5.1. Infinite Sequences of Constants in Chapter 2, We Developed

Home , Laurent series, Removable singularity

CHAPTER 5

Power Series and Laurent series

5.1. Infinite Sequences of Constants In Chapter 2, we developed properties of analytic functions. In Chapter 4, we developed many more properties. Here, in Chapter 5, infinite series yield even more properties of analytic functions. An infinite sequence of constants is a complex-valued function f whose domain is the set of positive integers Z+. It assigns a complex number f(n) to each n Z+. We often list the sequence in the form 2 f(1), f(2), . . . , f(n), . . . The elements of the sequence are called its terms: first term, second term, nth term, etc. We usually replace f(1) by c1, f(2) by c2, etc. to get

c1, c2, . . . , cn, . . . For a more compact form we use braces:

cn = c1, c2, . . . , cn, . . . Example. { } ( 1)n + in 1 i 1 1 + i 1 i 1 = 1 + i, 0, , , , 0, , , . . . n 3 2 5 7 4 ⇢ Definition (5.1). A sequence cn has limit L if given any ✏ > 0 there exists an integer N such that c {L }< ✏ for all n > N. | n | When L is the limit of a sequence c , we write { n} lim cn = L, n !1 and we say the sequence converges to L.

115 116 5. POWER SERIES AND LAURENT SERIES Theorem (5.1). If sequences c and d have limits C and D, then { n} { n} (i) kc has limit kC; { n} (ii) c d have limits C D; { n} ± { n} ± (iii) c d has limit CD; { n n} (iv) c /d has limit C/D provided D = 0 and none of the d = 0. { n n} 6 n

Theorem (5.2). Suppose cn = xn + yni and L = X + Y i. Then

lim cn = L lim xn = X and lim yn = Y. n n n !1 () !1 !1 ✓ni ⇥i Theorem (5.3). Suppose cn = rne and L = Re , where ✓n and ⇥ are principal values of arguments. Then

lim cn = L if lim rn = R and lim ✓n = ⇥. n n n !1 !1 !1 Theorem (5.4). lim cn = 0 lim cn = 0. n n !1 () !1| | ( 1)n + in Problem (Page 204 # 3). Find lim . n n !1 ⇢ Soution. For any n, ( 1)n + in has one of 4 values: 0, 2, 1 i. Then ± ( 1)n + in ( 1)n + in lim = 0 = lim = 0 n n ) n n !1 !1 ⇢ by using Theorem5.4. ⇤ n2 + i Problem (Page 204 # 5). Find lim sinh . n 1 n2i !1 ⇢ ✓ ◆ Solution. n2 + i 1 lim sinh = sinh = sinh i = (sin 1)i. n 1 n2i i !1 ⇢ ✓ ◆ ✓ ◆ ⇤ 5.2. INFINITE SERIES OF CONSTANTS 117 Example (5.3). Show lim zn = 0 if z < 1, and that the limit does not n | | exist when z > 1. What can!1you say if z = 1? | | | | solution. If z < 1, | | lim zn = lim z n = 0 = lim zn = 0. n n Th.5.4 n !1| | !1| | ) !1 If z > 1, then | | lim zn = lim z n = = lim zn DNE. n n n !1| | !1| | 1 ) !1 If z = 1, then lim zn = 1. n !1 Now suppose z = 1, but z = 1. By part (ii) of Theorem 5.1, if a sequence | | 6 cn converges, then lim (cn cn 1) = 0. By Theorem 5.4, we then get n { } !1 lim cn cn 1 = 0. It follows that if lim cn cn 1 = 0, then the sequence n | | n | | 6 !1c cannot converge. That is the case here:!1 { n} n n 1 n 1 lim z z = lim z z 1 = lim z 1 = z 1 = 0. n n n ⇤ !1| | !1| || | !1| | | | 6 5.2. Infinite Series of Constants When the terms of an infinite sequence are added together, we obtain an infinite series 1 c = c + c + + c + . n 1 2 · · · n · · · n=1 X If the sequence of partial sums S , where { n} S = c + c + + c , n 1 2 · · · n has limit S, we say the series has sum S and that the series converges to S, and write 1 cn = S. n=1 X If lim Sn does not exist, we say the series does not have a sum or that it n diverges.!1 118 5. POWER SERIES AND LAURENT SERIES Example. The nth partial sum of the geometric series 2 n 1 a + az + az + + az + · · · · · · is 2 n 1 S = a + az + az + + az . n · · · Multiplying the equation by z, we get zS = az + az2 + az3 + + azn. n · · · Subtracting these equations, we get S zS = a azn n n which can be solved for a(1 zn) S = , n 1 z provided z = 1. Using the results of Example 5.3, 6 a , z < 1 lim Sn = 1 z | | = n !1 (diverges, z 1 ) | | a 1 n 1 2 , z < 1 az = a + az + az + = 1 z | | . · · · n=1 (diverges, z 1 X | | Theorem (5.5). A complex series converges the series of real and () imaginary parts converge. Precisely, if cn = xn +yni and S = X +Y i, then

1 1 1 c = S x = X and y = Y. n () n n n=1 n=1 n=1 X X X 5.2. INFINITE SERIES OF CONSTANTS 119 1 ( 1)n Example. Does z converge where z = 2 + i ? n n n2 n=1 X n Solution. xn = 2 converges since it is a constant sequence. ( 1) takes ( 1)n on the values 1. Thus y = converges to 0, so by Theorem 5.5, ± n n2 1 ( 1)n 2 + i = 2. n2 ⇤ n=1 X ✓ ◆ 1 1 Theorem (5.6). If cn = C and dn = D, then n=1 n=1 X X 1 (i) (c d ) = C D, n ± n ± n=1 X 1 (ii) kcn = kC (where k is a constant). n=1 X 120 5. POWER SERIES AND LAURENT SERIES 1 Theorem (5.7). If cn converges, then lim cn = 0. n n=1 !1 X 1 Proof. If cn converges to S, its sequence of partial sums n=1 X S1, S2, . . . , Sn . . . has limit S, But then the sequence ⇤

0, S1, S2, . . . , Sn 1, . . . also converges to S. When we subtract these sequences term-by-term, the sequence S1 0, S2 S1, . . . , Sn Sn 1, . . . has limit S S = 0 by Theorem 5.1. Since cn = Sn Sn 1, lim cn = 0. n !1 Corollary (1 – nth term test). If lim cn = 0 or does not exist, then n !1 6 1 cn diverges. n=1 X 1 Definition (5.2). A complex series cn is said to be absolutely con- n=1 X 1 vergent (or converge absolutely) if c converges. | n| n=1 X 5.2. INFINITE SERIES OF CONSTANTS 121 Theorem 1 (5.8). If a complex series converges absolutely, then it converges.

1 Proof. Suppose the series cn converges absolutely where cn = an +bni. n=1 X 1 1 This means c = a2 + b2 has a sum, call it S. | n| n n n=1 n=1 X X p 1 Now each term in the series a is nonnegative and is less than or equal to | n| n=1 X 1 2 2 the corresponding term in an + bn . n=1 X p 1 1 1 It follows that a converges = the real series a of real parts of c | n| ) n n n=1 n=1 n=1 converges absolutelyX . X X

1 Similarly, the series of imaginary parts bn converges absolutely. n=1 X 1 By Theorem 5.5, cn converges. ⇤ n=1 X 1 By this theorem, a complex series cn converges when its series of moduli n=1 X 1 c converges. Since this latter series is a series of nonnegative reals, we | n| n=1 canX use the tests for such series in real variable theory. However, we will phrase them in the context of complex series and absolute convergence. 122 5. POWER SERIES AND LAURENT SERIES Recall. (1) For c = 1, 6 1 1 cn+1 1 cn = lim = if c < 1, n 1 c 1 c | | n=0 !1 X 1 and cn diverges if c 1. | | n=0 X 1 1 (2) The harmonic series diverges. n n=1 X 1 1 (3) The p-series converges for p > 1. np n=1 X 1 1 (4) The p-series diverges for p 1. np  n=1 X 1 Theorem (5.9 – Comparison Test). If c a for all n N and a | n|  n n n=1 X 1 converges, then cn converges absolutely. n=1 X 5.2. INFINITE SERIES OF CONSTANTS 123 1 Theorem (5.10 – Limit Comparison Test). If bn is a convergent series n=1 of positive real numbers, and X c lim | n| = l, 0 l < , n !1 bn  1 1 then cn converges absolutely. n=1 X Theorem (5.11 – Integral Test). If cn = f(n) and f(x) is a continuous, decreasing function for x N such that| | 1 f(x) dx < , 1 ZN 1 then cn converges absolutely. n=1 X cn+1 Theorem (5.12 –Limit Ratio Test). Suppose cn = 0 and lim = L. 6 n cn Then !1 1 (i) cn converges absolutely if L < 1; n=1 X 1 cn+1 (ii) cn diverges if L > 1 or if lim = ; n cn 1 n=1 !1 X 1 (iii) cn may converge or diverge if L = 1. n=1 X 124 5. POWER SERIES AND LAURENT SERIES n Theorem (5.13 – Limit Root Test). Suppose lim cn = L. Then n !1 | | 1 p (i) cn converges absolutely if L < 1; n=1 X 1 n (ii) cn diverges if L > 1 or if lim cn = ; n | | 1 n=1 !1 X p 1 (iii) cn may converge or diverge if L = 1. n=1 X 1 5 n Example. Does converge or diverge? 3 + 2i n=1 X ⇣ ⌘ 5 Solution. This is a geometric series with common ratio . Since 3 + 2i 5 5 = > 1, the series diverges. 3 + 2i p ⇤ 13 1 2n Problem (Page 210 # 11). Does converge or diverge? (n + 1)in n=1 X Solution. We use the limit ratio test. n+1 2 n+1 n (n+2)in+1 2 (n + 1)i L = lim n = lim n 2 n (n + 2)in+1 · 2n !1 (n+1)in !1 2 n + 1 = lim = 2 > 1. n i · n + 2 !1 Thus the series diverges. ⇤ 5.3. COMPLEX POWER SERIES 125 n 1 i Problem (Page 210 # 15). Does 2 + converge or diverge? n n=1 X ✓ ◆ Solution. We use the limit root test. i n i L = lim n 2 + = lim 2 + = 2 > 1. n n n n !1 s✓ ◆ !1 Thus the series diveges. ⇤ 5.3. Complex Power Series A complex power series is a series of the form

1 ( ) a (z z )n = a + a (z z ) + a (z z )2 + ⇤ n 0 0 1 0 2 0 · · · n=0 X where z0 is a point in the complex plane and the an are complex constants. Such series are called power series in z z or power series about the 0 point z0. When a complex number is substituted into ( ) for z, we get a series of complex numbers which may converge or diverge. There⇤ are exactly three possibilities as far as values of z for which a power series yields a convergent series of constants:

(i) the power series converges only for z = z0; (ii) the power series converges absolutely for all z; (iii) there exists a number R > 0 such that the power series converges absolutely for z z0 < R, diverges for z z0 > R, and may or may not converge for z | z = |R. | | | 0| 126 5. POWER SERIES AND LAURENT SERIES If a power series converges at a point z1, then it converges absolutely for all points z closer to z0 than z1.

1 Theorem (5.14). If a (z z )n converges for z , then it converges n 0 1 n=0 for all z in the circle zXz < z z . | 0| | 1 0|

Based on (iii) above, we call R the radius of convergence of the power series. In (i) and (ii), we say R = 0 and R = . 1 If 0 < R < , the series converges absolutely for all points in the circle of 1 convergence z z0 < R, diverges for all points outside the circle, and may converge or diverge| for| points on the circle. For many power series, the radius of convergence R can be calculated using a 1 R = lim n or lim . n a n a 1/n !1 n+1 !1 n | | 1 Problem (Page 223 # 3). Find the circle of convergence for n2in(z+i)n. n=1 X Solution. Since n2in n 2 1 R = lim = lim = 1, n (n + 1)2in+1 n n + 1 · i !1 !1 ⇣ ⌘ the circle of convergence is z + i <1. ⇤ | | 5.3. COMPLEX POWER SERIES 127 Problem (Page 223 # 5). Find the circle of convergence for 1 1 (z + 3i)n. nn n=1 X Solution. 1 R = lim = lim n = . n (1/nn)1/n n | | 1 !1 !1 Thus the series converges for all z. ⇤ Problem (Page 223 # 13). Find the circle of convergence for 1 1 (z i)2n+1. (2 + i)n n=0 X Solution. Let w = (z i)2, so that 1 1 1 1 (z i)2n+1 = (z i) wn. (2 + i)n (2 + i)n n=0 n=0 X X Then 1/(2 + i)n Rw = lim = lim 2 + i = p5. n 1/(2 + i)n+1 n | | !1 !1 Thus 1/4 Rz = Rw = 5 , and the circle of convergence is z i < 51/4. | p| ⇤ 128 5. POWER SERIES AND LAURENT SERIES 1 ( 1)n+13n Problem (Page 224 # 23). Find the sum of (z + i)2n. 5n+2 n=0 X 1 Solution. This is a geometric series with initial term and 25 3(z + i)2 r = . Then 5 1 ( 1)n+13n 1/25 1 (z + i)2n = = , 5n+2 2 5(2 + 6iz + 3z2) n=0 1 3(z + i) /5 X 3(z + i)2 ⇥ 5⇤ It is valid for r = < 1 or z + i < . 5 | | 3 ⇤ r 1 If f is the sum of the powerseries a (z z )n, what properties does f n 0 n=0 possess? X 5.3. COMPLEX POWER SERIES 129 Theorem (5.15). Suppose f is the sum of a complex power series

1 a (z z )n inside its circle of convergence z z < R, where R > 0. n 0 | 0| n=0 ThenX (i) f is an analytic function in the circle of convergence. (ii) The derivative of f can be obtained by term-by-term di↵erentiation of the power series for f,

1 n 1 f 0(z) = na (z z ) , n 0 n=0 X valid in the same circle of convergence, (iii) If C is any piecewise smooth curve (in the circle of convergence) joining point A to point B, then B 1 1 (z z )n+1 f(z) dz = a (z z )n dz = a 0 . n 0 n n + 1 C n=0 C n=0 A Z X Z X ⇢ (iv) Antiderivatives of f can be obtained by term-by-term antidi↵erentiation of the power series for f, 1 a f(z) dz = n (z z )n+1 + C , n + 1 0 C n=0 Z X valid on the same circle of convergence. Corollary (5.15.1). The sum of a power series with inﬁnite radius of convergence is an entire function. 130 5. POWER SERIES AND LAURENT SERIES

1 n 1 Problem (Page 224 # 25). Find the sum of n(z + 2i) . n=1 X Solution. The radius of convergence of the series is

n 1 n 1 R = lim = 1. Let S(z) = n(z + 2i) . Then n n + 1 !1 n=1 X 1 z + 2i z + 2i S(z) dz = (z + 2i)n + C = + C = + C. 1 (z + 2i) 1 2i z n=1 Z X Di↵erentiation then gives (1 2i z) (z + 2i)( 1) 1 S(z) = = , (1 2i z)2 (1 2i z)2 valid for z + 2i < 1. | | ⇤ Recall. 1 ( ) = 1 + z + z2 + , z < 1 ⇤ 1 z · · · | | Problem (Page 224 # 39). Use ( ) to ﬁnd the power series of the function 1 ⇤ f(z) = about z = i. 1 + z 0 Solution. 1 1 1 = = 1 + z (1 + i) + (z i) (1 + i) 1 + z i 1+i n n 1 1 z i 1 ( 1) n = = (z i) . 1 + i 1 + i (1 + i)n+1 n=0 n=0 X ✓ ◆ X z i This is valid for < 1 or z i < p2. 1 + i | | ⇤ 5.4. TAYLOR AND MACLAURIN SERIES 131 5.4. Taylor and Maclaurin Series We learned in Section 5.3 that every convergent power series sums to an analytic function in its circles of convergence. We now look at the issue of expanding analytic functions to power series. Theorem (5.18 –Taylor Expansion Theorem). Let f be analytic in a domain D and z0 be a point in D. Then f can be expanded in a power series f 00(z0) 2 f(z) = f(z ) + f 0(z )(z z ) + (z z ) + , 0 0 0 2! 0 · · · valid in all circles z z < r containing only points of D. | 0|

The expansion 1 f (n)(z ) ( ) f(z) = 0 (z z )n ⇤ n! 0 n=0 X is called the Taylor series of f about z0. The special case in which z0 = 0 1 f (n)(0) ( ) (z) = zn ⇤⇤ n! n=0 X is called the Maclaurin series of f.

The circle z z0 < r in which the Taylor series converges to the function is called the circle| of| convergence for the Taylor series. Corollary (5.18.1). Every power series representation of (or, Taylor series for) an entire function has infinite radius of convergence. 132 5. POWER SERIES AND LAURENT SERIES Four important Taylor (Maclaurin) series. z2 z3 1 1 (#1) ez = 1 + z + + + = zn, z < . 2! 3! · · · n! | | 1 n=0 X z3 z5 z7 1 ( 1)n (#2) sin z = z + + = z2n+1, z < . 3! 5! 7! · · · (2n + 1)! | | 1 n=0 X z2 z4 z6 1 ( 1)n (#3) cos z = 1 + + = z2n, z < . 2! 4! 6! · · · (2n)! | | 1 n=0 X 1 1 (#4) = 1 + z + z2 + z3 + = zn, z < 1. 1 z · · · | | n=0 X 1 z Example (5.13). Find the Maclaurin series for . 1 4z n=0 X solution. We use ( ). ⇤⇤ f(0) = 0, (1 4z) z( 4) 1 f 0(0) = 2 = 2 = 1, (1 4z) z=0 (1 4z) z=0 | | 2(4) f 00(0) = 3 = 2(4) (1 4z) z=0 | 2 2(3)(4 ) 2 f 000(0) = 4 = 3!4 . (1 4z) z=0 | (n) n 1 The pattern emerging is f (0) = n!4 for n 1, and therefore z 1 n 1 n = 4 z , 1 4z n=1 X 4n 1 1 with radius of convergence R = lim = . n 4n 4 !1 5.4. TAYLOR AND MACLAURIN SERIES 133 1 Alternatively, because may be regarded as the sum of an infinite geo- 1 4z metric series with first term equal to 1 and common ratio 4z,

z 2 1 n 1 n 1 = z 1+(4z)+(4z) + = 4 z , where 4z < 1 or z < . 1 4z · · · | | | | 4 ⇤ n=1 ⇥ ⇤ X Theorem (5.19). If f has a power series expansion about a point z0 with nonzero radius of convergence, it must be the Taylor series about z0. Theorem (5.20). The radius of convergence of the Taylor series for a function f(z) about a point z0 is the distance from z0 to the nearest singularity of f(z). Problem (Page 240 # 3). Find the Taylor series and radius of convergence z for about z = 0. (1 z)2 0 1 1 Solution. We know = zn, valid for z < 1. Di↵erentiating, 1 z | | n=0 X 1 1 n 1 = nz (1 z)2 n=0 Thus X z 1 1 = nzn = nzn, z < 1. (1 z)2 | | ⇤ n=0 n=1 X X 1 1 Theorem (5.21). If f(z) = a (z z )n and g(z) = b (z z )n n 0 n 0 n=0 n=0 have positive radii of convergencXe, then X

1 f(g) g(z) = (a b )(z z )n, ± n ± n 0 n=0 X valid for every z that is common to the circles of convergence of the two series. 134 5. POWER SERIES AND LAURENT SERIES Example (5.18). Find the Taylor series for Log z about z = 1. Solution. We begin with 1 1 1 1 1 = = = (1 z)n = ( 1)n(z 1)n, z (z 1) + 1 1 (1 z) n=0 n=0 X X valid for z 1 < 1. Term by term integration gives | | 1 ( 1)n Log z = (z 1)n+1 + C, z 1 < 1. n + 1 | | n=0 X Since Log 1 = 0, we must choose C = 0 = ) 1 ( 1)n 1 1 Log z = (z 1)n+1 = (z 1) (z 1)2 + (z 1)3 , n + 1 2 3 · · · n=0 X for z 1 < 1. | | ⇤ Going further, the above series holds for any branch of log z for which log 1 = 0 and which does not have a branch cut passing through the interior of the circle of convergence. Then

we may write n 1 1 ( 1) log z = (z 1)n, z 1 < 1, n | | n=1 X 3⇡ ⇡ provided < < . 2 2 5.4. TAYLOR AND MACLAURIN SERIES 135 1 Problem (Page 240 # 17). Find the Taylor series for about z = 3. 1 + 2z 0 Solution. 1 1 1 = = 1 + 2z 7 + 2(z 3) 2(z 3) 7 1 + 7 n h 1 1 i 2(z 3) 1 ( 1)n2n = = (z 3)n, 7 7 7n+1 n=0 n=0 X  X 2(z 3) 7 valid for < 1 or z 3 < . 7 | | 2 ⇤ We next look at an alternate method for determining the order of a zero. Theorem (5.23). A function f has a zero of order m at z0 if and only if it can be written in the form (#) f(z) = (z z )mg(z), 0 valid in some circle z z < R, where g is analytic at z and g(z ) = 0. | 0| 0 0 6 Corollary. The zeros of a nonconstant analytic function are isolated; that is, every zero has a neighborhood inside of which it is the only zero.

Proof. If z0 is a zero of f of order m, then f(z) can be written as in (#) where g is analytic at z and g(z ) = 0. Since g is thus continuous at z , 0 0 6 0 there must be a neighborhood of z0 in which g(z) = 0. In this neighborhood, f(z) = 0 except at z = z . 6 6 0 ⇤ Theorem (5.24 – L’Hopital’s Rule). Let f and g be functions which are analytic in a domain D but not identically equal to zero.

If f(z0) = g(z0) = 0 at a point z0 in D, then f(z) f (z) lim = lim 0 n n !1 g(z) !1 g0 (z) provided the latter limit exists. 136 5. POWER SERIES AND LAURENT SERIES z Problem (Page 240 # 19). Find the Taylor series for about (1 3z)2 z0 = 1. Solution. The series 1 1 1 = (3z)n = 3nzn 1 3z n=0 n=0 X X 1 converges for 3z < 1 or z < . Di↵erentiating, | | | | 3

3 1 n n 1 = n3 z . (1 3z)2 n=0 z X Then multiply by to get 3

z 1 n 1 n 1 n 1 n = n3 z = n3 z , (1 3z)2 n=0 n=1 X X 1 valid for z < . | | 3 ⇤ z2 + 1 Problem (Page 240 # 27). Identify all zeros and their orders for . z2(z + 1) Solution. Zeros of the function are z = i. We express the function in the form ± z + i (z i) , z2(z + 1)  and noting that the expression in brackets is analytic at z = i and is not 0 z2 + 1 there, by Theorem 5.23, has a zero of order 1 at z = i. z2(z + 1) We get a similar result for z = i. ⇤ 5.5. LAURRNT SERIES 137 5.5. Laurrnt Series Laurent series represent complex functions near isolated singularities rather than at nice points where a function is analytic. Theorem (5.25 – Laurent Expansion Theorem). Let f be analytic in an annulus D : r < z z < R. Then f(z) can be expressed in the form | 0| 1 ( ) f(z) = a (z z )n ⇤ n 0 n= X1 which converges and represents f(z) in D. Coecients can be calculated with the formula 1 f(⇣) ( ) a = d⇣, ⇤⇤ n 2⇡i (⇣ z )n+1 IC 0 where C is any simple, closed, piecewise smooth curve in D which contains z0 in its interior.

( ) is called the Laurent expansion of f in the annulus D. ⇤ 138 5. POWER SERIES AND LAURENT SERIES

In the particular case that f is analytic for z z0 r as well as in D, we f(⇣) | |  ﬁnd that for n < 0, is analytic in z z < R, and (⇣ z )n+1 | 0| 0 f(⇣) = 0, (⇣ z )n+1 IC 0 and for n 0, the generalized Cauchy integral formula gives 1 f(⇣) f (n)(z ) a = d⇣ = 0 . n 2⇡i (⇣ z )n+1 n! IC 0 Thus, in this case, the Laurent series reduces to the Taylor series. Corollary. If f has a Laurent series valid in an annulus r < z z < R, then it has only one such expansion. | 0|

Because of this corollary, we seldom use ( ) to calculate the coecients an in the Laurent expansion. Often, we use prop⇤⇤ erties of geometric and/or Taylor series. 2 Example. Find the Laurent series expansion for f(z) = (z + 1)(z + 3) in each of the following domains: Solution. First, using partial fractions 2 1 1 1 1 f(z) = = = (z + 1)(z + 3) z + 3 z + 1 3 ( z) 1 ( z). (a) z < 1 = z < 1. | | ) | | 1 1 = 1 z + z2 z3 + = ( 1)nzn, z < 1. 1 + z · · · | | n=0 X Since z < 1 = z < 3, | | ) | | 1 1 1 1 1 z n 1 z z2 z3 = = ( 1)n = + + . 3 + z 3 1 + z 3 3 3 9 27 81 · · · 3 n=0 ✓ ◆ X ⇣ ⌘ 5.5. LAURRNT SERIES 139 Subtracting 1 1 z n 1 2 8 26 80 f(z) = ( 1)n ( 1)nzn = + z z2 + z3 . 3 3 3 9 27 81 · · · n=0 n=0 X ⇣ ⌘ X (b) 1 < z < 3 | | For z > 1, | | 1 1 1 1 1 1 n 1 1 1 1 = = ( 1)n = + + = z + 1 z 1 + 1 z z z z2 z3 z4 · · · ) z n=0 ✓ ◆ X ⇣ ⌘ 1 1 1 1 1 z z2 z3 f(z) = + + + + + · · · z4 z3 z2 z 3 9 27 81 · · · 1 1 z n 1 1 1 n = ( 1)n ( 1)n . 3 3 z z n=0 n=0 X ⇣ ⌘ X ⇣ ⌘ (c) z > 3 = z > 1. | | ) | | 1 1 1 1 1 3 n 1 3 9 27 = = ( 1)n = + + = z + 3 z 1 + 3 z z z z2 z3 z4 · · · ) z n=0 ✓ ◆ X ⇣ ⌘ 2 8 26 f(z) = + + . z2 z3 z4 · · · (d) 0 < z + 1 < 2. | | 1 1 1 1 1 1 z + 1 n = = = ( 1)n z + 3 2 + (z + 1) 2 1 + z+1 2 2 2 n=0 ✓ ◆ X ⇣ ⌘ 1 z + 1 (z + 1)2 (z + 1)3 = + + = 2 4 8 16 · · · ) 1 1 z + 1 (z + 1)2 (z + 1)3 f(z) = + + + 1 + z 2 4 8 16 · · · ⇤ 140 5. POWER SERIES AND LAURENT SERIES z Example. Find the Laurent series for for 0 < z + 2 < 1. (z + 1)(z + 2) | | Solution. f has singularities at z = 1 and z = 2. Let z +2 = u. Then z u 2 2 1 2 1 = = = + (z + 1)(z + 2) (u 1)u u u 1 u 1 u 2 1 2 1 = + un = + (z + 2)n u z + 2 n=0 n=0 2 X X = +1+(z+2)+(z+2)2+(z+2)3+ , 0 < u < 1 or 0 < z+2 < 1. z + 2 · · · | | | | ⇤ 1 Example. Find the Laurent series for (z 3) sin for z + 2 > 0. z + 2 | | ✓ ◆ Solution. Let z + 2 = u = z = u 2. Then ) 2n+1 1 1 1 ( 1)n 1 (z 3) sin = (u 5) sin = (u 5) z + 2 u (2n + 1)! u n=0 ✓ ◆ ✓ ◆ X ✓ ◆ 1 1 1 5 1 5 1 5 = (u 5) + = 1 + + u 3!u3 5!u5 · · · u 3!u2 3!u3 5!u4 5!u5 · · · ⇢ 5 1 5 1 1 = 1 + + z + 2 6(z + 2)2 6(z + 2)3 120(z + 2)4 24(z + 2)5 · · · for u > 0 or z + 2 > 0. | | | | ⇤ Theorem (5.26). When the Lauent series of a function

1 f(z) = a (z z )n, valid in an annulus r < z z < R, is di↵er- n 0 | 0| n= X1 entiated term-by-term, the resulting series converges to f 0(z) in the same annulus. 5.6. CLASSIFICATION OF SINGULARITIES 141 5.6. Classiﬁcation of Singularities

When z0 is an isolated singularity of an analytic function f, there exists an annulus 0 < z z0 < R centered at z0 in which f is analytic, and therein f has a Laurent| expansion. | Definition (5.4). Suppose an analytic function f has an isolated singularity at z0, and 1 f(z) = a (z z )n n 0 n= X1 is the Laurent expansion of f valid in some annulus 0 < z z < R. Then | 0| (i) if an = 0 for all n < 0, z0 is called a removable singularity;

(ii) if an = 0 for n < m, m a ﬁxed positive integer, but a m = 0, z0 is called a pole of order m; 6

(iii) if an = 0 for an inﬁnity of negative integers n, z0 is called an essential singularity6 . Example (5.28). sin z (a) Classify the singularities of f(z) = . z Solution. z3 z5 sin z = z + , z < = 3! 5! · · · | | 1 ) sin z z2 z4 = 1 + z 3! 5! · · · sin z is the Lauent series for in the annulus 0 < z < = z = 0 is a z | | 1 ) sin z removable singularity by (i), and it is clear that lim = 1. Then z 0 ! z 1, z = 0 g(z) = is an entire function. f(z), z = 0 ( 6 142 5. POWER SERIES AND LAURENT SERIES ⇤ (b) Classify the singularities of f(z) = e1/z. Solution. z2 z3 ez = 1 + z + + + , z < = 2! 3! · · · | | 1 ) 1 1 1 e1/z = 1 + + + + , 0 < z < . z 2!z2 3!z3 · · · | | 1 This is the Laurent series of e1/z about z = 0, which is an essential singularity by (iii) ⇤ z sin z (c) Classify the singularities of f(z) = . z6 Solution. The Laurent series about z = 0 is z sin z 1 z3 z5 1 1 = z z + = + = z6 z6 3! 5! · · · 3!z3 5!z · · · )  ⇣ ⌘ z = 0 is a pole of order 3. ⇤ The following theorem allows us to ﬁnd poles and their orders quickly.

Theorem (5.27). A function f has a pole of order m at z0 if and only if it can be written in the form g(z) f(z) = (z z )m , 0 valid in some annulus 0 < z z0 < R, where g is analytic at z0 and g(z ) = 0. | | 0 6 cos 5z Example. Classify the singularities of f(z) = . (z 3)5 Solution. Since cos 5z is entire and cos 15 = 0, z = 3 is a pole of order 6 5. ⇤ 5.6. CLASSIFICATION OF SINGULARITIES 143 z2 + z 2 Example. Classify the singularities of f(z) = . z4 3z3 + 3z2 z Solution. The factored form z2 + z 2 (z 1)(z + 2) z + 2 f(z) = = = z4 3z3 + 3z2 z z(z 1)3 z(z 1)2 z + 2 has isolated singularities at z = 0 and z = 1. Since g(z) = is analytic z z + 2 at z = 1 and g(1) = 3 = 0, z = 1 is a pole of order 2. Since h(z) = 6 (z 1)2 is analytic at z = 0 and h(0) = 2 = 0, z = 0 is a pole of order 1 (also called a 6 simple pole. ⇤

Behavior of Functions Near Isolated Singularities The behavior of a ﬁnction near a removable singularity, a pole, and an essential singularilty are very di↵erent.

Theorem (5.28). If a function has a removable singularity at z0, then:

(a) f(z) can be deﬁned, or redeﬁned, at z0 so that the new function is analytic at z0; and

(b) f(z) has a limit as z approaches z0; and

Conversely, if a function fz) has an isolated singularity at z0, and has any one of those properties, then z0 is a removable singularity.

This theorem shows that a function is well-behaved near a removable singularity; only the value at the singularity is a problem.

Theorem (5.29). An isolated singularity z0 of a function f(z) is a pole if, and only if, lim f(z) = . z z ! 0| | 1 144 5. POWER SERIES AND LAURENT SERIES

Theorem (5.30). A function f(z) has an essential singularity at z0 if, and only if, lim f(z) does not exist, and lim f(z) = . z z z z ! 0| | ! 0| | 6 1 This result describes what a function does not do near an essential singularity; the following theorem describes what it does do. Theorem (5.31 – Picard’s Theorem). If a function f(z) has an essential singularity at z0, then the function takes on every possible value, with possibly one exception, in every neighborhood of z0. Example (5.30). Show that e1/z has an essential singularity at z = 0 and conﬁrm the result of Picard’s Theorem. solution. The ﬁrst part was done in Example 5.28(b). Since e1/z is never 0, 0 must be the exception in Picard’s Theorem. Given any other value k C, we need to show that in any circle z < r, there is a solution of the equation2 | | 1 1 e1/z = k = = log k = z = . ) z ) ln k + (Arg k + 2n⇡)i | | Then, by choosing n suciently large, the modulus of ln k + (Arg k + 2n⇡)i 1 | | can be made greater than . Then z < r = z is in the circle z < r. r | | ) | | ⇤

Theorem (5.32). If z0 is an essential singularity of a function f(z), and the Laurent series of f(z) around z0 is valid in the annulus 0 < z z0 < R, then f(z) is unbounded in every circle z z < r < R. | | | 0| Singularities at infinity: Definition (5.5). If a function f(z) is analytic for z > R, for some R, it is said to have an isolated singularity at the point of infinit| | y. It is a removable singularity, a pole, or an essential singularity of f(z) if z = 0 is a removable 1 singularity, a pole, or an essential singularity of f . z ⇣ ⌘ 5.6. CLASSIFICATION OF SINGULARITIES 145 Example. 1 1 3 (a) f(z) = z2 3z + 1 is entire = w(z) = f = + 1 = w has a ) z z2 z ) pole of order 2 at z = 0 = f has a pole of order⇣ ⌘2 at infinity. ) 1 1 z3 (b) f(z) = is analytic for z > 3 = w(z) = f = is z z3 | | ) z z2 1 analytic at z =0 = f(z) has a removable sngularity at infinit⇣ ⌘y. ) 1 (c) w(z) = sin has an essential singularity at z = 0 = f(z) = sin z has z ) an essential singularit⇣ ⌘ y at infinity. (d) There is no annulus z > R in which csc z is analytic (singularities at z = n⇡). Thus we do not|regard| the point at infinity as an isolated singularity for this function.