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4 Laurent's Theorem for Complex Functions LAURENT’S THEOREM FOR COMPLEX FUNCTIONS 487 4 LAURENT’S THEOREM FOR COMPLEX FUNCTIONS 1 Consider the function f : C n f2g ! C given by f(z) = z¡2 . By Taylor’s theorem, f has a Taylor series centered at z0 = 0 with neighborhood of con- vergence N2(0). That is, 1 X f (j)(0) f(z) = zj on N (0). j! 2 j=0 But this function f is also defined for jzj > 2, so it is natural to ask if f can be represented by some (other) series expansion centered at z0 = 0 and convergent for jzj > 2. To explore this, note that à ! 1 1 1 1 1 1 X ³2´j X 2j f(z) = = = = , z ¡ 2 z 1 ¡ 2 z z zj+1 z j=0 j=0 ¯ ¯ ¯ 2 ¯ and this series converges for z < 1, i.e., for jzj > 2. This type of power series, which involves powers of 1 , or equivalently, negative powers of z¡z0 (z ¡ z0), is known as a Laurent series. More generally, the form of a Laurent series centered at a point z0 involves both positive and negative powers of (z ¡ z0), that is, X1 X1 b f(z) = a (z ¡ z )j + j , (9.21) j 0 (z ¡ z )j j=0 j=1 0 and so it consists of two infinite series. The first series in (9.21) is often referred to as the analytic part and the second series in (9.21) is often referred to as the singular part, or the principal part of the Laurent series. For the Laurent series expansion to exist at a particular z 2 C, both the analytic part and the singular part must be convergent at z. In fact, the analytic part consisting of positive powers of (z ¡ z0) will converge for all z inside some circle of radius R, while the singular part consisting of negative powers of (z ¡ z0) will converge for all z outside some circle of radius r, as in our example. It is the overlap of the two regions of convergence associated with the analytic part and the singular part that comprises the region of convergence of the Laurent series. This region of overlap is typically the annulus centered at z0 R and denoted by Ar (z0) = fz 2 C : r < jz ¡ z0j < Rg. See Figure 9.3 for an illustration. The situation R = 1 and r = 0 corresponds to the case where the Laurent series converges everywhere in C except possibly at z0. (In our 1 particular example, we have convergence in the annulus A2 (0).) Note that since the annulus so described excludes the point z0, it is not necessary that f be differentiable at z0 for it to have a Laurent series expansion centered there. (In fact, the function f need not be differentiable anywhere within Nr(z0).) R1 R If 0 < r < r1 < R1 < R we will refer to the annulus Ar1 (z0) ½ Ar (z0) R as a proper subannulus of Ar (z0). In a manner similar to that of a convergent 488 TAYLOR SERIES, LAURENT SERIES, AND THE RESIDUE CALCULUS R Ar (z0) r R z0 Figure 9.3 The region of convergence of a Laurent series. Taylor series on subsets of its neighborhood of convergence, when a function R f has a convergent Laurent series expansion on an annulus Ar (z0), it will R R1 R1 converge absolutely on Ar (z0) and uniformly on Ar1 (z0) where Ar1 (z0) is R any proper subannulus of Ar (z0). To establish this, we must consider the convergence of a given Laurent series a bit more carefully. Since the analytic part of the Laurent series is a power series, it will converge absolutely on its neighborhood of convergence NR(z0). Recall that the convergence is uniform on NR1 (z0) for any 0 < R1 < R. To analyze the singular part, we define ½ ¾ X1 b r ´ inf jz ¡ z j : j converges . 0 (z ¡ z )j j=1 0 If r fails to exist then the singular part never converges. If r exists then we C will show that the singular part converges absolutely for (Nr(z0)) , and uni- C formly on (Nr1 (z0)) for any r1 > r. To see this, choose r1 > r. Then as indi- P1 bj cated in Figure 9.4 there exists z1 such that r · jz1¡z0j < r1 and j j=1 (z1¡z0) converges. Therefore, there exists M ¸ 0 such that ¯ ¯ ¯ ¯ ¯ bj ¯ ¯ j ¯ · M for all j ¸ 1. (z1 ¡ z0) C Now, for z 2 (Nr1 (z0)) we have ¯ ¯ ¯ ¯ µ ¶ µ ¶ ¯ ¯ ¯ ¯ j j ¯ bj ¯ ¯ bj ¯ jz1 ¡ z0j jz1 ¡ z0j ¯ j ¯ = ¯ j ¯ · M . (z ¡ z0) (z1 ¡ z0) jz ¡ z0j r1 P1 bj Since jz1¡z0j < r1 this establishes via the Weierstrass M-test that j j=1 (z¡z0) C C converges absolutely on (Nr(z0)) and uniformly on (Nr1 (z0)) . Since r1 > r was arbitrary, this result holds for any r1 > r. We leave it to the reader to P1 bj show that j diverges on Nr(z0). As with a Taylor series, the points j=1 (z¡z0) on the boundary Cr(z0) must be studied individually. LAURENT’S THEOREM FOR COMPLEX FUNCTIONS 489 r r z0 1 Figure 9.4 The regions of convergence and divergence of the singular part of a Laurent series. P1 bj I 9.42 As claimed above, show that j diverges on Nr(z0). j=1 (z¡z0) The above discussion and exercise establish the following result. Proposition 4.1 Suppose f : D ! C has a Laurent series expansion f(z) = P1 j P1 bj R aj(z ¡ z0) + j on the annulus A (z0) ½ D (where r ¸ 0 and R j=0 j=1 (z¡z0) r R1 R may be 1). Then for any proper subannulus Ar1 (z0) ½ Ar (z0) the given Laurent R1 R1 series expansion for f converges absolutely on Ar1 (z0) and uniformly on Ar1 (z0). This result, in turn, implies the following. It will be instrumental in proving part of the key result of this section. P1 bj Proposition 4.2 Let j be the singular part of a Laurent series expan- j=1 (z¡z0) R P1 bj sion for f : D ! C on A (z0) ½ D. Then j represents a continuous r j=1 (z¡z0) C C function on (Nr(z0)) , and for any contour C ½ (Nr(z0)) we have Z µ 1 ¶ 1 Z X bj X bj j dz = j dz. (z ¡ z0) (z ¡ z0) C j=1 j=1 C I 9.43 Prove the above proposition. It is also true, and of great practical value as we will see, that when a Laurent series expansion exists for a function it is unique. To see this, suppose f is R differentiable on the annulus Ar (z0) and suppose too that it has a convergent Laurent series there given by 1 1 X X b f(z) = a (z ¡ z )k + k . (9.22) k 0 (z ¡ z )k k=0 k=1 0 490 TAYLOR SERIES, LAURENT SERIES, AND THE RESIDUE CALCULUS Recall that we established the uniqueness of the Taylor series representation of a function centered at a point by differentiating the series term-by-term; here, we will establish the uniqueness of the Laurent series representation by R integrating term-by-term. Let C be any simple closed contour in Ar (z0) with R1 R nC (z0) = 1, and note that C ½ Ar1 (z0) ½ Ar (z0) for some proper subannulus R1 R Ar1 (z0) ½ Ar (z0). Also note that for any k ¸ 0, I I 1 I 1 f(³) X X b d³ = a (³ ¡ z )j¡k¡1 d³ + j d³ k+1 j 0 j+k+1 (³ ¡ z0) (³ ¡ z0) C C j=0 C j=1 I I X1 X1 b = a (³ ¡ z )j¡k¡1 d³ + j d³ (9.23) j 0 j+k+1 (³ ¡ z0) j=0 C j=1 C = 2¼i ak nC (z0) = 2¼i ak. Note in (9.23) above that the integral of each summand corresponding to aj for j 6= k, and for bj for j ¸ 1, vanishes due to the integrand having an antiderivative within NR1 (z0). Also, in integrating the singular part term- by-term, we have applied Proposition 4.2. This shows that each ak in (9.22) is uniquely determined by f and z0. A similar argument can be made for the uniqueness of each b for k ¸ 1 in (9.22) by considering the integral H k f(³) d³ for any fixed k ¸ 1. We leave this to the reader. (³¡z )¡k+1 C 0 I 9.44 Establish the uniqueness of the bk terms in the Laurent series representation H f(³) (9.22) by carrying out the integral ¡k+1 d³ for any fixed k ¸ 1. (³¡z0) C With these important facts about an existing Laurent series established, we now state and prove Laurent’s theorem, the key result of this section. It is a kind of sibling to Taylor’s theorem in complex function theory. It gives conditions under which a function f : D ! C is guaranteed a Laurent series R representation convergent on an annulus Ar (z0) ½ D. Theorem 4.3 (Laurent’s Theorem) R Let f : D ! C be differentiable on the annulus Ar (z0) = fz : r < jz ¡ z0j < Rg ½ D (where r ¸ 0 and R may be 1). Then f(z) can be expressed uniquely by X1 X1 b f(z) = a (z ¡ z )j + j , j 0 (z ¡ z )j j=0 j=1 0 R R on Ar (z0).
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