5. Taylor and Laurent Series Complex Sequences and Series an Infinite

Home , Analytic function, Laurent series, Power series

5. Taylor and Laurent series

Complex sequences and series

An inﬁnite sequence of complex numbers, denoted by zn , can be { } considered as a function deﬁned on a set of positive integers into n + 1 the unextended complex plane. For example, we take zn = 2n 1+ i 2+ i 3+ i so that the complex sequence is = . zn , 2 , 3 , { } 2 2 2 ··· Convergence of complex sequences

Given a complex sequence zn , if for each positive quantity ǫ, there { } exists a positive integer N such that

zn z <ǫ whenever n>N, | − | then the sequence is said to converge to the limt z. We write

lim z = z. n n →∞

1 In general, the choice of N depends on ǫ. The deﬁnition implies • that every ǫ-neighborhood of z contains all but a ﬁnite number of members of the sequence. The limit of a convergent sequence is unique. If the sequence fails to converge, it is said to be divergent.

Suppose we write zn = xn + iyn and z = x + iy, then • xn x zn z xn x + yn y , | − | ≤ | − |≤| − | | − | yn y zn z xn x + yn y . | − | ≤ | − |≤| − | | − |

Suppose lim = x and lim y = y exist, then for any ǫ > 0, there n n n →∞ →∞ exist Nx and Ny such that ǫ xn x < whenever n > Nx | − | 2 ǫ yn y < whenever n > Ny. | − | 2 Choose n> max(Nx, Ny), then ǫ ǫ zn z xn x + yn y < + = ǫ. | − |≤| − | | − | 2 2

2 Using the above inequalities, it is easy to show that • lim z = z lim x = x and lim y = y. n n n n n n →∞ ⇐⇒ →∞ →∞ Therefore, the study of the convergence of a complex sequence is equivalent to the consideration of two real sequences.

The above theorem enables us to write

lim (x + iy )= lim x + i lim y n n n n n n n →∞ →∞ →∞ whenever we know that both limits on the right exist or the one on the left exists.

3 For example, the sequence 1 zn = + i, n = 1, 2, , n3 ··· 1 converges to i since lim and lim 1 exist, so n 3 n →∞ n →∞ 1 1 lim + i = lim + i lim 1=0+ i 1= i. n 3 n 3 n →∞ n →∞ n →∞ ·

One can show that for each positive number ǫ 1 zn i <ǫ whenever n> . | − | √3 ǫ

4 Inﬁnite series of complex numbers

An infinite series of complex numbers z ,z ,z , is the infinite • 1 2 3 ··· sum of the sequence zn given by { } n z + z + z + = lim z . 1 2 3 ··· n  k →∞ kX=1   To study the properties of an infinite series, we define the se- • quence of partial sums Sn by { } n Sn = zk. kX=1

If the limit of the sequence Sn converges to S, then the series • { } is said to be convergent and S is its sum; otherwise, the series is divergent.

The consideration of an inﬁnite series is relegated to that of an • inﬁnite sequence of partial sums.

5 Remainder after n terms

Suppose an inﬁnite series converges. We deﬁne the remainder after n terms by

Rn = S Sn − and obviously lim R = 0. n n →∞ Conversely, suppose lim R = 0, then for any ǫ > 0, there exists n n →∞ N(ǫ) such that Rn <ǫ for n > N(ǫ). This is equivalent to | | Sn S <ǫ for n > N(ǫ), | − | hence S = lim S . n n →∞

6 A necessary condition for the convergence of a complex series is that lim z = 0. n n →∞ This is obvious since

lim z = lim (R R )= lim R lim R = 0. n n n n 1 n n n 1 n n →∞ →∞ − − →∞ − − →∞

Comparison Test

∞ If Mj is a convergent series with real nonnegative terms and for jX=1 all j larger than some number J, z M , then the series ∞ z | j| ≤ j | j| jX=1 n converges also. This is easily seen since Sn = z is a bounded | j| jX=1 increasing sequence, so it must have a limit.

7 Absolute convergence

∞ ∞ The complex series zn is absolutely convergent if zn con- | | nX=1 nX=1 2 2 verges. Note that zn = x + y and since | | n n q 2 2 2 2 xn xn + yn and yn xn + yn, | |≤ q | |≤ q then from the comparison test, the two series

∞ ∞ xn and yn | | | | nX=1 nX=1 must converge. Thus, absolute convergence in a complex sequence implies convergence in that sequence.

The converse may not hold. If Σzn converges but Σ zn does not, | | the series Σzn is said to be conditionally convergent. For example, einθ Log(1 eiθ)= ∞ , θ = 0, − − n 6 nX=1 is conditionally convergent. 8 Example

Show that the series ∞ (3 + 2i)/(j + 1)j converges. jX=1 Solution

We compare the series 3 + 2i (3 + 2i) (3 + 2i) ∞ = + + (A) (j + 1)j 9 64 ··· jX=1 with the convergent geometric series 1 1 1 1 ∞ = + + + . (B) 2j 2 4 8 ··· jX=1

Since 3 + 2i = √13 < 4, one can easily verify that for j 3 | | ≥ 3 + 2i 4 1 < . (j + 1)j (j + 1)j ≤ 2j

The terms of (B) dominate those of (A), hence (A) converges.

9 Limit superior

Consider a sequence xn of real numbers, and let S denote the set { } of all of its limit points. The limit superior of xn is the supremum { } n (least upper bound) of S. For example, xn = 3+( 1) ,n = 1, 2, , − ··· the limit points are 2 and 4 so that

lim x = max(2, 4) = 4. n n →∞

Root test

1/n Suppose the limit superior of zn equals L, the series Σzn {| | } converges absolutely if L < 1 and diverges if L > 1. The root test fails when L = 1.

10 Ratio test

zn+1 Suppose lim converges to L, then Σzn is absolutely conver- n zn →∞ gent if L < 1 and divergent if L > 1. When L = 1, the ratio test fails.

Gauss’ test

z Suppose n+1 admits the following asymptotic expansion zn

z k α n+1 = 1 + n + 2 , zn − n n ···

where αn is bounded for all n > N for some suﬃciently large N, | | then Σzn converges absolutely if k > 1, and diverges or converges conditionally if k 1. ≤

11 Proof of the Ratio Test

z Suppose lim n+1 converges to L,L < 1, then we can choose an n zn →∞ integer N such that for n N, we have

≥ z n+1 r, where L

Now, we have

z r z , z r z r2 z , and so forth, | N+1|≤ | N | | N+2|≤ | N+1|≤ | N | z + z + z (r + r2 + ). | N+1| | N+2| ···≤| N | ··· ∞ Thus, zn converges absolutely by virtue of the comparison test. | | nX=1

12 Sequences of complex functions

Let f (z), , fn(z), , denoted by fn(z) , be a sequence of com- 1 ··· ··· { } plex functions of z that are deﬁned and single-valued in a region R in the complex plane.

For some point z R, fn(z ) becomes a sequence of complex 0 ∈ { 0 } numbers. Supposing fn(z ) converges, the limit is unique. The { 0 } value of the limit depends on z0, and we write

f(z )= lim f (z ). 0 n n 0 →∞

If this holds for every z R, the sequence fn(z) deﬁnes a complex ∈ { } function f(z) in R. We write

f(z)= lim f (z). n n →∞ This is usually called pointwise convergence.

13 Deﬁnition

A sequence of complex functions fn(z) defined in a region R is { } said to converge to a complex function f(z) defined in the same region if and only if, for any given small positive quantity ǫ, we can find a positive integer N(ǫ; z) [in general, N(ǫ; z) depends on ǫ and z] such that

f(z) fn(z) <ǫ for all n > N(ǫ; z). | − | The region R is called the region of convergence of the sequence of complex functions.

In general, we may not be able to ﬁnd a single N(ǫ) that works for all point in R. However, when this is possible, fn(z) is said to { } converge uniformly to f(z) in R.

14 Convergence of series of complex functions

An inﬁnite series of complex functions

f (z)+ f (z)+ f (z)+ = ∞ f (z) 1 2 3 ··· k kX=1 is related to the sequence of partial sum Sn(z) { } n Sn(z)= fk(z). kX=1 The inﬁnite series is said to be convergent if

lim S (z)= S(z), n n →∞ where S(z) is called the sum; otherwise the series is divergent.

15 Many of the properties related to convergence of complex functions can be extended from their counterparts of complex numbers. For example, a necessary but not suﬃcient condition for the inﬁnite series of complex functions to converge is that

lim fk(z) = 0, k →∞ for all z in the region of convergence.

Example

Consider the complex series sin kz ∞ , k2 kX=1 show that it is absolutely convergent when z is real but it becomes divergent when z is non-real.

16 Solution

(i) When z is real, we have sin kz 1 , for all positive integer values of k. k2 ≤ k2

Since ∞ 1/k2 is known to be convergent, then ∞ sin kz/k2 is kX=1 kX=1 absolutely convergent for all z by virtue of the comparison test.

(ii) When z is non-real, we let z = x + iy,y = 0. From the relation 6 sin kz e kyeikx ekye ikx = − − − , k2 2k2i we deduce that sin kz ek y e k y | | − − | | as k . k2 ≥ 2k2 →∞ →∞

Since sin kz/k 2 is unbounded as k , the series is divergent. →∞

17 Uniform convergence of an inﬁnite series of complex functions

n Let Rn(z)= S(z) f (z)= S(z) Sn(z). − k − kX=1

The inﬁnite series Σfk(z) converges uniformly to S(z) in some region R iﬀ for any ǫ > 0, there exists N which is independent of z such that for all z R ∈ Rn(z) <ǫ whenever n>N. | |

Weierstrass M-test

If f (z) M where M is independent of z in and the series | k | ≤ k k R ΣM converges, then Σf (z) is uniformly convergent in . k k R

18 Proof of the Weierstrass M-test

The remainder of the series fk(z) after n terms is P Rn(z)= f (z)+ f (z)+ . n+1 n+2 ··· Now,

Rn(z) f (z) + f (z) + M + M + . | |≤| n+1 | | n+2 | ···≤ n+1 n+2 ··· Since M converges, M + M + can be made less than k n+1 n+2 ··· ǫ > 0P by choosing n > N for some N = N(ǫ). As N is clearly independent of z, we have

Rn(z) <ǫ for n>N, | | so the series is uniformly convergent. We also have absolute convergence of the series.

19 Test for uniform convergence using the M-test

2 ∞ 2 Take Mn = and note that converges. n2 n2 nX=3

20 Example

Prove that the series

z(1 z)+ z2(1 z)+ z3(1 z)+ − − − ··· converges for z < 1 and ﬁnd its sum. | | Solution

n Sn(z) = z(1 z)+ + z (1 z) − ··· − = z z2 + z2 z3 + + zn zn+1 = z zn+1. − − ··· − − n+1 n+1 For z < 1, consider Sn(z) z = z = z < ǫ. In order | | | − | |− | | | that this is true, we choose n such that ln ǫ ln ǫ (n + 1) ln z < ln ǫ so that n + 1 > or n> 1,z = 0. | | ln z ln z − 6 | | | | When z = 0,Sn(0) = 0 so Sn(0) 0 <ǫ for all n. | − |

Hence, lim S (z)= z for all z such that z < 1. n n →∞ | | 21 Questions

1 1. Does the series converge uniformly to z for z ? | |≤ 2 1 ln ǫ 1 Yes. If z , the largest value of 1 occurs when z = | |≤ 2 ln z − | | 2 and is given by | | ln ǫ 1. 1 − ln 2 ln ǫ Take N(ǫ) to be the largest integer smaller than 1. It then 1 − ln 2 follows that Sn(z) z < ǫ for n > N where N depends only on | − | 1 ǫ and not on the particular z in z . Hence, we have uniform | |≤ 2 1 convergence of the series for z . | |≤ 2

22 2. Does the series converge uniformly for z 1? | |≤ The same argument given in Qn (1) serves to show uniform convergence for z 0.9 or z 0.99 by using | |≤ | |≤ ln ǫ ln ǫ N = ﬂ 1 and N = ﬂ 1 , respectively. ln 0.9 − ln 0.99 −

However, if we apply the argument to z 1, this would require ln ǫ | |≤ N = ﬂ 1 , which is inﬁnite. The series does not converge ln 1 − uniformly for z 1. | |≤

23 Example

1 Show that the geometric series ∞ zn converges uniformly to 1 z nX=0 − on any closed subdisk z r < 1 of the open unit disk z < 1. | |≤ | | Solution

∞ ∞ n Since Mn = r is convergent if 0 r < 1, we conclude that ≤ nX=1 nX=0 ∞ the series zn converges uniformly for z r. | |≤ nX=1

24 For uniformly convergent inﬁnite series of complex functions, properties such as continuity and analyticity of the continuous functions fk(z) are carried over to the sum S(z). More precisely, suppose f (z), k = 1, 2, , are all continuous (analytic) in the region of k ··· convergence, then fk(z) is also continuous (analytic) in the same region. X

Further, an uniform convergent inﬁnite series allows for termwise diﬀerentiation and integration, that is,

∞ ∞ fk(z) dz = fk(z) dz C C Z kX=1 kX=1 Z d ∞ ∞ f (z)= f ′ (z). dz k k kX=1 kX=1

25 Power series

n The choice of fn(z)= an(z z ) leads to the power series expanded − 0 at z = z0. A power series deﬁnes a function f(z) for those points z at which it converges.

∞ n Given a power series an(z z0) , there exists a non-negative real n=1 − number R, R can beX zero or inﬁnity, such that the power series converges absolutely for z z < R, and diverges for z z > R. | − 0| | − 0|

R is called the radius of convergence

z z < R is called the circle of convergence. | − 0| Convergence of the power series must be determined for each point on the circle of convergence.

26 Ratio test as applied to the inﬁnite power series

an The radius of convergence R is given by lim , given that the n →∞ an+1 limit exists.

n+1 a z z an Consider the ratio | n+1|| − 0| and suppose lim exists, n n an z z0 →∞ an+1 | | | − | then the power series converges absolutely when satisﬁes z z 0 | − | a an lim n+1 z z < 1 z z < lim . n 0 0 n →∞ an | − | ⇔ | − | →∞ an+1

27 a (i) R = when lim n+1 = 0. The series converges in the whole n ∞ →∞ an

plane.

an (ii) R =0 when lim = 0. The series does not converges for n →∞ an+1

any z other than z . 0

Root test as applied to the inﬁnite power series

The radius of convergence can also be found by 1 R = , which is a consequence of the root test. n limn an →∞ | | Note that q

n 1 lim an z z0 < 1 z z0 < . n n →∞ | || − | ⇔ | − | lim an q n | | →∞ q

28 Example Find the circle of convergence for each of the following power series:

1 (a) ∞ (z i)k, k − kX=1 (b) ∞ kln k(z 2)k, − kX=1 z k (c) ∞ , k kX=1 2 1 k (d) ∞ 1+ zk. k kX=1

29 Solution

(a) By the ratio test, we have 1/k R = lim = 1; k 1/(k + 1) →∞ so the circle of convergence is z i = 1. | − |

(b) Using the root test, we have

1 k k ln k = lim ak = lim k . R k | | k →∞ q →∞ p To evaluate the limit, we consider the logarithm, 2 k k (ln k) ln lim kln k = lim ln kln k = lim = 0; k k k k →∞ →∞ →∞ so p p 1 k = lim kln k = e0 = 1. R k →∞ p The circle of convergence is z 2 = 1. | − |

30 (c) By the ratio test, we have

1 k 1 k = lim k = lim ( + 1) 1+ = ; R k+1 k k 1 k k ∞ →∞ k+1 →∞ so the circle of convergence is the whole complex plane.

(d) By the root test, we have

2 1 k 1 k 1 k = lim v 1+ = lim 1+ = e, R k u k k k →∞ u →∞ t so the circle of convergence is z = 1/e. | |

31 Theorem

If z is a point inside the circle of convergence z z = R of a 1 | − 0| power series

∞ n an(z z ) − 0 nX=0 then the power series must be uniformly convergent in the closed disk z z R , where R = z z . | − 0|≤ 1 1 | 1 − 0| Some useful results

Let S(z) denote the sum of the inﬁnite power series inside the circle of convergence. Then

(i) S(z) represents a continuous function at each point interior to its circle of convergence.

∞ n 1 (ii) S′(z) = nan(z z0) − at each point z inside the circle of n=1 − convergence.X

32 Power series of f(z)

Suppose a power series represents the function f(z) inside the circle of convergence, that is,

∞ n f(z)= an(z z ) . − 0 nX=0 It is known that a power series can be diﬀerentiated termwise so that

∞ n 1 f (z) = nan(z z ) ′ − 0 − nX=1 ∞ n 2 f (z) = n(n 1)an(z z ) , . ′′ − − 0 − ··· nX=2 Putting z = z0 successively, we obtain (n) f (z0) an = , n = 0, 1, 2, n! (n) ∞ f (z0) n f(z) = (z z0) . n! − nX=0

33 A power series represents an analytic function inside its circle of convergence. Can we expand an analytic function in Taylor series and how is the domain of analyticity related to the circle of convergence?

Taylor series theorem

Let f(z) be analytic in a domain with boundary ∂ and z . D D 0 ∈ D Determine R such that

R = min z z , z ∂ . {| − 0| ∈ D} Then there exists a power series ∞ a (z z )k which converges to k − 0 kX=0 f(z) for z z < R. The coeﬃcients a are given by | − 0| k (k) 1 f(ζ) f (z0) ak = dζ = , k = 0, 1, 2, , 2πi C (ζ z )k+1 k! ··· I − 0 C is any closed contour around z and lying completely inside . 0 D

34 Proof

Take a point z such that z z = r < R. | − 0|

A circle is drawn around z0 with radius R1, where r < R1 < R. Since z lies inside C1, by virtue of the Cauchy Integral Theorem, we have 1 f(ζ) f(z)= dζ. 2πi C ζ z I 1 −

The circle C lies completely inside . 1 D 35 f(ζ) z z The trick is to express the integrand in powers of − 0. Recall ζ z ζ z − − 0 1 un+1 =1+ u + u2 + + un + . 1 u ··· 1 u − −

1 1 1 = ζ z ζ z z z0 0 1 ζ−z − − − − 0 n+1 n z z0 1 z z0 (z z0) ζ−z = 1+ + + + − 0 .  − − n z z  ζ z0 ζ z0 ··· (ζ z0) 1 − 0 −  − − − ζ z0   −    f(ζ) We then multiply by and integrate along C1. 2πi

36 By observing the property

1 f(ζ) f (k)(z ) dζ = 0 , 2πi C (ζ z )k+1 k! I 1 − 0 we obtain n (k) f (z) k f(z)= (z z0) + Rn, k! − kX=0 where the remainder is given by n+1 1 f(ζ) z z0 Rn = − dζ. 2πi C ζ z ζ z ! I 1 − − 0 To complete the proof, it suﬃces to show that

lim R = 0. n n →∞

(i) f(ζ) M for ζ C since f(z) is continuous inside . | |≤ ∈ 1 D (ii) ζ z = (ζ z ) (z z ) ζ z z z = R r | − | | − 0 − − 0 |≥| − 0|−| − 0| 1 −

37 so that n+1 +1 f(ζ) z z M r n − 0 .

ζ z ζ z0 ≤ R1 r R1! − − −

Finally

+1 +1 1 M r n MR r n R 2πR = 1 0 as n . n 1 | |≤ 2πi R1 r R1! R1 r R1! → →∞ − arc length −

| {z } The Taylor coeﬃcients are given by 1 f(ζ) ak = dζ, k = 0, 1, 2, 2πi C (ζ z )k+1 ··· I 1 − 0 1 f(ζ) = dζ, 2πi C (ζ z )k+1 I − 0 where C1 is replaced by C and C is any simple closed contour en- closing z and lying completely inside [by virtue of Corollary 3 of 0 D the Cauchy-Goursat Theorem].

38 Example

1 Consider the function , the Taylor series at z = 0 is given by 1 z − 1 = ∞ zn, z < 1. 1 z | | − nX=0

The function has a singularity at z = 1. The maximum distance • from z = 0 to the nearest singularity is one, so the radius of convergence is one.

Alternatively, the radius of convergence can be found by the • ratio test, where a R = lim k = 1.

k ak+1 →∞

39 If we integrate along the contour C inside the circle of convergence z < 1 from the origin to an arbitrary point z, we obtain | | 1 dζ = ∞ ζn dζ C 1 ζ C Z − nX=0 Z zn+1 zn Log(1 z) = ∞ = ∞ . − − n + 1 n nX=0 nX=1 The radius of convergence is again one (checked by the ratio test). y

xz C x x 0

40 Example 1 Consider the Taylor series of the real function: 1+ x2 1 = 1 x2 + x4 x6 + 1+ x2 − − ··· What is the interval of convergence?

The Taylor series converges only for x < 1. This is because the • 1 | | complex extension has singularities on the circle z = 1 so 1+ z2 | | that the radius of convergence of the inﬁnite series ∞ ( 1)nz2n n=0 − is one. The above inﬁnite power series diverges forX points out- side the circle of convergence, including points on the real axis, where x > 1. | | When x = 1, the series becomes • | | 1 1 + 1 − ··· with alternating terms. It is known to be divergent.

41 Example

1 Find the Taylor expansion of f(z)= at z = 1. The function 1+ z2 is analytic inside z 1 < √2. | − | 1 1 1 1 = 1+ z2 2i z i − z + i − 1 1 1 1 1 = 2i 1 i z 1 − 1+ i z 1  1+ 1− i 1+ 1+− i − −  1 1 1 = ∞ ( 1)n (z 1)n. − 2i "(1 i)n+1 − (1 + i)n+1# − nX=0 − By observing 1 i = √2e iπ/4 and 1+ i = √2eiπ/4, we obtain − − 1 ei(n+1)π/4 e i(n+1)π/4 = ∞ ( 1)n − − (z 1)n 1+ z2 − (2i)2(n+1)/2 − nX=0 sin(n + 1)π = ∞ ( 1)n 4 (z 1)n. − 2(n+1)/2 − nX=0

42 Example

Find the Maclaurin expansion of f(z) = (z + 1)1/2, where the principal branch of the function is used. Where is the expansion valid?

Solution

1Log( +1) The required branch is identical to e2 z , whose derivative is given by 1/2 1Log( +1) 1 (z + 1) e2 z = . 2(z + 1) 2(z + 1) Deductively, we have

1 1 1 1 1 1 2 f ′(z)= (z + 1)2− , f ′′(z)= 1 (z + 1)2− , , 2 2 2 − ··· (n) 1 1 1 1 n f (z)= 1 (n 1) (z + 1)2− , n 1, 2 2 − ··· 2 − − ≥ 1 1Log(z+1) 1 n (z + 1)2 e2 where (z + 1)2− = = . (z + 1)n (z + 1)n

43 The principal branch of Log(z+1) is taken to have branch cut along the negative real axis with branch points at z = 1 and z = . − ∞ 1Log 1 When z = 0, e2 = 1 so that the Maclaurin coeﬃcients are 1 1 1 1 1 c0 = 1, cn = 1 2 n 1 , n 1. n! 22 − 2 − ··· 2 − − ≥

The singularity of (z +1)1/2 nearest to the origin is the branch point z = 1. Hence, the circle of convergence is z < 1. − | | Remark

An analytic branch of a multi-valued function can be expanded in a Taylor series about any point within the domain of analyticity of the branch, provided that one takes care to use this branch consistently in obtaining the coeﬃcients of the series.

44 Laurent series

Consider an inﬁnite power series with negative power terms

∞ n bn(z z ) , − 0 − nX=1 1 how to ﬁnd the region of convergence? Set w = , the series z z − 0 bn becomes ∞ b wn, a Taylor series in w. Suppose R = lim n ′ n n=1 →∞ bn+1 X

∞ n 1 exists, then bnw converges for w < R z z0 > . | | ′ ⇔| − | R nX=1 ′

45 1 Special cases: (i) R′ = 0 and (ii) = 0. R′

n bn 1. When R = lim = 0, the inﬁnite series b (z z ) n ′ n n 0 − →∞ bn+1 1 − nX does not converge for any z, not even at z = z . − 0

1 b 2. When = lim n+1 = 0, we consider the ratio of successive n R′ →∞ bn

∞ n terms in bnw and observe that nX=1 b lim n+1 w < 1 n →∞ bn | |

is satisfied for all (except = since the infinite series w w n ∞ Σbn(z z ) is not defined at z = z ). By virtue of the ra- − 0 − 0 tio test, the region of convergence is the whole complex plane except at z = z , that is, z z > 0. 0 | − 0|

46 For the more general case (Laurent series at z0)

∞ n ∞ n an(z z ) + bn(z z ) , − 0 − 0 − nX=0 nX=1 principal part | {z }

an bn suppose R = lim and R = lim exists, and RR > 1, n ′ n ′ →∞ an+1 →∞ bn+1 then inside the annular domain

1 z : < z z0 < R R′ | − | the Laurent series is convergent.

When RR′ 1, the intersection of the two regions: z z0 < • ≤ 1 | − | R and z z0 > is the empty set. This is because the | − | R ′ 1 combination of z z0 > and RR′ 1 implies z z0 R, | − | R′ ≤ | − | ≥ which contradicts with z z < R. | − 0|

47 The annulus degenerates into

(i) hollow plane if R = ∞ (ii) punctured disc if R = . ′ ∞

48 Remarks

∞ n 1. When R = 0, an(z z ) does not converge for any z other − 0 nX=0 than the trivial point z0. However, z = z0 is a singularity for the principal part. Actually, when R = 0, RR′ > 1 can never be satisﬁed.

2. A Laurent series deﬁnes a function f(z) in its annular region of convergence. The Laurent series theorem states that a function analytic in an annulus can be expanded in a Laurent series expansion.

49 Laurent series theorem

Let f(z) be analytic in the annulus A : R < z z < R , then f(z) 1 | − 0| 2 can be represented by the Laurent series,

f(z)= ∞ c (z z )k, k − 0 k= X−∞ which converges to f(z) throughout the annulus. The Laurent co- eﬃcients are given by 1 f(ζ) ck = dζ, k = 0, 1, 2, , 2πi C (ζ z )k+1 ± ± ··· I − 0 where C is any simple closed contour lying completely inside the annulus and going around the point z0.

50 Remarks

1. Suppose f(z) is analytic in the full disc: z z < R (without | − 0| 2 the punctured hole), then the integrand in calculating ck for negative k becomes analytic in z z < R . Hence, c = 0 for | − 0| 2 k k = 1, 2, . − − ··· The Laurent series is reduced to a Taylor series.

1 2. When k = 1, c 1 = f(ζ) dζ. We may ﬁnd c 1 by any − − 2πi C − means, so a contour integralI can be evaluated without resort to direct integration.

51 Example

The Laurent expansion of e1/z at z = 0 is given by 1 e1/z = ∞ . n!zn nX=0 The function e1/z is analytic everywhere except at z = 0 so that the annulus of convergence is z > 0. We observe | | b 1/(n + 1)! 1 lim n+1 = lim =0 so that = 0. n n →∞ bn →∞ 1/n! R′

1 Lastly, we consider e /z dz, where the contour C is z = 1. Since C | | C lies completely insideI the punctured disc z > 0, we have | | 1 e1/z dz = 2πi(coeﬃcient of in Laurent expansion) = 2πi IC z

52 Example

Find the Laurent expansion of 1 f(z) = sin z . − z The function has a singularity at z = 0 so that the annulus of convergence is z > 0. The Laurent coeﬃcient cn is given by | | 1 1 sin z c = − z dz, n = 0, 1, 2, n n+1 2πi IC z ± ± where C is chosen to be the unit circle z = 1. | |

53 Take z = eiθ so that dz = ieiθ dθ, then 1 π sin(2i sin θ)ieiθ c = dθ n i(n+1)θ 2πi Z π e 1 −π = i sinh(2 sin θ)(cos nθ i sin nθ) dθ. 2π π − Z−

Note that sinh(2 sin θ) is an odd function in θ, hence π sinh(2 sin θ) cos nθ dθ = 0. π Z− Finally, 1 π cn = sinh(2 sin θ) sin nθdθ, n = 0, 1, 2, . 2π π ± ± ··· Z−

54 Example

Find the Laurent expansion of the function 1 f(z)= z k − that is valid inside the domain z > k , where k is real and k < 1. | | | | | | Using the Laurent expansion, deduce that k cos θ k2 ∞ kn cos nθ = − , 1 2k cos θ + k2 nX=1 − k sin θ ∞ kn sin nθ = . 1 2k cos θ + k2 nX=1 −

55 Solution

For z > k , where k < 1, we have | | | | | | 1 1 = z k z 1 k − − z 1 k k2 k = 1+ + + for < 1. 2 z z z ··· ! z

y z=eiq

x |k |

56 Take the point z = eiθ, which lies inside the region of convergence of the above Laurent series. Substituting into the inﬁnite series

1 1 iθ 2 2iθ n inθ = (1 + ke− + k e− + + k e− + ). eiθ k eiθ ··· ··· − Rearranging the terms eiθ(e iθ k) 1 k(cos θ + i sin θ) (1 2k cos θ + k2) − − 1 = − − − (eiθ k)(e iθ k) − 1 2k cos θ + k2 − − − − k cos θ k2 ik sin θ = − − 1 2k cos θ + k2 − = ∞ kn cos nθ i ∞ kn sin nθ. − nX=1 nX=1 Equating the real and imaginary parts, we obtain the desired results.

57 Example

By evaluating the contour integral 1 2n dz z + , z =1 z z I| | show that 2π 1 3 5 (2n 1) cos2n θ dθ = 2π · · ··· − . 0 2 4 6 2n Z · · ··· Solution

1 2n dz On z = 1,z = eiθ, z + = 22n cos2n θ, = i dθ. | | z z 1 2n dz 2π I = z + = 22ni cos2n θ dθ. z =1 z z 0 I| | Z On the other hand, the integrand can be expanded as 1 1 z2n + C z2n 2 + + C + + C , 0 < z < . 2n 1 − 2n n 2n 2n 2n z ··· ··· z | | ∞

58 Since the integrand is analytic everywhere except at z = 0, the above expansion is the Laurent series of the integrand valid in the annulus: z > 0. | | 1 2n dz 1 z + = 2πi coeﬃcient of in the Laurent series z =1 z z z I| | = 2πi2nCn. Hence,

2π Cn cos2n θ dθ = 2π2n 2n Z0 2 (2n)! = 2π (n!)222n 1 3 (2n 1)(2nn!) = 2π · ··· − (2nn!)(2nn!) 1 3 (2n 1) = 2π · ··· − . 2 4 6 2n · · ···

59 The shaded region is z > 0. The circle C : z = 1 lies completely | | | | inside the annulus of convergence and goes around z =0. This is a punctured complex plane with a single deleted point.

60 Example

Find all the possible Taylor and Laurent series expansions of 1 f(z)= at z0 = 0. (z i)(z 2) − − Specify the region of convergence (solid disc or annulus) of each of the above series.

61 Solution

There are two isolated singularities, namely, at z = i and z = 2. The possible circular or annular regions of analyticity are

(i) z < 1, (ii) 1 < z < 2, (iii) z > 2. | | | | | | (i) For z < 1 | | 1 i z zn = z = i 1+ + + + for z < 1 z i 1 " i ··· in ! # | | − − i 1 1 1 1 z z n = = 1+ + + + for z < 2 z 2 −2 1 z −2 2 ··· 2 ··· | | − − 2 1 1 1 1 z n 1 z n f(z) = = i ∞ + ∞ i 2 z i − z 2 i 2  i 2 2  − − − − nX=0 nX=0 1 1 n 1 1   = ∞ − + zn. i 2 " i 2n+1# − nX=0 This is a Taylor series which converges inside the solid disc z < 1. | |

63 Example

Find all possible Laurent series of 1 f(z)= at the point α = 1. 1 z2 − − Solution

The function has two isolated singular points at z = 1 and z = 1. − There exist two annular regions (i) 0 < z + 1 < 2 and (ii) z + 1 > | | | | 2 where the function is analytic everywhere inside the respective region.

64 (i) 0 < z + 1 < 2 | | 1 1 = 1 z2 2(z + 1) 1 z+1 − − 2 1 (z + 1)k = ∞ 2(z + 1) 2k kX=0 1 1 1 1 = + + (z +1)+ (z + 1)2 + . 2(z + 1) 4 8 16 ··· z + 1 The above expansion is valid provided < 1 and z + 1 = 0.

2 6

Given that 0 < z + 1 < 2, the above requirement is satisﬁed. | | For any simple closed curve C lying completely inside z +1 < 2 1 | | and encircling the point z = 1, we have − 1 1 1 c 1 = dz = . − 2πi C 1 z2 2 I 1 −

65 (ii) z + 1 > 2 | | 1 1 = 1 z2 −(z + 1)2 1 2 − − z+1 1 2k 2 = ∞ since < 1 2 k −(z + 1) =0 (z + 1) z + 1 kX

1 2 4 = + + + . − "(z + 1)2 (z + 1)3 (z + 1)4 ··· #

For any simple closed curve C encircling the circle: z + 1 = 2, 2 | | we have 1 1 c1 = dz = 0. 2πi C 1 z2 I 2 −

66 Example

1 Suppose f(z) is analytic inside the annulus: r < z < ,r < 1 and | | r 1 satisﬁes f = f(z). Write the Laurent expansion of f(z) at z = 0 z ∞ k as f(z)= akz . Show that k= X−∞ (a) ak = a k, − (b) f(z) is real on z = 1. | |

67 Solution

(a) Consider

1 ∞ k ∞ k f = akz− = a kz z k= k= − X−∞ X−∞ ∞ k f(z) = akz k= X−∞ 1 f = f(z) ak = a k. z ⇒ − (b) It suffices to show f(z) = f(z) when z satisfies zz = 1. This is 1 obviously satisfied from the given property f = f(z), for all z 1 z lying inside the annulus: r < z < . | | r

68 Example

From the Maclaurin series expansion of ez, it follows that the Lau- rent series expansion of e1/z about 0 is 1 e1/z = ∞ , z > 0. znn! | | nX=0 The Laurent coeﬃcients are 1 e1/z bn = dz 2πi z1 n IC − where is chosen to be the circle z = 1. Such choice of is C | | C feasible since the circle z = 1 lies completely inside the annulus | | region of convergence of e1/z and it goes around the point z = 0. By comparing the coeﬃcients, we obtain

e1/z 2πi dz = , n N. z1 n n! ∈ IC −

69 On z = 1,z = eiθ, π θ π, so that the complex integral can be | | − ≤ ≤ expressed as

π iθ π exp(e− ) iθ (cos θ i sin θ+inθ) 2πi ie dθ = ie − dθ = π e(1 n)iθ π n! Z− − Z− giving π 2π ecos θ[cos(nθ sin θ)+ i sin(nθ sin θ)] dθ = . π − − n! Z− Comparing the real parts on both sides, we obtain π 2π ecos θ cos(nθ sin θ) dθ = , for all n N. π − n! ∈ Z− Equating the imaginary parts gives π ecos θ sin(nθ sin θ) dθ = 0. π − Z− The above result is obvious since the integrand ecos θ sin(nθ sin θ) − is an odd function.

70 Classiﬁcation of singularities and residue calculus

Singular point of a function f(z) is a point at which f(z) is not analytic.

Isolated singularity Existence of a neighborhood of z0 in which z0 is the only singular point of f(z).

1 e.g. f(z)= has i as isolated singularities. z2 + 1 ±

Singularities of csc z are all isolated and they are simply the zeros of sin z, namely, z = kπ,k is any integer.

Non-isolated singularities f(z)= z is nowhere analytic so that every point in C is a non-isolated singularity.

71 π Consider f(z) = csc , all points z = 1/n,n = 1, 2, , are iso- z ± ± ··· lated singularities, while the origin is a non-isolated singularity.

Suppose z0 is an isolated singularity, then there exists a positive number R such that f(z) is analytic inside the deleted neighborhood 0 < z z < R. | − 0| This forms an annular domain within which the Laurent series theorem is applicable, where

∞ n ∞ n f(z)= an(z z ) + bn(z z ) , 0 < z z < R. − 0 − 0 − | − 0| nX=0 nX=1

The isolated singularity z0 is classiﬁed according to the property of the principal part of the Laurent series expanded in the deleted neighborhood of z0 where f is analytic. A singular point must be either a removable singularity, an essential singularity or a pole.

72 Removable singularity

The principal part vanishes and the Laurent series is essentially a Taylor series. The series represents an analytic function in z z < | − 0| R.

1 cos z For example, − is undeﬁned at z = 0. The Laurent expansion z2 1 cos z of − in a deleted neighborhood of z =0 is z2 1 cos z 1 z2 z4 − = 1 1 + z2 z2 " − − 2! 4! ··· !# 1 z2 z4 = + +, 0 < z < , 2! − 4! 6! ··· | | ∞ 1 cos z 1 Note that lim − = . The singularity at z = 0 can be re- z 0 z2 2! → 1 (1 cos z)/z2, z = 0 moved by deﬁning f(0) = . The function f(z)= − 6 2 ( 1/2, z = 0 admits the above Taylor series expansion valid for z < . | | ∞

73 Essential singularity

The principal part has inﬁnitely many non-zero terms. For example, consider z2e1/z. Inside the annular region 0 < z < , the Laurent | | ∞ series of z2e1/z is 1 1 1 1 1 z2e1/z = z2 + z + + + + , 0 < z < . 2! 3! z 4!z2 ··· | | ∞

Pole of order k

The Laurent expansion in the deleted neighborhood of z0 is b b b f(z)= ∞ a (z z )n + 1 + 2 + + k , n 0 2 k − z z0 (z z0) ··· (z z0) nX=0 − − − with bk = 0. It is called a simple pole when k = 1. For example, 6 1 z = i are simple poles of . ± 1+ z2

74 Example

Observe that 1 cos z 1 1 z2 z4 − = + , 0 < z < , z5 z3 "2! − 4! 6! −··· # | | ∞ so that (1 cos z)/z5 has a pole of order 3 at z = 0. − Example 1 The function f(z) = has a pole of order 2 at z = 1. It z(z 1)2 admits the following Laurent− expansion 1 1 1 1 = + + , z 1 > 1. z(z 1)2 (z 1)3 − (z 1)4 (z 1)5 −··· ··· | − | − − − −

It is incorrect to conclude that z = 1 is an essential singularity by observing that the above Laurent expansion has inﬁnitely many negative power terms. This is because the annulus of convergence of the series is not a deleted neighborhood of z = 1.

75 Example

The Laurent expansion zn ∞ z n + ∞ − 2n+1 nX=1 nX=0 contains inﬁnitely many negative power terms of z. Is z = 0 an essential singularity of the function represented by the series?

1/z 1 The ﬁrst sum converges to = for z > 1 and the 1 1/z z 1 | | 1/2 − 1 − second sum converges to = for z < 2. 1 z/2 2 z | | − − Hence, the Laurent series converges to 1 1 1 f(z)= + = in 1 < z < 2. z 1 2 z −z2 3z + 2 | | − − − The annulus of convergence is not a deleted neighborhood of z = 0. Hence, we cannot make any claim about whether z = 0 is an essential singularity of f(z) simply through the examination of the above series. 76 Simple method of ﬁnding the order of a pole

If z0 is a pole of order k, then

lim (z z )kf(z)= b , b = 0. z z 0 k k → 0 − 6 In general, if we multiply f(z) by (z z )m and take the limit z z , − 0 → 0 then

bk m = k m lim (z z0) f(z)=  0 m > k . z z0 − →  m < k ∞ 1 cos z  For example, take f(z)= − .  z5 1 m = 3 m1 cos z 2 lim z − =  0 m> 3 . z 0 z5 →  m< 3 ∞ Hence, z = 0 is a pole of order 3 of (1 cos z)/z5. −

77 Example

Suppose z = 0 is a pole of order n and m, respectively, of the functions f1(z) and f2(z), ﬁnd the possible order of z =0 as a pole for f (z)+ f (z). Without loss of generality, we assume n m. 1 2 ≥

(i) When n > m n (1) k ∞ (1) k (1) f (z) = b z + a z , bn = 0, 1 k − k 6 kX=1 kX=0 m (2) k ∞ (2) k (2) f (z) = b z + a z , bm = 0, 2 k − k 6 kX=1 kX=0 n so f1(z)+ f2(z) must contain negative power terms up to z− .

(ii) When n = m, in general, z = 0 is a pole of order n for f1(z)+ 1 f2(z). However, it is possible that f1(z)= f2(z)+ , 0 ℓ n. − zℓ ≤ ≤ In this case, z = 0 is a pole of order ℓ. When ℓ = 0,z =0 is not an isolated singularity.

78 Behavior of f near isolated singularities

1. If z0 is a pole of f, then

lim f(z)= . z z → 0 ∞ To show the claim, suppose f has a pole of order m at z0, then φ(z) f(z)= (z z )m − 0 where φ(z) is analytic and non-zero at z0. Since m m lim (z z0) 1 (z z0) z z 0 lim = lim − = → 0 − = = 0, z z0 f(z) z z0 φ(z) lim φ(z) φ(z ) → → z z 0 → 0 hence lim f(z)= . z z → 0 ∞

79 2. If z0 is a removable singularity of f, then f admits a Taylor series in some deleted neighborhood 0 < z z <δ of z , so it | − 0| 0 is analytic and bounded in that neighborhood.

3. In each deleted neighborhood of an essential singularity, the function assumes values that come arbitrarily close to any given number.

Casorati-Weierstrass Theorem

Suppose that z0 is an essential singularity of a function f, and let w0 be any complex number. Then, for any positive number ǫ,

f(z) w <ǫ (A) | − 0| is satisﬁed at some point z in each deleted neighborhood of z0.

80 Proof

Take any deleted neighborhood of z0. Since z0 is an isolated singularity, the neighborhood itself or a subset of which will have the property that f is analytic inside 0 < z z < δ for some value | − 0| of δ. Assume that inequality (A) is not satisfied for any z in the neighborhood 0 < z z <δ (inside which f is analytic). Thus, | − 0| f(z) w ǫ when 0 < z z < δ. | − 0|≥ | − 0| Define 1 g(z)= , 0 < z z0 < δ, f(z) w | − | − 0 which is bounded and analytic. Hence, z0 is a removable singularity ⋆ of g . We let g be defined at z0 such that it is analytic there.

⋆ We use the following result: Suppose f is analytic and bounded in 0 < z z < ǫ. If f is not analytic at z , then it has a | − 0| 0 removable singularity at z0.

81 We consider the following two cases:

(i) If g(z ) = 0, then 0 6 1 f(z)= + w0, 0 < z z0 < δ, (B) g(z) | − | 1 becomes analytic at z0 if f(z0) = + w0. This means that g(z0) z0 is a removable singularity of f, not an essential one. We have a contradiction.

(ii) If g(z0) = 0, then g must have a zero of some ﬁnite order m at z0 since g(z) is not identically equal to zero in the neighborhood z z < δ. Write g(z) = (z z )mψ(z ), where ψ(z ) = 0. In | − 0| − 0 0 0 6 this case, f has a pole of order m since (z z )m 1 lim f(z)(z z )m = lim − 0 = z z 0 z z → 0 − → 0 g(z) ψ(z0) which is ﬁnite. Again, there is a contradiction.

82 Example

Classify the zeros and singularities of the function sin(1 z 1). − − Solution

Since the zeros of sin w occur only when w is an integer multiple of π, the function sin(1 z 1) has zeros when − − 1 z 1 = nπ, − − i.e., at 1 z = , n = 0, 1, 2, . 1 nπ ± ± ··· −

83 Furthermore, the zeros are simple because the derivative at these points is d 1 sin(1 1) = cos(1 1) z− 2 z− dz − 1 z − 1 z=(1 nπ)− z=(1 nπ)− − 2 − = (1 nπ) cos nπ = 0. − 6

The only singularity of sin(1 z 1) appears at z = 0. Suppose we let − − z approach 0 through positive values, sin(1 z 1) oscillates between − − 1. Such behavior can only characterize an essential singularity. ±