ( = = = ( ( n+ n (where > Hence, on the yellow annulus A, (, the function f can be expressed as the following Laurent series: ( ( 2 = 2 = ( n+2. Example 4.6. Find the Laurent series about of the function: defined on C\{}. 2 e Solution First recall that the Taylor series for e w is: e w = w n n! Substitute w =, where =, we get: e = for any w C. n! n,

4.2. Laurent Series 93 and hence: 2 e = 2 = Exercise 4.. Express the function: n! n n! n 2 = 2 + + 2 + 3! + 4! 2 + ( ( 2 as a Laurent series about in each of the following annuli: A, (, A 2, (, A,2 (. Also, express the function as a Laurent series about in each of the following annuli: A, (, A, (. [Hint: First expand f into partial fractions.] Exercise 4.2. Find all possible Laurent (or Taylor series about for the function: 2 2. For each series, state the annulus or ball on which it converges. Exercise 4.3. Find all possible Laurent (or Taylor series about each below for the function. (a = (b = (c = i For each series, state the annulus or ball on which it converges. Exercise 4.4. Show that for any w such that w <, we have: [Hint: use Exercise 4.5] ( + w 3 = n n(n ( w n 2. 2 n=2 Hence, find all possible Laurent or Taylor series about i for the function: 3. For each series, state the annulus or ball on which it converges.

94 4. Taylor and Laurent Series Exercise 4.5. Find the Laurent series about on the annulus A, ( for the functions: sin and g( = cos. Hence, find the Laurent series about on A, ( for: h( = sin. Exercise 4.6. What s wrong with the following argument? By subtraction, we get: = n = + 2 + 3 + = = n = 2 = + 2 + + + + 2 + = n. n= 4.2.2. Existence Theorem of Laurent Series. We have learned how to express a function into a Laurent series through examples. Next, we proved a general existence theorem of Laurent series for any holomorphic function on any annular region. Theorem 4.4 (Laurent Theorem. Let f be a holomorphic function defined on an annulus A R,r ( := {r < < R} where R, r [, ], then f can be expressed as a Laurent series about on the annulus A R,r ( : for some complex numbers c n s. c n ( n n= Proof. The proof is similar to that of Taylor s series, but is a bit trickier since an annulus is not simply-connected and so Cauchy s integral formula cannot be applied directly. Fix A R,r (, we first consider a simple closed curve Γ in A R,r ( which encloses both and (just like in the proof of Taylor s Theorem. However, we cannot apply Cauchy s integral formula on the integral: f (ξ dξ Γ ξ since f is not holomorphic on B R (. However, we can construct a key-hole contour: C = Γ + L γ L where γ is the clockwise circle, and L is a straight-path as shown in the figure below. We can pick Γ to be the circle with radius slightly smaller than R, and γ with radius slightly bigger than r so that C encloses.

4.2. Laurent Series 95 Γ R r γ L L Under such a construction, the contour C = Γ + L γ L is a simple closed curve and the region enclosed by C becomes simply connected. We can then apply Cauchy s integral formula: (4.2 C = ( = Γ f (ξ ξ dξ + Γ L γ f (ξ ξ dξ L f (ξ ξ dξ γ f (ξ ξ dξ. The key idea of the proof is to express the integral over Γ as a series of non-negative powers, and the integral over γ as a series of negative powers. When ξ Γ, we have < ξ, so: ξ = (ξ ( = ξ Hence, the first integral becomes: f (ξ Γ ξ dξ = Γ = ξ ξ ( f (ξ n dξ. ξ ξ ( ξ In order to switch the infinite summation and the integral sign, we justify that the series converges uniformly on ξ Γ. Suppose Γ has radius R, then for any ξ Γ: ( f (ξ n ( n ξ ξ R R sup f. Γ Note that sup Γ f is finite by compactness of Γ. Since < R, the geometric series ( n R R sup f Γ converges. By Weierstrass s M-test, the series f (ξ ( ξ ξ n n

96 4. Taylor and Laurent Series converges uniformly on ξ Γ, so one can switch the summation and integral signs and get: ( f (ξ (4.3 Γ ξ dξ = f (ξ dξ ( Γ (ξ n+ n. The second integral can be handled similarly. The difference is that when ξ γ, we have ξ < instead. We instead write: ξ = (ξ ( = ξ = ( ξ n Hence, f (ξ γ ξ dξ = ( f (ξ ξ n dξ. γ We leave it as an exercise for readers to argue that the series converges uniformly on ξ γ so that we can switch the integral and summations signs: ( f (ξ (4.4 γ ξ dξ = f (ξ(ξ n dξ γ ( n+. Combining (4.2, (4.3 and (4.4, we obtain: ( f (ξ(ξ n dξ n= γ ( n }{{} + (4.4 ( f (ξ dξ ( Γ (ξ n+ n }{{} (4.3 It completes the proof by defining c n = f (ξ(ξ n dξ for n =, 2, 3, γ c n = f (ξ Γ (ξ n+ dξ for n =,, 2, 3, Exercise 4.7. Justify the claim in the above proof that the series: f (ξ ( ξ n converges uniformly on ξ γ (and, are considered to be fixed. Remark 4.5. Although from the proof of Theorem 4.4 one can express the coefficient c n s of a Laurent series in terms of contour integrals, we do not usually find the coefficients this way since these contour integrals may not be easy to compute. 4.2.3. Laurent Series with Remainders. Similar to Taylor series, one can refine Theorem 4.4 a bit by deriving the remainder terms. Using the remainder terms, one can argue that for a holomorphic function f defined on an annulus A R,r (, the Laurent series converges uniformly to f on every smaller annulus A R,r ( (where r < r < R < R. This result is remarkable as it allows us to integrate a Laurent s series term-by-term.

4.2. Laurent Series 97 Proposition 4.6. Let f be a holomorphic function on the annulus A R,r (, where r < R. Then, for each positive integer N and A R,r (, we have: N ( f (ξ(ξ n dξ n= γ ( n + ( f (ξ ξ N dξ γ ξ }{{} + N ( f (ξ dξ Γ (ξ n+ ( n + =:r N ( ( f (ξ N dξ Γ ξ ξ }{{} =:R N ( where Γ and γ are any pair of circles in A R,r ( centered at such that is bounded between Γ and γ. Proof. We leave the proof of Proposition 4.6 as an exercise. It is very similar to the proof of Proposition 4.2 for Taylor series. Readers should first digest the whole proof of Proposition 4.2, then write up a coherent proof for this proposition. Exercise 4.8. Prove Proposition 4.6. Using this, show that the Laurent series about for f converges uniformly on every smaller annulus A R,r ( where r < r < R < R. [Hint: show that both remainders R N ( and r N ( converge uniformly to on A R,r ( as N.] One practical use of uniform convergence is term-by-term integration. For example, consider the function 2 e, which can be expressed as a Laurent series: 2 e = 2 + + 2 + 3! + 4! 2 + Then, to integrate f ( over the circle =, we can integrate the Laurent series term-by-term: 2 e d = = = 2 d + d + = = 2 = d + 3! d + = 4! 2 d + = + + + + + + = πi 3! 6. Recall that for any simple closed γ enclosing the origin, the contour integral n d is non-ero only when n =. γ From the above example, we see the significance of expressing a function as a Laurent series. To compute a contour integral, it often amounts to finding the coefficient c of the Laurent series. It leads to the develop of residue theory to be discussed in the next section.

98 4. Taylor and Laurent Series 4.3. Residue Calculus In this section we discuss both theory and applications of an important topic in Complex Analysis: residue calculus. It has many powerful applications on evaluations of some complicated real integrals that physicists and engineers often encounter. 4.3.. Classification of Singularities. A singular point, or singularity, refers to a point at which a function f fails to be complex differentiable. For instance, and 2 are singularities of the function: ( ( 2. It is possible for a function to have infinitely many singularities, such as: g( = sin whose singularities are, ±π, ±2π, etc. Some functions even have singularities that form a cluster. For instance, consider: h( = sin which is singular when { π n : n Z} {}. The singular set { π n : n Z} {} form a cluster around, meaning there is no way to find an annulus A R, ( centered at such that h is holomorphic on A R, (. Hence, it is not possible to analye the function h by a Laurent series about on A R, (. In order to utilie Laurent series, we focus on those singularities that can be isolated from others. We have the following terminology: Definition 4.7 (Isolated Singularity. A point is said to be an isolated singularity for a function f ( if there exists ε > such that f is holomorphic on A ε, ( = B ε ( \{ }. For the function g( =, all singularities are isolated as depicted in the sin diagram below: y x Around every isolated singularity of a function f (, it is possible (thanks to Theorem 4.4 to express the function f as a Laurent series on a small annulus A ε, ( : c n ( n. n= Depending on the smallest n such that c n =, we have the following terminology: If c = c 2 = c 3 = =, then is said to be a removable singularity of f. For instance, is such a singularity for the function: sin = 2 3! + 4 5! 6 7! +

4.3. Residue Calculus 99 If k is a positive integer such that c k = while c (k+ = c (k+2 = =, then is said to be a pole of order k of f. For instance, is a pole of order 3 for the function: sin 4 = 3 3! + 5! 3 7! + Moreover, a pole of order is usually called a simple pole. If c n = for infinitely many negative integers n, then is said to be an essential singularity. For instance, is such a singularity for the function: e = + + 2! 2 + 3! 3 + If is a removable singularity for f : B R ( \{ } C, then one can define f ( := c so that f extends to become a holomorphic function on B R (. That s why we can removable. Similarly, if is a pole of order n for f : A R, ( C, then ( n f ( extends to become a holomorphic function on B R (. However, a function with an essential singularity cannot be extended to become a holomorphic function in a similar way (that s why we call it essential. To determine the order of a pole, we may simply find its Laurent series expansion. However, sometimes it is not easy to do so, such as for the function sin. An alternative way to find the order of a pole is to consider the it: If k is an integer such that: ( k f (. ( k f ( exists and is non-ero, then the order of the pole is k. For example, since: = =, sin is a pole of order for the function. Hence, one can express this function as a sin Laurent series on a small annulus A ε, (: sin = c + c + c + c 2 2 + Multiplying on both sides, we get: sin = c + c + c 2 + c 2 3 + and by letting, we can also conclude that c =. Therefore, if γ is a simple close curve enclosing in this small annulus A ε, (, then we have: ( γ sin γ d = + c + c + c 2 2 + d = d + + + =. γ Exercise 4.9. Find all isolated singularities of each function below, and classify the nature of these singularities. For poles, state also their orders. (a e (b g( = Log( ( 3 5 (c h( = 423 cos

4. Taylor and Laurent Series 4.3.2. Residues. As illustrated in many examples, the coefficient c of a Laurent series plays a special role in evaluating a contour integral. It is special in a sense that for an integer n, { ( n if n = d = =ε otherwise Hence, to integrate a Laurent series, one only needs to integrate the term c, which can be done by Cauchy s integral formula. In view of the special role of c, we define: Definition 4.8 (Residues. Let be an isolated singularity of f ( such that the Laurent series about for f on some annulus A ε, ( is given by: c n ( n, n= then we denote and define the residue of f at by: Res( f, := c. Example 4.7. Find the residue of the function at each of its isolated singularity. Solution 2 2 ( + 2 ( 2 + 4 The denominator has roots, 2i and 2i, hence they are isolated singularities of f. In this solution, we will decompose f ( into partial fractions. It may not be a pleasant way finding residues, but we will later provide an easier way. Note that f is a rational function, we can break it into partial fractions: A ( + 2 + B + + C 2i + D + 2i. We leave it as an exercise for readers to determine the value of A, B, C and D. One should be able to get: 3 5 ( + 2 + 4 7+i 5 + + 25 2i + 7 i 25 + 2i }{{} holomorphic near 7+i 7 i 25 On a small annulus A ε, ( about, the last two terms 2i + 25 + 2i are holomorphic. Therefore, if one express them as a Laurent series about, only non-negative powers of + will appear, and the coefficient of + will not be affected. Therefore, we have: Res( f, = 4 5. By a similar reason, we have: Res( f, 2i = 7 + i 25 and Res( f, 2i = 7 i 25..

4.3. Residue Calculus Exercise 4.2. Determine all isolated singularities of the function 2 + ( + ( 2. and find the residue at each isolated singularity. It is no doubt that partial fraction decompositions are time-consuming and not fun (it may remind you the computational nightmare you might have encountered in MATH 4. Fortunately, there is a better way for finding residues for poles (does not work for essential singularity. If we know already that is a pole of order (i.e. simple pole of a function f (, then c + c + c ( + c 2 ( 2 + It is then easy to see that c = ( f (. Therefore, in order to find Res( f, for a simple pole, we simply need to compute the above it. Now consider the case if is a pole of order k for f, then its Laurent series about is given by: c k ( k + c (k ( k + + c + c + c ( + c 2 ( 2 + Our goal is to find c. By multiplying both sides by ( k, we can get: ( k c k + c (k ( + + c ( k + c ( k + By differentiating both sides for k times, all terms involving ( n with n < k will disappear: d k d k ( k c (k! + c ( + c ( 2 + We have used the fact that dk d k ( k = (k!, and c, c,... are some complex numbers (which we do not need to know their values. By letting, we get: d k d k ( k c (k! which provides a good way to find c without expanding a Laurent series: Proposition 4.9. Suppose is a pole of order k < for a function f, then we have: Res( f, = In particular, for a simple pole, we have: (k! d k d k ( k f (. Res( f, = ( f (. Proof. See the preceding paragraph.

2 4. Taylor and Laurent Series Example 4.8. Find the residue of the function 2 2 ( + 2 ( 2 + 4 of each isolated singularity using Proposition 4.9. Solution As discussed before, the isolated singularities are, 2i and 2i. Observe that: ( + 2 2 2 2 + 4 = 3 =. 5 Hence is a pole of order 2. From Proposition 4.9, we have: Res( f, = (2! d = 2 2 d 2 + 4 d 2 d 2 ( + 2 f ( 2 = 2 + 8 8 ( 2 + 4 2 = 4 25. Both 2i and 2i are simple poles, so we have: Res( f, 2i = ( 2i 2 2 2i 2i ( + 2 ( + 2i = 7 + i 25 Res( f, 2i = ( + 2i 2 2 2i 2i ( + 2 ( 2i = 7 i 25 Example 4.9. Find the residue at of each function below: e sin g( = e sin h( = e sin 2 Solution For each function, we first determine whether is a pole, and find out its order. For f (, we consider: sin e = e = =. Hence is a simple pole for f, and Res( f, =. For g(, note that: + 2 2! g( = + 3 3! + 3 3! + 5 5! = + 2! + 2 3! + 2 3! + 4 5! = <. Hence is a removable singularity of g(, and there is no -term in the Laurent series, and so Res(g, =. For h(: 2 h( = ( 2 e = =. sin

4.3. Residue Calculus 3 Hence is a pole of order 2 for h. By Proposition 4.9, we can find: Res(h, =! d d d 2 h( = 2 e d sin 2 sin 2 (2e = + 2 e 2 e 2 sin cos sin 4 [ = 2 e ( ] sin cos sin 2 + 2e sin 3 ( ( 3 = + 2e 3! + 5 5! 2 2! + 4 4! sin 3 ( ( = + 2e 2! 3! 3 ( 4! 5! 5 + sin 3 (( = + 2e 2! 3 ( 3! sin 3 4! 5 5! sin 3 + ( = + e 2! 3! + + + =. Exercise 4.2. For each function below, find its residue at each isolated singularity using any method: 2 3 ( 2 + e sin 6 2 + 8 + 9 e 2i sin 2 e / Exercise 4.22. Compute the following residues: ( (a Res 2 cos 2 + 2, ( 2n (b Res ( n, 997 247 e 2i sin sin 2 ( π 3 4.3.3. Residue Theorem. The residue Res( f, of an isolated singularity determines the value of a contour integral f ( d where γ is a tiny simple closed curve so γ that is the only singularity it encloses. Namely, we have f ( d = Res( f,. If a simple closed curve γ encloses more than one isolated singularities {,, N }, then we may first express the contour integral over γ as the sum of contour integrals: f ( d = f ( d + + f ( d γ γ γ N where each γ j is a small simple-closed curve so that j is the only singularity it encloses. Then, each γ j -integral is given by Res( f, j, and hence we have the following theorem: γ

4 4. Taylor and Laurent Series Theorem 4. (Residue Theorem. Let f : Ω C be a complex-valued functions whose singularities are all isolated. Let γ be a simple closed curve, and,, N Ω be all the singularities enclosed by γ. Then, we have: γ f ( d = N j= Res( f, j. Proof. Let ε > be sufficiently such that each circle { j = ε}, denoted by γ j, encloses j as the only singularity of f (see figure below. γ γ 2 γ 3 2 3 Then, by the standard hole-drilling argument, we have: f ( d = f ( d + + f ( d γ γ γ N Each γ j encloses j as the only singularity of f. Express f as a Laurent series on A ε, ( j : c n ( j n. n= Recall that ( j n d = only when n =, and by uniform convergence of γ j Laurent series, we get: c = Res( f, j. γ j Therefore, we have: completing the proof. γ f ( d = γ N j= Res( f, j,

4.3. Residue Calculus 5 Example 4.. Use Residue Theorem to evaluate the contour integral: 2 2 ( + 2 ( 2 + 4 d where R is in the range of: (a < R < (b < R < 2 (c 2 < R. Solution =R Denote 2 2 ( + 2 ( 2. The singularities of f are, 2i and 2i. We + 4 have calculated in Example 4.8 that Res( f, = 4 5 Res( f, 2i = 7 + i 25 y Res( f, 2i = 7 i 25 2i (a (b (c x 2i (a When < R <, the circle = R does not enclose any singularities, hence f ( d =. =R (b When < R <, the circle = R encloses the singularity only, hence f ( d = Res( f, = 28πi 5 =R (c When R > 2, the circle = R encloses all three singularities, hence f ( d = (Res( f, + Res( f, 2i + Res( f, 2i =R ( = 4 5 + 7 + i 25 + 7 i = 56πi 25 75.

6 4. Taylor and Laurent Series Example 4.. Let N be a positive integer and γ N be the square contour with vertices ±(N + 2 ± (N + 2 i. Use Residue Theorem to show: n 2 = π2 3 + π cot π d. γ N 2 Hence, deduce that: Solution N n= + 2 2 + 3 2 + = π2 6. Denote f ( := π π cos π cot π = 2 2. Its singularities are the set of all integers n. sin π First we observe that π 3 cos π =, sin π so is a pole of order 3 for f. By Proposition 4.9, its residue is given by: Res( f, = 2! d 2 d 2 3 2 π = 2 (π cos π sin π sin 3 π = π2 d 2 π cot π d2 ( π( π2 2 2! + (π π3 3 3! + sin 3 π = π 2 π33 3 + higher-order terms sin 3 = π2 π 3. For any non-ero integer n, observe that: π( n cos π ( n n n 2 ( n = ( n sin(π( n n }{{} 2 ( n = n 2. =sin π Hence, n is a simple pole of f (for any n =, and Res( f, n = n 2. Now consider the contour γ N. The singularities it encloses are:, ±, ±2, ±N. y (N + 2 i (N + 2 N + 2 x (N + 2 i

4.3. Residue Calculus 7 By Residue Theorem, we have: N f ( d = Res( f, n γ N n= N ( = Res( f, + 2 ( + 2 2 + + N ( 2 = π2 N 3 + 2 n= n 2. By rearrangement, we have the desired result: n 2 = π2 6 + N n= The remaining task is to show: N We do so by estimating the contour integral: γ N γ N f ( d. π cot π d =. 2 When γ N, we have N + 2 > N. It is also possible to show that cot π < 2 for any γ N (this is left as an exercise. Therefore, on γ N, we have the bound: π 2 cot π 2π N 2. The length of γ N is 8N + 4. By Lemma 3.6, we get: γ N completing the proof. π cot π d 2 2π (8N + 4 as N, N2 Exercise 4.23. Complete the detail of the above example that: cot π < 2 for any γ N. [Hint: Write = x + yi, and find an expression for cot π in terms of x and y. Then, maximie cot π on each side of the contour γ N.] Exercise 4.24. Use Residue Theorem to evaluate the following contour integrals: (a =3 2 + d 3 + 3 + (b =2 4 5 2 d e + (c i =2 ( 4 d sin (d d i =2 ( i 423 (e tan π d where γ is the rectangle contour with vertices: γ ( 2,, (2,, (2,, ( 2,.

8 4. Taylor and Laurent Series Exercise 4.25. Let γ N be the square contour with vertices ±(N + 2 π ± (N + 2 πi where N is a positive integer. Show that: γ N 2 csc d = 6 + 2 N ( π n 2 n= n 2. Hence, show that: 2 2 + 3 2 4 2 + = π2 2. Exercise 4.26. Determine the residues of all isolated singularities of the function: (2 sin π. By considering a suitable contour integral of f, show that: 3 + 5 7 + = π 4. 4.3.4. Evaluation of Real Integrals. Residues are often used to evaluate some difficult real integrals that physicists and engineers may encounter. Example 4.2. Evaluate the real definite integral: 2π a b cos θ dθ where a and b are real numbers such that < b < a. Solution The key trick is to express the real integral as a complex integral of the circle contour =, which is parametried by = e iθ where θ 2π. When = e iθ is on the contour { = }, we have cos θ = eiθ + e iθ = ( + 2 2 d = ie iθ dθ = dθ = ie iθ d = i d Therefore, the real integral can be written as a complex integral as: 2π a b cos θ dθ = = a b 2 ( + i d = 2i = b 2 2a + b d. We can then use residue theory to evaluate the complex integral. The singularities of the integrand are roots of the quadratic equation b 2 2a + b =, which are: ω = a a 2 b 2 and ω 2 = a + a 2 b 2. b b Note that a > b, so both roots are real. We further observe that: ω 2 > a + > and ω b ω 2 =, and so ω <. Therefore, ω is the only singularity enclosed by the contour =. As ω and ω 2 are distinct, they are simple poles, and so the contour

4.3. Residue Calculus 9 integral is given by: = b 2 2a + b = d = = Hence, the real integral is given by: 2π = b( ω ( ω 2 d b( ω 2 ω d = = b(ω ω 2 = πi a 2 b. 2 πi dθ = 2i a b cos θ a 2 b 2 = [ ] b( ω 2 2π a 2 b 2. Before we proceed to the next example, let s first prove a useful observation which will come in handy later on. =ω Exercise 4.27. Show that the function e i is bounded on the upper-half plane, i.e. there exists M > such that e i M whenever Im(. On the other hand, show that the function = cos is unbounded on the upper-half plane. Example 4.3. Evaluate the following real integral: cos x + x 2 dx. Solution Let s consider the following semi-circle contour: y C R i R L R R x Denote C R to be the (open semi-circle with radius R, L R to be the straightpath from R to R, and γ R to be the closed semi-circular path C R + L R. We consider this contour because cos x dx = + x2 R + R R cos x dx = + x2 R + L R cos + 2 d. Note that: cos γ R + 2 d = cos L R + 2 d + cos C R + 2 d. The γ R -integral can be computed using residues. If we are able to show the C R -integral tends to as R +, then one can determine our desired it cos R + L R + 2 d. Unfortunately, it is not possible to bound cos as cos is unbounded according to Exercise 4.27. One trick to get around with this issue is to consider + 2 the

4. Taylor and Laurent Series following function instead: ei + 2. When L R, we have = x + i and so: eix cos x + i sin x = + x2 + x 2 = L R f ( d = R R cos x + i sin x + x 2 dx. If we are able to find out f ( d, then one can recover the value of R + L R cos x by simply taking the real-part of f ( d. + x2 R + L R By considering the contour γ R = C R + L R, we have: f ( d = f ( d + f ( d. γ R L R C R The only singularity enclosed by γ R is i (when R is sufficiently large, so: f ( d = Res( f, i = γ R 2ie = π e. Next we show the C R -integral converges to as R. From Exercise 4.27, the term e i is bounded on the upper-half plane, and so whenever C R, we have: e i + 2 M + 2 M 2 = M R 2, where M is an upper bound of e i on the upper-half plane. Therefore, by Lemma 3.6, we get the estimate: e CR i + 2 d πr M R 2 as R +. Therefore, we get: f ( d = f ( d f ( d R L R R γ R R C R cos x + i sin x + x 2 dx = π e = π e. This shows: cos x + x 2 = π e. Before we give another example, we recall some fundamental facts that: For any =, the principal argument Arg( is in ( π, π]. Log( = ln + iarg( for any = Log( is holomorphic on C\(, ]. Therefore, if we apply Cauchy s integral formula or residue theory for an integral involving Log(, then we need to make sure the closed curve γ lies in C\(, ]. As such, we cannot apply residue methods with a semi-circle contour as in the previous example. Nonetheless, this kind of semi-circle contour is very useful when dealing with real integrals over (,. To get around with this issue, we can define a different branch of logarithm by the following. For any =, we let (4.5 Log π/2 ( := ln + iθ(

4.3. Residue Calculus where θ( is the unique angle in [ π 2, 3π 2 such that = eiθ(. By doing so, we still have e Log π/2 ( =. The notable difference from Log( is that now Log π/2 ( is holomorphic on C\{ + yi : y }, the yellow region below. y C R i C ε R ε ε R x Figure 4.. Domain of Log π/2 ( Exercise 4.28. Determine the value of Log π/2 ( when: (a = i (b = x + i where x > (c = x + i where x < Example 4.4. Let α be a real constant in (,. Evaluate the real integral: Solution I := First observe that for any x >, we have: x α ( + x 2 dx. x α = e α ln x = e αlog π/2 (x where Log π/2 is the special branch of logarithm defined in (4.5. It prompts us to consider a contour integral of the function: e αlog π/2 ( ( + 2. We pick a contour as shown in Figure 4., where C R and C ε are semi-circles with radii R and ε respectively. Since the closed contour γ R,ε := [ R, ε] + C ε + [ε, R] + C R lies completely inside the domain of Log π/2 (, by Residue Theorem, we have: γ R,ε e αlog π/2 ( ( + 2 d = Res( f, i = e αlog π/2 ( ( + i = 2ie α(ln i + π 2 i = π e απ 2 i. On the other hand, the γ R,ε -integral can break down into: (4.6 γ R,ε e αlog π/2 ( ( + 2 d = ( ε R R + + + C ε ε C R =i e αlog π/2 ( ( + 2 d

2 4. Taylor and Laurent Series Hence, Hence, When = x + i [ε, R], the integrand is simply: R ε R ε e αlog π/2 (x ( + x 2 = x α ( + x 2. e αlog π/2 ( ( + 2 d = When = x + i [ R, ε], the integrand becomes: x α ( + x 2 dx =: I. e αlog π/2 (x ( + x 2 = e α(ln x +πi ( + x 2 = e απi x α ( + x 2. ε R ε R e αlog π/2 ( ( + 2 d = e απi x α ( + x 2 }{{} even function dx = I e απi. We are left to analye the two semi-circular integrals. We will show that they tend to as ε and R. When C ε, we have: e αlog π/2 ( ( + 2 e α(ln +iθ( = 2 = e α ln ε ε 2 = ε α ε 2. By Lemma 3.6, we get: C ε e αlog π/2 ( ( + 2 d ε α ε πε α πε = as ε. 2 ε2 Similarly when C R, we have: e αlog π/2 ( ( + 2 e α(ln +iθ( 2 = R α R 2. By Lemma 3.6, we have the estimate: C ε e αlog π/2 ( ( + 2 d R α πr α R 2 πr = R 2 as R. Finally, by letting ε and R on both sides of (4.6, we get: Solving for I, we get: I = π e απ 2 i ( + e απi = π = e απ 2 i I + I e απi. π e απ 2 i + e απ 2 i = π 2 cos απ 2 = π 2 sec απ 2.

4.3. Residue Calculus 3 Exercise 4.29. Evaluate the following real integrals using residue methods: (a (b (c (d (e (f (g (h 2π 2π dθ where a > b >. (a + b cos θ 2 dθ where a R and a = ±. 2a cos θ + a2 x 2 (x 2 + a 2 dx where a >. 2 (x 2 dx where n N. + n cos ax x 2 dx where a and b are positive real numbers + b2 sin ax x(x 2 dx where a is a positive real number. + ln x x 2 dx where a >. + a2 x α ( + x 4 dx where α (,. Exercise 4.3. Show that for any t R, we have: Exercise 4.3. Show that: ( + x 2 e itx x 2 + dx = πe t. (2n!! dx = π. n+ (2n!!