Example Example demonstrates that every complex number is a nonisolated singularity

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1 4.3. The Residue Theorem This section presents another landmark result in basic complex function theory, namely the Residue Theorem. We begin our preparations by classifying the various ways in which a function can fail to be analytic. Definition (i) A complex number z is said to be a singularity of a function f if f is not analytic at z. (ii) Suppose that z is a singularity of f. Wesaythatz is an isolated singularity if there is some r> such that f is analytic at every point in the punctured disc D (z ; r). A singularity is said to be nonisolated if it is not isolated. Example Example demonstrates that every complex number is a nonisolated singularity of the function z. In what follows we shall be concerned exclusively with isolated singularities. We begin by classifying them on a finer scale. Definition Suppose that z is an isolated singularity of a function f. Let r be a positive number such that f is analytic at every point in the punctured disc {z : < z z <r} (in practice one usually picks the largest such r available). Theorem 4.2. asserts that f can be expanded in a Laurent series within this punctured disc. Let us write a n (z z ) n + b n (z z ) n, < z z <r. (4.3.) (i) We say that z is a removable singularity if each b n in (4.3.) is zero. (ii) Let m be a positive integer. The singularity z is said to be a pole of order m if, in the expansion (4.3.), we have b m andb n =foreveryn>m. (iii) We say that z is an essential singularity if, in the expansion (4.3.), b n for infinitely many values of n. (iv) The coefficient of (z z ) in the expansion (4.3.), namely the number b, is called the residue of f at z ;wedenoteitbyres(f;z ). Example (i) The complex number z = is the only singularity of the function sin(z)/z; in particular it is an isolated singularity. Moreover, the third identity in Example 4..5 implies that the Laurent expansion of f in the punctured complex plane is given by sin z z = ( ) n z 2n (2n +)!, < z <. This Laurent expansion has no principal part, hence z is a removable singularity. (ii) Let /(z )(z 2). The number z = is a singularity of f, andfis analytic in the punctured disc D (z ; ) (which is, in fact, the largest such disc available). Therefore z is an isolated singularity of f. Furthermore, equation (4.2.8) shows that b = andb n =forevery n 2. Thus z is a pole of order (a simple pole), and Res(f; z )=. (iii) Let f be as in (ii) above. The number z = 2 is also a singularity of f, and, since f is analytic in the punctured disc D (z ; ), we find that z is an isolated singularity. However, the Laurent series in Example 4.2.3(v) cannot be used to determine the nature of the singularity at z, for that

2 expansion is valid in the region {z :< z 2 < }, andnot in a punctured disc centred at z. The required Laurent expansion in D (z ;) is given by (z 2)(z ) = (z 2)( + z 2) = ( ) n (z 2) n (z 2) = ( ) n (z 2) n, < z 2 <, from which we determine that z is a simple pole, and also that Res(f; z )=. (iv) The number z = is the only singularity, hence an isolated one, of the function exp(/z). Example 4.2.3(i) reveals that the singularity is essential, and that Res(f; z )=. (v) The number z = is the only singularity, hence an isolated one, of the function sin(z)/z 2. Example 4.2.3(ii) shows that z is a simple pole, and that Res(f; z )=. A convenient method of determining poles and their orders is the subject of this next result. Lemma Suppose that f is a function and that z. Let m be a positive integer. The following are equivalent: (i) The point z is a pole of order m of f. (ii) There is a positive number r such that f admits the following (local) representation in the punctured disc D (z ; r): φ(z) (z z ) m, < z z <r, (4.3.2) where φ is analytic in D(z ; r) and φ(z ). Proof. If f has a pole of order m at z, then there is some r>such that f admits the following Laurent expansion in D (z ; r): a n (z z ) n + b n (z z ) n, < z z <r, (4.3.3) with b m. Multiply both sides of (4.3.3) by (the nonzero number) (z z ) m to get Define (z z ) m φ(z) := a n (z z ) n+m + b n (z z ) m n, < z z <r. (4.3.4) b n (z z ) m n + a n (z z ) n+m, z D(z ;r). Then φ is analytic in the disc D(z ; r), because it is the sum of a convergent power series in that disc (Theorem 4..). Moreover, φ(z )=b m, and (4.3.4) is equivalent to (4.3.2). onversely, assume that f admits the local representation (4.3.2) in some punctured disc D (z ; r). Theorem 4..4 assures us that φ can be expanded in a Taylor series in D(z ; r): φ(z) = φ (n) (z ) (z z ) n, z z <r. (4.3.5) 2

3 Substituing (4.3.5) in (4.3.2) leads to the relation φ (n) (z ) (z z ) n m, < z z <r, (4.3.6) and the uniqueness of Laurent series expansions ensures that (4.3.6) is, in fact, the Laurent expansion of f in the punctured disc D (z ; r). The principal part of this series is given by m φ (n) (z ) (z z ) n m, (4.3.7) from which we see that b n =foreveryn>m, and also that b m = φ(z ). Thusz is a pole of order m. Remark (i) Suppose that f has a pole of order m at z,andletrand φ be given by the previous lemma. Equations (4.3.6) and (4.3.7) reveal that Res(f; z )=b = φ(m ) (z ) (m )! (ii) Let z, φ, etc. be as above. As φ is analytic, in particular continuous, at z,andφ(z ), there exists a positive number ρ<rsuch that φ(z) φ(z ) < φ(z ) /2 whenever z z ρ.so for every such z we have φ(z) = φ(z ) (φ(z ) φ(z)) φ(z ) φ(z ) φ(z) > φ(z ) /2; in particular φ(z) foreveryz D(z ;ρ). Example (i) Let /(z )(z 2). Defining φ(z) :=/(z ), we may write φ(z), < z 2 <. (z 2) As φ is analytic throughout the disc D(2; ) and φ(2) =, we see that f admits a local representation of the form (4.3.2) with m =andr=. onsequently, Lemma asserts that the complex number 2 is a simple pole of f, whereas Remark 4.3.6(i) shows that Res(f;2) =. This confirms our findings in Example 4.3.4(iii). (ii) The function cos(πz) f(z) := (z ) 2 (z +) has two singularities: z := andz :=. Put. [cos(πz)/(z )2 ] z + =: φ (z) z +, < z+ <2. Then φ is analytic in D(z ;2) and φ (z )=cos( π)/( ) 2 = /4. Therefore f has a simple pole at z and Res(f; z )=φ (z )= /4. Turning to the other singularity, let us write [cos(πz)/(z +)] (z ) 2 =: φ (z), < z < 2. (z ) 2 3

4 The function φ is analytic throughout the open disc D(z ;2) and φ (z )=cos(π)/( + ) = /2. Therefore Lemma implies that z is a pole of order 2. Moreover, φ cos(πz) (z) = π(z+)sin(πz) (z +) 2, < z <2, so that via Remark 4.3.6(i). Res(f; z )=φ ()/! = /4 We are now ready for the main attraction, to wit, the Residue Theorem. The reason for its name will soon become obvious. Theorem (Residue Theorem) Let be a (positively oriented) simple closed contour enclosing a region D. Suppose that f is analytic at every point in D, except for finitely many singularities z,...,z m, all of which lie in D. Then f(z)dz = m Res(f; z k ). Proof. Let k m. As z k D, an open set, there is a positive number R k such that D(z k ; R k ) D. Let R := min{r k : k m}, ρ := min{ z j z k : j, k m, j k}, and r := (/2) min{r, ρ}. Then D(z k ;r) D for every k m, andd(z j ;r) D(z k ;r)= whenever j k. Asfis analytic in the punctured disc D (z k ; r) for every k {,...,m},itadmits a Laurent expansion there. Let us write a n,k (z z k ) n + b n,k (z z k ) n, < z z k <r, k m, and note for future reference that Res(f; z k )=b,k, k m. (4.3.8) Let k, k m, denote the circle of radius r/2, centred at z k, and traversed once in the positive direction. As f is analytic throughout ( D) \{z,...,z m }, an extension of the ontour Transfer Principle (see remark immediately following Lemma 3.4.) asserts that f(z) dz = f(z) dz. (4.3.9) k Recalling from Theorem 4.2. that b,k = f(z) dz, k m, k we find that the required result obtains via (4.3.9) and (4.3.8). Example (i) Let denote the (positively oriented) boundary of the square with vertices ±2 ± 2i. Theorem and Example 4.3.7(ii) combine to show that cos(πz) (z ) 2 dz =( /4+/4) =. (z +) 4

5 (ii) Let l be a fixed positive integer. The complex number is the only singularity of the function exp(/z l ). Moreover, the Laurent expansion of f in the punctured complex plane is given by whence we conclude that ( ) exp z l = Res(f;)=, < z <, ()zln {, if l =;, otherwise. onsequently, if is any (positively oriented) simple closed contour enclosing the origin, then the Residue Theorem asserts that ( ) {, if l =; exp z l dz =, otherwise. (iii) Let be a (positively oriented) simple closed contour enclosing a region D. Suppose that f is analytic throughout D, andletw Dbe a fixed point. Define g(z) := f(z) (z w) k+, where k is a fixed nonnegative integer. Then g is analytic at every point in ( D) \{w},sothe Residue Theorem affirms that f(z) dz =Res(g;w). (z w) k+ (4.3.) As w belongs to the open set D, there is some positive number r such that D(w; r) D. The analyticity of f in D(w; r) and Theorem 4..4 combine to provide the representation whence g(z) = f (n) (w) (z w) n, z D(w;r), f (n) (w) (z w) n k, < z w <r. The series given above must be the Laurent expansion of g in D (w; r) (because of uniqueness); in particular, the residue of g at w is the coefficient of (z w) in that series. Thus Res(g; w) = f(k) (w), k! and combining this with (4.3.), we see that we have deduced the auchy Integration Formulae from the Residue Theorem. (iv) Let n 3 be a fixed positive integer. Let us compute the infinite integral I(n) := 5 dt +t n.

6 Let R be a large positive number, and let R be the contour comprising the following three pieces: the straight-line segment L from z =toz=r, followed by the circular arc γ R (traversed counterclockwise) from z = R to z = R exp(/n), followed by the line segment L 2 from z = R exp(/n) toz=. onsider the function f(z) := +z n, which has n singularities (at the n points where z n + = ). These are given by ( ( π z k := exp i n + 2kπ )), k n, n of which only z lies within R. Let r be a positive number such that z k / D(z ; r) for every k n. As n +z n = (z z k )=:(z z )p(z), z, we may write k= /p(z) z z =: φ(z) z z, < z z <r, and conclude, via Lemma and Remark 4.3.6(i), that z is a simple pole of f, andthat Res(f; z )=φ(z )= p(z ). An appeal now to the Residue Theorem leads us to the conclusion that f(z) dz = Res(f; z )= p(z ). (4.3.) R Parametrize L by the function t t, t R,and L 2 by the function t t exp(/n), t R. Then (4.3.) can be rewritten as follows: R dt +t n + γ R dz R +z n e /n +(te /n ) n dt = which, in turn, gives the equation ( R e /n) dt +t n = p(z ) γ R p(z ), (4.3.2) dz +z n. (4.3.3) Now if z γ R,then +z n z n =R n, so the M-L Theorem provides the bound dz +z n 2πR R n, γ R and the right-hand side of this inequality approaches zero as R tends to infinity. Using this in (4.3.3), we obtain I(n) = ( ) e /n p(z ) = π csc ( ) π ( ) n = n n ( e /n (z z k ) ). e i2kπ/n 6