Complex Series. Chapter 20

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1 hapter 20 omplex Series As in the real case, representation of functions by infinite series of simpler functions is an endeavor worthy of our serious consideration. We start with an examination of the properties of sequences and series of complex numbers and derive series representations of some complex functions. Most of the discussion is a direct generalization of the results of the real series. A sequence {z k } k= of complex numbers is said to converge to a limit z if sequence, lim k z z k =0. Inotherwords,foreachpositivenumberε there must exist an integer N such that z z k < ε whenever k>n. The reader may show that the real (imaginary) part of the limit of a sequence of complex numbers is the limit of the real (imaginary) part of the sequence. Series can be converted into sequences by partial summation. For instance, to study the infinite series k= z k, we form the partial sums Z n n k= z k and investigate the sequence {Z n } n=.wethussaythattheinfiniteseries k= z k converges to Z if lim n Z n = Z. Example Aseriesthatisusedofteninanalysisisthegeometricseries Z = zk.letusshowthatthisseriesconvergesto/( z) for z <. For a partial sum of n terms, we have n Z n z k =+z + z z n. Multiply this by z and subtract the result from the Z n sum to get (see also Example 9.3.3) Z n zz n = z n+ Z n = zn+. z We now show that Z n converges to Z =/( z). We have Z Z n = z zn+ z = z n+ z = z n+ z and z n+ lim Z Zn = lim n n z = lim z n z n+ =0 for z <. Thus, zk =/( z) for z <. convergence to a limit, partial sums, and series

2 56 omplex Series absolute convergence If the series z k converges, both the real part, x k, and the imaginary part, y k, of the series also converge. From hapter 9, we know that a necessary condition for the convergence of the real series x k and y k is that x k 0andy k 0. Thus, a necessary condition for the convergence of the complex series is lim k z k =0. Thetermsofsucha series are, therefore, bounded. Thus, there exists a positive number M such that z k <M for all k. Acomplexseriesissaidtoconverge absolutely, ifthereal series z k = x 2 k + y2 k converges. learly, absolute convergence implies convergence. 20. Power Series power series We now concentrate on the power series which, as in the real case, are infinite sums of powers of (z z 0 ). It turns out as we shall see shortly that for complex functions, the inclusion of negative powers is crucial. Theorem If the power series a k(z z 0 ) k converges for z (assumed to be different from z 0 ), then it converges absolutely for every value of z such that z z 0 < z z 0.Similarlyifthepowerseries b k/(z z 0 ) k converges for z 2 z 0,thenitconvergesabsolutely for every value of z such that z z 0 > z 2 z 0. Proof. We prove the first part of the proposition; the second part is done similarly. Since the series converges for z = z,alltheterms a k (z z 0 ) k are smaller than a positive number M. We,thereforehave a k (z z 0 ) k = a k(z z 0 ) k (z z 0) k (z z 0 ) k = a k (z z 0 ) k z z 0 k z z 0 MB k = M B k = M B, where B (z z 0 )/(z z 0 ) is a positive real number less than. Since the RHS is a finite (positive) number, the series of absolute values converges, and the proof is complete. The essence of Theorem 20.. is that if a power series with positive powers converges for a point at a distance r from z 0,thenitconvergesfor all interior points of a circle of radius r centered at z 0.Similarly,ifapower series with negative powers converges for a point at a distance r 2 from z 0, then it converges for all exterior points of a circle of radius r 2 centered at z 0 (see Figure 20.).

3 20. Power Series 57 z 0 r z 0 r 2 (a) (b) Figure 20.: (a) Power series with positive exponents converge for the interior points of a circle. (b) Power series with negative exponents converge for the exterior points of acircle. Box When constructing power series, positive powers are used for points inside a circle and negative powers for points outside it. The largest circle about z 0 such that the first power series of Theorem 20.. converges is called the circle of convergence of the power series. It circle of follows from Theorem 20.. that the series cannot converge at any point convergence outside the circle of convergence. (Why?) Let us consider the power series S(z) a k (z z 0 ) k (20.) which we assume to be convergent at all points interior to a circle for which z z 0 = r. This implies that the sequence of partial sums {S n (z)} converges. Therefore, for any ε > 0, there exists an integer N ε such that S(z) S n (z) < ε whenever n>n ε. In general, the integer N ε may be dependent on z; thatis,fordifferent values of z, wemaybeforcedtopickdifferent N ε s. When N ε is independent of z, we say that the convergence is uniform. We state the following result without proof: Theorem The power series S(z) = a n(z z 0 ) n is uniformly convergent for all points within its circle of convergence, and S(z) is an analytic function of z there. Furthermore, such a series can be differentiated and integrated term by term: ds(z) dz = na n (z z 0 ) n, n= γ S(z) dz = a n (z z 0 ) n dz, γ uniform convergence explained apowerseriesis uniformly convergent and analytic; it can be differentiated and integrated term by term.

4 58 omplex Series at each point z and each path γ located inside the circle of convergence of the power series. By substituting the reciprocal of (z z 0 )inthepowerseries,wecanshow that if b k/(z z 0 ) k is convergent in the annulus r 2 < z z 0 <r,then it is uniformly convergent for all z in that annulus, and the series represents acontinuousfunctionofz there Taylor and Laurent Series omplex series, just as their real counterparts, find their most frequent utility in representing well-behaved functions. The following theorem, which we state without proof, is essential in the application of complex analysis. Theorem Let and 2 be circles of radii r and r 2,bothcentered at z 0 in the z-plane with r >r 2. Let f(z) be analytic on and 2 and throughout S, theannularregionbetweenthetwocircles. Then,ateachpoint z of S, f(z) is given uniquely by the Laurent series f(z) = n= a n (z z 0 ) n, where a n = f(ξ) dξ, 2πi (ξ z 0 ) n+ and is any contour within S that encircles z 0. When r 2 =0,theseriesis called Taylor series. Inthatcasea n =0for negative n and a n = f (n) (z 0 )/n! for n 0. Maclaurin series We can see the reduction of the Laurent series to Taylor series as follows. The Laurent expansion is convergent as long as r 2 < z z 0 <r.inparticular, if r 2 =0,andifthefunctionisanalyticthroughouttheinteriorofthe larger circle, then f(ξ)/(ξ z 0 ) n+ will be analytic for negative integer n, and the integral will be zero by the auchy Goursat theorem. Therefore, a n will be zero for n =, 2,... Thus, only positive powers of (z z 0 )willbe present in the series, and we obtain the Taylor series. For z 0 =0,theTaylorseriesreducestotheMaclaurin series: f(z) =f(0) + f (0)z + = f (n) (0) z n. n! Box 9..4 tells us that we can enlarge and shrink 2 until we encounter a point at which f is no longer analytic. Thus, we can include all the possible analytic points by enlarging and shrinking 2. Example Let us expand some functions in terms of series. For entire functions there is no point in the entire complex plane at which they are not analytic. For a proof, see Hassani, S. Mathematical Physics: A Modern Introduction to Its Foundations, Springer-Verlag,999,Section9.6.

5 20.2 Taylor and Laurent Series 59 Thus, only positive powers of (z z 0)willbepresent,andwewillhaveaTaylor expansion that is valid for all values of z. (a) We expand e z around z 0 =0. Thenth derivative of e z is e z.thus,f (n) (0) =, and the Taylor (Maclaurin) expansion gives e z = f (n) (0) z n = n! z n n!. (b) The Maclaurin series for sin z is obtained by noting that { d n dz sin z 0 if n is even, n = z=0 ( ) (n )/2 if n is odd, and substituting this in the Maclaurin expansion: sin z = ( ) (n )/2 z n n! = ( ) k z 2k+ (2k +)!. n odd Similarly, we can obtain cos z = ( ) k z 2k (2k)!, sinh z = z 2k+ (2k +)!, cosh z = z 2k (2k)!. It is seen that the series representation of all these functions is obtained by replacing the real variable x in their real series representation with a complex variable z. (c) The function /(+z) is not entire, so the region of its convergence is limited. Let us find the Maclaurin expansion of this function. Starting from the origin (z 0 =0), the function is analytic within all circles of radii r<. At r =weencountera singularity, the point z =. Thus, the series converges for all points z for which z <. 2 For such points we have f (n) (0) = dn dz [( + n z) ] =( ) n n!. z=0 Thus, +z = f (n) (0) z n = n! ( ) n z n. The Taylor and Laurent series allow us to express an analytic function as apowerseries. ForaTaylorseriesoff(z) the expansion is routine because the coefficient of its nth term is simply f (n) (z 0 )/n!, where z 0 is the center of the circle of convergence. However, when a Laurent series is applicable in a given region of the complex plane, the nth coefficient is not, in general, easy to evaluate. Usually it can be found by inspection and certain manipulations of other known series. Then the uniqueness of Laurent series expansion assures us that the series so obtained is the unique Laurent series for the function in that region. 3 2 As remarked before, the series diverges for all points outside the circle z =. This does not mean that the function cannot be represented by a series for points outside the circle. On the contrary, we shall see shortly that the Laurent series, with negative powers is designed precisely for such a purpose. 3 See Hassani, S. Mathematical Physics: A Modern Introduction to Its Foundations, Springer-Verlag, 999, p there is only one Laurent series for agivenfunction defined in a given region.

6 520 omplex Series we can add, subtract, and multiply convergent power series. As in the case of real series, Box We can add, subtract, and multiply convergent power series. Furthermore, if the denominator does not vanish in a neighborhood of a point z 0,thenwecanobtaintheLaurentseriesoftheratiooftwopower series about z 0 by long division. Thus converging power series can be manipulated as though they were finite sums (polynomials). Such manipulations are extremely useful when dealing with Taylor and Laurent expansions in which the straightforward calculation of coefficients may be tedious. The following examples illustrate the power of infinite-series arithmetic. In these examples, the following equations are very useful: z = z n, +z = ( ) n z n, z <. (20.2) Example To expand the function f(z) = 2+3z z 2 + z 3 about z =0,rewriteitas f(z) = z 2 ( ) 2+3z = ( 3 ) ( = +z z 2 +z z 2 3 in a Laurent series ) ( ) n z n = z 2 (3 +z z2 + z 3 )= 2 z 2 + z +z z2 +. This series converges for 0 < z <. We note that negative powers of z are also present. This is a reflection of the fact that the function is not analytic inside the entire circle z =;itdivergesatz =0. Example The function f(z) =z/[(z )(z 2)] has a Taylor expansion around the origin for z <. To find this expansion, we write 4 f(z) = z + 2 z 2 = z z/2. Expanding both fractions in geometric series (both z and z/2 are less than ), we obtain f(z) = zn (z/2)n.addingthetwoseriesyields f(z) = ( 2 n )z n for z <. This is the unique Taylor expansion of f(z) withinthecircle z =. 4 We could, of course, evaluate the derivatives of all orders of the function at z =0and use the Maclaurin formula. However, the present method gives the same result much more quickly.

7 20.2 Taylor and Laurent Series 52 For the annular region < z < 2wehaveaLaurentseries. Thiscanbeseen by noting that f(z) = /z /z z/2 = ( ) z /z z/2. Since both fractions on the RHS are analytic in the annular region ( /z <, z/2 < ), we get f(z) = ( ) n ( z ) n = z n 2 n z n z z 2 = n= z n 2 n z n = n= a nz n, where a n = forn<0anda n = 2 n for n 0. This is the unique Laurent expansion of f(z) in the given region. Finally, for z > 2wehaveonlynegativepowersofz. Weobtaintheexpansion in this region by rewriting f(z) as follows: f(z) = /z /z + 2/z 2/z. Expanding the fractions yields f(z) = z n + 2 n+ z n = (2 n+ )z n. This is again the unique expansion of f(z) intheregion z > 2. The example above shows that a single function may have different series representations in different regions of the complex plane, each series having its own region of convergence. Example Define f(z) as { ( cos z)/z 2 for z 0, f(z) = for z =0. 2 We can show that f(z) isanentirefunction. Since cos z and z 2 are entire functions, their ratio is analytic everywhere except at the zeros of its denominator. The only such zero is z =0. Thus,f(z) is analytic everywhere except possibly at z = 0. Toseethebehavior off(z) atz = 0, we look at its Maclaurin series: cos z = ( ) n z 2n (2n)! which implies that cos z z 2 = ( ) n+ z 2n 2 (2n)! n= = 2 z2 4! + z4 6!. The expansion on the RHS shows that the value of the series is,which,bydefinition, is f(0). Thus, the series converges for all z, andbox20..2saysthatf(z) is 2 entire.

8 522 omplex Series ALaurentseriescangiveinformationabouttheintegralofafunction around a closed contour in whose interior the function may not be analytic. In fact, the coefficient of the first negative power in a Laurent series is given by Thus, a = 2πi f(ξ) dξ. (20.3) Box To find the integral of a (nonanalytic) function around a closed contour surrounding z 0,writetheLaurentseriesforthefunction and read off a,thecoefficient of the /(z z 0 ) term. The integral is 2πia. Example As an illustration of this idea, let us evaluate the integral I = dz/[z2 (z 2)], where is a circle of radius centered at the origin. The function is analytic in the annular region 0 < z < 2. We can, therefore, expand it as a Laurent series about z =0inthatregion: z 2 (z 2) = 2z 2 = 2 ( ) = ( z ) n z/2 2z 2 2 ) ( ) 4 z 8. ( z 2 Thus, a = 4,and dz/[z2 (z 2)] = 2πia = iπ/2. Any other way of evaluating the integral is nontrivial Problems 20.. Expand sinh z in a Taylor series about the point z = iπ Let be the circle z i =3integratedinthepositivesense. Find the value of each of the following integrals using the IF or the derivative formula (9.0): (a) (d) e z z 2 + π 2 dz. (b) dz (z 2 +9) 2. (e) For 0 <r<, show that sinh z (z 2 + π 2 dz. (c) ) 2 cosh z (z 2 + π 2 dz. (f) ) 3 dz z z 2 3z +4 z 2 4z +3 dz. r k cos kθ = r cos θ +r 2 2r cos θ and r k sin kθ = r sin θ +r 2 2r cos θ.

9 20.3 Problems Find the Taylor expansion of /z 2 for points inside the circle z 2 < Use mathematical induction to show that d n ( + z) dzn =( ) n n!. z= Find the (unique) Laurent expansion of each of the following functions in each of its regions of analyticity: (a) (z 2)(z 3). (b) z cos(z2 ). (c) z 2 ( z). sinh z z (d) z 4. (e) ( z) 3. (f) z 2. (g) z2 4 z 2 9. (h) (z 2 ) 2. z (i) z Show that the following functions are entire: e 2z (a) f(z) = z 2 2 { z for z 0, sin z (b) f(z) = z for z 0, 2 for z =0. for z =0. cos z (c) f(z) = z 2 π 2 for z ±π/2, /4 /π for z = ±π/ Obtain the first few nonzero terms of the Laurent-series expansion of each of the following functions about the origin by approximating the denominator by a polynomial and using the technique of long division of polynomials. Also find the integral of the function along a small simple closed contour encircling the origin. (a) (e) sin z. (b) cos z. (c) z cosh z. (d) z 2 z sin z. e z. (f) z 2 sin z. (g) z 4 6z + z 3 6sinhz Obtain the Laurent-series expansion of f(z) = sinhz/z 3 about the origin.