Complex Series

Chapter 20

Complex 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 {zk }k=1 of complex numbers is said to converge to a limit z if limk→∞ |z − zk | = 0. In other words, for each positive number ε there must exist an integer N such that |z − zk | < 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, !n to study ∞ z , we form the partial sums Z ≡ the infinite series k n k and k=1 k=1 !z∞ ∞ investigate the sequence {Zn }n=1 . We thus say that the infinite series k=1 zk converges to Z if limn→∞ Zn = Z.

Example 20.0.1. A series that is used often in analysis is the geometric series ! k Z= ∞ k=0 z . Let us show that this series converges to 1/(1 − z) for |z| < 1. For a partial sum of n terms, we have Zn ≡

n "

k=0

zk = 1 + z + z2 + · · · + zn .

Multiply this by z and subtract the result from the Zn sum to get (see also Example 9.3.3) 1 − z n+1 . Zn − zZn = 1 − z n+1 ⇒ Zn = 1−z We now show that Zn converges to Z = 1/(1 − z). We have # # # # # 1 |z|n+1 1 − z n+1 ## ## z n+1 ## = = |Z − Zn | = ## − # # # 1−z 1−z 1−z |1 − z|

and

1 |z|n+1 = lim |z|n+1 = 0 lim |Z − Zn | = lim n→∞ n→∞ |1 − z| |1 − z| n→∞ ! k for |z| < 1. Thus, ∞ k=0 z = 1/(1 − z) for |z| < 1.

!

sequence, convergence to a limit, partial sums, and series

516

absolute convergence

Complex Series !∞ !∞ If the series !k=0 zk converges, both the real part, k=0 xk , and the ∞ imaginary part, y , of the series also converge. From Chapter k k=0 !∞9, we know! that a necessary condition for the convergence of the real series k=0 xk ∞ and k=0 yk is that xk → 0 and yk → 0. Thus, a necessary condition for the convergence of the complex series is limk→∞ zk = 0. The terms of such a series are, therefore, bounded. Thus, there exists a positive number M such that |zk | < M for all k. A complex series is said to converge absolutely, if the real series ∞ ∞ # " " |zk | = x2k + yk2 k=0

k=0

converges. Clearly, absolute convergence implies convergence.

20.1 power series

Power Series

We now concentrate on the power series which, as in the real case, are infinite sums of powers of (z − z0 ). It turns out—as we shall see shortly—that for complex functions, the inclusion of negative powers is crucial. ! k Theorem 20.1.1. If the power series ∞ k=0 ak (z − z0 ) converges for z1 (assumed to be different from z0 ), then it converges absolutely for every value !∞ of z such that |z − z0 | < |z1 − z0 |. Similarly if the power series k=0 bk /(z − z0 )k converges for z2 ̸= z0 , then it converges absolutely for every value of z such that |z − z0 | > |z2 − z0 |. Proof. We prove the first part of the proposition; the second part is done similarly. Since the series converges for z = z1 , all the terms |ak (z1 − z0 )k | are smaller than a positive number M . We, therefore have $ ∞ ∞ $ k $ " " $ k k (z − z0 ) $ $ |ak (z − z0 ) | = $ak (z1 − z0 ) (z1 − z0 )k $ k=0

=

k=0 ∞ " k=0

=M

$ $ ∞ $ z − z0 $k " $ ≤ |ak (z1 − z0 )k | $$ M Bk z1 − z0 $

∞ "

k=0

k=0

Bk =

M , 1−B

where B ≡ |(z − z0 )/(z1 − z0 )| is a positive real number less than 1. Since the RHS is a finite (positive) number, the series of absolute values converges, and the proof is complete. The essence of Theorem 20.1.1 is that if a power series—with positive powers—converges for a point at a distance r1 from z0 , then it converges for all interior points of a circle of radius r1 centered at z0 . Similarly, if a power series—with negative powers—converges for a point at a distance r2 from z0 , then it converges for all exterior points of a circle of radius r2 centered at z0 (see Figure 20.1).

20.1 Power Series

517

z0

z0 r1

r2

(b)

(a)

Figure 20.1: (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 a circle.

Box 20.1.1. When constructing power series, positive powers are used for points inside a circle and negative powers for points outside it. The largest circle about z0 such that the first power series of Theorem 20.1.1 converges is called the circle of convergence of the power series. It follows from Theorem 20.1.1 that the series cannot converge at any point outside the circle of convergence. (Why?) Let us consider the power series S(z) ≡

∞ !

k=0

ak (z − z0 )k

circle of convergence

(20.1)

which we assume to be convergent at all points interior to a circle for which ∞ |z − z0 | = r. This implies that the sequence of partial sums {Sn (z)}n=0 converges. Therefore, for any ε > 0, there exists an integer Nε such that |S(z) − Sn (z)| < ε

whenever n > Nε .

In general, the integer Nε may be dependent on z; that is, for different values of z, we may be forced to pick different Nε ’s. When Nε is independent of z, we say that the convergence is uniform. We state the following result without proof: "∞ Theorem 20.1.2. The power series S(z) = n=0 an (z − z0 )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) ! = nan (z − z0 )n−1 , S(z) dz = an (z − z0 )n dz, dz γ γ n=1 n=0

uniform convergence explained a power series is uniformly convergent and analytic; it can be differentiated and integrated term by term.

518

Complex 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 − z0 ) in the power series, we can show ∞ that if k=0 bk /(z − z0 )k is convergent in the annulus r2 < |z − z0 | < r1 , then it is uniformly convergent for all z in that annulus, and the series represents a continuous function of z there.

20.2

Taylor and Laurent Series

Complex series, just as their real counterparts, find their most frequent utility in representing well-behaved functions. The following theorem, which we state without proof,1 is essential in the application of complex analysis. Theorem 20.2.1. Let C1 and C2 be circles of radii r1 and r2 , both centered at z0 in the z-plane with r1 > r2 . Let f (z) be analytic on C1 and C2 and throughout S, the annular region between the two circles. Then, at each point z of S, f (z) is given uniquely by the Laurent series f (z) =

∞ "

n=−∞

an (z − z0 )n ,

where

an =

1 2πi

#

C

f (ξ) dξ, (ξ − z0 )n+1

and C is any contour within S that encircles z0 . When r2 = 0, the series is called Taylor series. In that case an = 0 for negative n and an = f (n) (z0 )/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 r2 < |z − z0 | < r1 . In particular, if r2 = 0, and if the function is analytic throughout the interior of the larger circle, then f (ξ)/(ξ − z0 )n+1 will be analytic for negative integer n, and the integral will be zero by the Cauchy–Goursat theorem. Therefore, an will be zero for n = −1, −2, . . . . Thus, only positive powers of (z − z0 ) will be present in the series, and we obtain the Taylor series. For z0 = 0, the Taylor series reduces to the Maclaurin series: ∞ " f (n) (0) n z . f (z) = f (0) + f (0)z + · · · = n! n=0 ′

Box 19.1.4 tells us that we can enlarge C1 and shrink C2 until we encounter a point at which f is no longer analytic. Thus, we can include all the possible analytic points by enlarging C1 and shrinking C2 . Example 20.2.2. 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. 1 For a proof, see Hassani, S. Mathematical Physics: A Modern Introduction to Its Foundations, Springer-Verlag, 1999, Section 9.6.

20.2 Taylor and Laurent Series

519

Thus, only positive powers of (z − z0 ) will be present, and we will have a Taylor expansion that is valid for all values of z. (a) We expand ez around z0 = 0. The nth derivative of ez is ez . Thus, f (n) (0) = 1, and the Taylor (Maclaurin) expansion gives ez =

∞ ∞ ! f (n) (0) n ! z n z = . n! n! n=0 n=0

(b) The Maclaurin series for sin z is obtained by noting that # " " 0 if n is even, dn " sin z " = (n−1)/2 dz n (−1) if n is odd, z=0 and substituting this in the Maclaurin expansion: sin z =

!

(−1)(n−1)/2

n odd

Similarly, we can obtain cos z =

∞ !

(−1)k

k=0

z 2k , (2k)!

∞

! zn z 2k+1 (−1)k = . n! (2k + 1)! k=0

sinh z =

∞ !

k=0

z 2k+1 , (2k + 1)!

cosh z =

∞ ! z 2k . (2k)! k=0

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 1/(1+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 (z0 = 0), the function is analytic within all circles of radii r < 1. At r = 1 we encounter a singularity, the point z = −1. Thus, the series converges for all points z for which |z| < 1.2 For such points we have " " dn −1 " [(1 + z) ] = (−1)n n!. f (n) (0) = " n dz z=0 Thus,

∞ ∞ ! 1 f (n) (0) n ! = z = (−1)n z n . 1+z n! n=0 n=0

!

The Taylor and Laurent series allow us to express an analytic function as a power series. For a Taylor series of f (z) the expansion is routine because the coefficient of its nth term is simply f (n) (z0 )/n!, where z0 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| = 1. 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, 1999, p. 258.

there is only one Laurent series for a given function defined in a given region.

520 we can add, subtract, and multiply convergent power series.

Complex Series As in the case of real series, Box 20.2.1. We can add, subtract, and multiply convergent power series. Furthermore, if the denominator does not vanish in a neighborhood of a point z0 , then we can obtain the Laurent series of the ratio of two power series about z0 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: ∞ ! 1 = zn, 1 − z n=0

∞ ! 1 = (−1)n z n , 1 + z n=0

Example 20.2.3. To expand the function f (z) = about z = 0, rewrite it as 1 f (z) = 2 z =

"

2 + 3z 1+z

#

1 = 2 z

"

1 3− 1+z

#

1 = 2 z

|z| < 1.

(20.2)

2 + 3z in a Laurent series z2 + z3 $

3−

∞ !

n n

(−1) z

n=0

%

1 2 1 (3 − 1 + z − z 2 + z 3 − · · · ) = 2 + − 1 + z − z 2 + · · · . z2 z z

This series converges for 0 < |z| < 1. 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| = 1; it diverges at z = 0. !

Example 20.2.4. The function f (z) = z/[(z − 1)(z − 2)] has a Taylor expansion around the origin for |z| < 1. To find this expansion, we write4 f (z) = −

2 1 1 1 + = − . z−1 z−2 1−z 1 − z/2

Expanding both in geometric series (both |z| and |z/2| are less than 1), we &fractions &∞ n n obtain f (z) = ∞ z − n=0 n=0 (z/2) . Adding the two series yields f (z) =

∞ !

(1 − 2−n )z n

n=0

for

|z| < 1.

This is the unique Taylor expansion of f (z) within the circle |z| = 1. 4 We could, of course, evaluate the derivatives of all orders of the function at z = 0 and use the Maclaurin formula. However, the present method gives the same result much more quickly.

20.2 Taylor and Laurent Series

521

For the annular region 1 < |z| < 2 we have a Laurent series. This can be seen by noting that ! " 1/z 1 1 1 1 f (z) = − =− − . 1/z − 1 1 − z/2 z 1 − 1/z 1 − z/2 Since both fractions on the RHS are analytic in the annular region (|1/z| < 1, |z/2| < 1), we get ∞ ! "n ∞ $ % ∞ ∞ # # # z n 1# 1 − =− z −n−1 − 2−n z n f (z) = − z n=0 z 2 n=0 n=0 n=0 =−

−∞ #

n=−1

zn −

∞ #

n=0

2−n z n = −

∞ #

an z n ,

n=−∞

−n

for n ≥ 0. This is the unique Laurent where an = −1 for n < 0 and an = −2 expansion of f (z) in the given region. Finally, for |z| > 2 we have only negative powers of z. We obtain the expansion in this region by rewriting f (z) as follows: f (z) = −

2/z 1/z + . 1 − 1/z 1 − 2/z

Expanding the fractions yields f (z) = −

∞ #

z −n−1 +

n=0

∞ #

2n+1 z −n−1 =

n=0

∞ #

(2n+1 − 1)z −n−1 .

n=0

This is again the unique expansion of f (z) in the region |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 20.2.5. Define f (z) as f (z) =

&

(1 − cos z)/z 2 1 2

for z ̸= 0, for z = 0.

We can show that f (z) is an entire function. Since 1 − 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. To see the behavior of f (z) at z = 0, we look at its Maclaurin series: ∞ # z 2n (−1)n 1 − cos z = 1 − (2n)! n=0

which implies that

∞ # 1 z2 z4 z 2n−2 1 − cos z = − + − ··· . = (−1)n+1 2 z (2n)! 2 4! 6! n=1

The expansion on the RHS shows that the value of the series is 12 , which, by definition, is f (0). Thus, the series converges for all z, and Box 20.1.2 says that f (z) is ! entire.

522

Complex Series A Laurent series can give information about the integral of a function 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 ! 1 f (ξ) dξ. (20.3) a−1 = 2πi C Thus, Box 20.2.2. To find the integral of a (nonanalytic) function around a closed contour surrounding z0 , write the Laurent series for the function and read off a−1 , the coefficient of the 1/(z − z0 ) term. The integral is 2πia−1 . Example 20.2.6. As an illustration of this idea, let us evaluate the integral I = "

dz/[z 2 (z − 2)], where C is a circle of radius 1 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 = 0 in that region: C

$ # ∞ 1 1 1 % & z 'n 1 = − = − z 2 (z − 2) 2z 2 1 − z/2 2z 2 n=0 2 # $ # $ 1 1 1 1 1 =− − −··· . − 2 z2 4 z 8 " 2 Thus, a−1 = − 14 , and C dz/[z (z − 2)] = 2πia−1 = −iπ/2. Any other way of evaluating the integral is nontrivial. !

20.3

Problems

20.1. Expand sinh z in a Taylor series about the point z = iπ. 20.2. Let C be the circle |z − i| = 3 integrated in the positive sense. Find the value of each of the following integrals using the CIF or the derivative formula (19.10): ! ! ! ez sinh z dz . dz. (b) dz. (c) (a) 2 + π2 2 + π 2 )2 2+9 z (z z C C C ! ! ! 2 dz cosh z z − 3z + 4 dz. (d) . (e) dz. (f) 2 + 9)2 2 + π 2 )3 2 (z (z C C C z − 4z + 3 20.3. For 0 < r < 1, show that ∞ %

k=0

rk cos kθ =

1 − r cos θ 1 + r2 − 2r cos θ

and

∞ %

k=0

rk sin kθ =

r sin θ . 1 + r2 − 2r cos θ

20.3 Problems

523

20.4. Find the Taylor expansion of 1/z 2 for points inside the circle |z −2| < 2. 20.5. Use mathematical induction to show that ! ! dn −1 ! (1 + z) = (−1)n n!. ! n dz z=0

20.6. Find the (unique) Laurent expansion of each of the following functions in each of its regions of analyticity: 1 . (z − 2)(z − 3) 1 . (e) (1 − z)3 z (i) . z−1

(a)

(b) z cos(z 2 ). (f)

z2

1 . −1

1 . − z) z2 − 4 (g) 2 . z −9

(c)

z 2 (1

sinh z − z . z4 1 (h) 2 . (z − 1)2

(d)

20.7. Show that the following functions are entire: ⎧ % ⎨ e2z − 1 − 2 for z ̸= 0, sin z 2 z z (b) f (z) = (a) f (z) = z ⎩2 1 for z = 0. ⎧ ⎨ 2 cos z2 for z ̸= ±π/2, (c) f (z) = z − π /4 ⎩−1/π for z = ±π/2.

for z ̸= 0, for z = 0.

20.8. 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. 1 . sin z 1 (e) z . e −1

(a)

1 . 1 − cos z 1 (f) 2 . z sin z (b)

z . 1 − cosh z z4 (g) . 6z + z 3 − 6 sinh z (c)

(d)

z2 . z − sin z

20.9. Obtain the Laurent-series expansion of f (z) = sinh z/z 3 about the origin.