student solutions manual to accompany advanced engineering mathematics vol 2

7/25/2019 Student Solutions Manual to Accompany Advanced Engineering Mathematics VOL 2

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P A R T D

ComplexAnalysis

Chap. 13 Complex Numbers and Functions.Complex Differentiation

Complex numbers appeared in the textbook before in different topics. Solving linear homogeneous ODEsled to characteristic equations, (3), p. 54 in Sec. 2.2, with complex numbers in Example 5, p. 57, andCase III of the table on p. 58. Solving algebraic eigenvalue problems in Chap. 8 led to characteristicequations of matrices whose roots, the eigenvalues, could also be complex as shown in Example 4, p. 328.

Whereas, in these type of problems, complex numbers appear almost naturally as complex roots ofpolynomials (the simplest beingx2 C 1 D 0),it is much less immediate to considercomplex analysisthesystematic study of complex numbers, complex functions, and complex calculus. Indeed, complexanalysis will be the direction of study in Part D. The area has important engineering applications inelectrostatics, heat flow, and fluid flow. Further motivation for the study of complex analysis is given onp. 607 of the textbook.

We start with the basics in Chap. 13 by reviewing complex numberszD x C yi in Sec. 13.1 andintroducing complex integration in Sec.13.3. Those functions that are differentiable in the complex, onsome domain, are calledanalyticand will form the basis of complex analysis. Not all functions areanalytic. This leads to the most important topic of this chapter, theCauchyRiemann equations(1),p. 625 in Sec. 13.4, which allow us to test whether a function is analytic. They are very short but you haveto remember them! The rest of the chapter (Secs. 13.513.7) is devoted to elementary complex functions(exponential, trigonometric, hyperbolic, and logarithmic functions).

Your knowledge and understanding of real calculus will be useful. Concepts that you learned in realcalculus carry over to complex calculus; however, be aware that there aredistinct differences between

real calculus and complex analysisthat we clearly mark. For example, whereas the real equation ex D 1has only one solution, its complex counterpartez D 1has infinitely many solutions.


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258 Complex Analysis Part D

Sec. 13.1 Complex Numbers and Their Geometric Representation

Much of the material may be familiar to you, but we start from scratch to assure everyone starts at thesame level. This section begins with the four basic algebraic operations of complex numbers (addition,subtraction, multiplication, and division). Of these, the one that perhaps differs most from real numbers is

division(or forming a quotient).Thus make sure that you remember how to calculate the quotient of twocomplex numbers as given in equation (7),Example 2, p. 610, andProb. 3. In (7) we take the number z2from the denominator and form its complex conjugateNz2and a new quotientNz2=Nz2. We multiply the givenquotient by this new quotientNz2=Nz2 (which is equal to1 and thus allowed):

zD z1z2

D z1z2

1 D z1z2

Nz2Nz2;

which we multiply out, recalling thati 2 D 1[see (5), p. 609]. The final result is a complex number in aform that allows us to separate its real (Re z/ and imaginary (Im z/ parts. Also remember that1= iD i(seeProb. 1), as it occurs frequently. We continue by defining the complex planeand use it to graphcomplex numbers (note Fig. 318, p. 611, and Fig. 322, p. 612). We use equation (8), p. 612, to go fromcomplex to real.

Problem Set. 13.1. Page 612

1. Powers ofi.We compute the various powers ofi by the rules of addition, subtraction,multiplication, and division given on pp. 609610 of the textbook. We have formally that

i 2 D i iD .0;1/.0; 1/ [by (1), p. 609]D .0 0 1 1; 0 1 C 1 0/ [by (3), p. 609]D .0 1; 0 C 0/ (arithmetic)D .1;0/D 1 [by (1)],

(I1)

where in (3), that is,multiplication of complex numbers, we usedx1D 0,x2D 0,y1D 1,y2D 1.

(I2) i 3 D i 2iD .1/ iD i:

Here we used (I1) in the second equality. To get (I3), we apply (I2) twice:

(I3) i 4 D i 2i 2 D .1/ .1/ D 1:

(I4) i 5 D i 4iD 1 iD i;

and the pattern repeats itself as summarized in the table below.We use (7), p. 610, in the following calculation:

(I5)1

iD 1

i

NiNiD

1

i

.i /

.i /D .1 C 0i/.0 i/

.0 C i/.0 i / D 1 0 C 0 1

02 C 12 C i0 0 1 1

02 C 12 D 0 iD i:


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Chap. 13 Complex Numbers and Functions. Complex Differentiation 259

By (I5) and (I1) we get

1

i 2D 1

i 1

iD .i/.i / D .1/i .1/iD 1 i 2 D 1;(I6)

1

i 3D 1i 2 1

i D .1/.i/ D i [from (I6) and (I5)];

1

i 4D

1

i

2 1

i

2D .1/.1/ D 1,

and the pattern repeats itself. Memorize thati 2 D 1and 1= iD i as they will appear quitefrequently.

i 8 i 9 : .

i 4 i 5 i 6 i 7

Start ! i 0 i i 2 i 31 i

1

i

1= i 4 1=i 3 1=i 2 1=i Start1= i 8 1=i 7 1=i 6 1=i 5

. . 1= i 10 1=i 9

Sec. 13.1. Prob. 1. Table of powers ofi

3. Division of complex numbersa.The calculations of (7), p. 610, in detail are

zD z1z2

D x1 C iy1x2 C iy2

(by definition ofz1and z2)

D x1

Ciy1

x2 C iy2 x2

iy2

x2 iy2 (N.B. corresponds to multiplication by1)D .x1 C iy1/.x2 iy2/

.x2 C iy2/.x2 iy2/D x1x2 x1iy2 C iy1x2 iy1iy2

x2x2 x2iy2 C iy2x2 iy2iy2(multiplying it out: (3) in notation (4), p. 609)

D x1x2 ix1y2 C ix2y1 i2y1y2

x22 ix2y2 C ix2y2 i 2y22(grouping terms, using commutativity)

D x1x2 ix1y2 C ix2y1 C y1y2x22C y22

(usingi 2 D 1and simplifying)

D x1x2 C Cy1y2x22C y22

C ix2y1 x1y2x22C y22

(breaking into real part and imaginary part).

b.A practical example using (7) is

26 18i6 2i D

.26 18i/.6 2i /

.6 C 2i /

.6 C 2i /D 26 6 C 26 2i 18 6i 18 2i 2

62 C 22

D 156 C 52i 108iC 3636 C 4 D

192 56i40

D 4:8 1:4i:


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5. Pure imaginary number a.IfzD x C iyis pure imaginary, thenNzD z:Proof.Let zD x C iybe pure imaginary. ThenxD 0, by definition on the bottom of p. 609.Hence

(A) zD iy and (B) NzD iy (by definition. of complex conjugate, p. 612).

If we multipy both sides of (A) by 1, we get

zD iy;

which is equal toNz, hencezD Nz:

b.IfNzD z thenzD x C iyis pure imaginary.Proof.Let zD x C iyso thatNzD x iy. We are given thatNzD z, so

NzD x iyD zD .x C iy/ D x iy:

By the definition of equality (p. 609) we know that the real parts must be equal and that theimaginary parts must be equal. Thus

Re NzD Re.z/;xD x;

2xD 0;xD 0;

and

Im NzD Im.z/;yD y;

which is true for anyy. Thus

zD x C iyD iy:But this means, by definition, thatz is pure imaginary, as had to be shown.

11. Complex arithmetic

z1 z2D .2 C 11i/ .2 i/D 2 C 11i 2 C iD .2 2/ C .11 C 1/iD 4 C 12i

.z1 z2/2 D .4 C 12i/.4 C 12i/ D 16 48i 48i 144 D 128 96i.z1 z2/2

16 D 128

16 96

16iD 8 16

16 2

5 324

D 8 6i:

Next consider z1

4 z2

4

2

:

We have

z1

4D 1

4.2 C 11i / D 2

4C 11

4i;

z2

4D 2

4 1

4i:


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Their difference is

z1

4 z2

4D 2

4 2

4C

11

4C 1

4

i D 1 C 3i:

Hence z1

4 z2

4

2 D .1 C 3i/.1 C 3i/ D 1 3i 3iC 9i 2 D 1 6i 9 D 8 6i;

which is the same result as before.

19. Real part and imaginary part ofz=Nz.For zD x C iy, we have by (7), p. 610,z

NzD z

NzNNzNNzD

z

Nzz

z

since the conjugate of the conjugate of a complex number is the complex number itself (which youmay want to prove!). Then

z

NzD z

2

NzzD .x C iy/

2

x2

Cy2

D x2 C 2ixy y2

x2

Cy2

D x2 y2

x2

Cy2

C i 2xyx2

Cy2

:

Hence we get the result as shown on p. A34 of the textbook:

Re

z

Nz

D x

2 y2x2 C y2 I Im

z

Nz

D 2xy

x2 C y2 :

Sec. 13.2 Polar Form of Complex Numbers. Powers and Roots

Polar coordinates, defined by (1) and (2) on p. 613, play a more important role in complex analysis thanin calculus. Their study gives a deeper understanding of multiplication and division of complex numbers(pp. 615616) and absolute values. More details are as follows.

The polar angle(taken counterclockwise, see Fig. 323, p. 614) of a complex number is determinedonly up to integer multiples of2 . While often this is not essential, there are situations where it matters.For this purpose, we introduce the concept of the principal valueArgz in (5), p. 614, and illustrate it in

Example 1,Probs. 9and 13.The triangle inequality defined in (6), p. 614, and illustrated in Example 2, p. 615, is very important

since it will be used frequently in establishing bounds such as in Chap. 15.Often it will be used in its generalized form (6*), p. 615, which can be understood by the following

geometric reasoning. Draw several complex numbers as little arrows and let each tail coincide with thepreceding head. This gives you a zigzaging line ofn parts, and the left side of (6*) equals the distancefrom the tail ofz1to the head ofzn. Can you see it? Now take your zigzag line and pull it taut; then youhave the right side as the length of the zigzag line straightened out.

In almost all cases when we use (6*) in establishing bounds, it will not matter whether or not the rightside of (6*) is much larger than the left. However, it will be essential that we have such an upper bound forthe absolute value of the sum on the left, so that in a limit process, the latter cannot go to infinity.

The last topic is roots of complex numbers, illustrated in Figs. 327329, p. 617, andProb. 21. Look atthese figures and see how, for differentn, the roots of unity (16), p. 617, lie symmetrically on the unit

circle.

Problem Set 13.2. Page 618

1. Polar form.SketchzD 1 C i to understand what is going on. Pointz is the point.1; 1/in thecomplex plane. From this we see that the distance ofz from the origin is jzj D

p2. This is the


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absolute value ofz. Furthermore,z lies on the bisecting line of the first quadrant, so that its argument(the angle between the positive ray of thex-axis and the segment from 0 toz) is 45 or=4.

Now we show how the results follow from (3) and (4), p. 613. In the notation of (3) and (4) wehavezD x C iyD 1 C i: Hence the real part ofz is xD 1and the imaginary part ofz is yD 1.From (3) we obtain

j z j D p12 C 12 D p2;as before. From (4) we obtain

tan D yxD 1; D 45 or

4:

Hence the polar form (2), p. 613, is

zDp

2cos

4C i sin

4

:

Note that here we have explained the first part ofExample 1, p. 614, in great detail.

1 + i1

y

x1

45o

Sec. 13.2 Prob. 1. Graph ofzD 1 C i in the complex plane

5. Polar form.We use (7), p. 610, in Sec. 13.1, to obtain

(A)

p2

C 13

i

p8 23 i D

p2

C 13

i

p8 23 i

p8

C 23

i

p8 C 23 i :The numerator of (A) simplifies top

2 C 13

i

p

8 C 23

i

D p

16 C

23

p2 1

3

p8

i 29D 38

9C

23

p2 1

32p

2

iD 389

:

The denominator of (A) is

p82

C 23

2 D 8 C 49D 72

9C 4

9D 76

9:

Putting them together gives the simplification of (A), that is,

p2 C 1

3i

p8 23 i D 38

9

769 D 38

9 9

76 D

38

76D 1

2:

HencezD 12

corresponds to1

2; 0

in the complex plane. Furthermore, by (3), p. 613,

jzj D rDp

x2 C y2 Dq1

2

2 C 02 D 12


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and by (4), p. 613,

tanD yxD 01

2

D 0I D 180 D :

Hence by (2), p. 613, the desired polar form is

zD r.cosC isin / D 12

.cosC isin / :

180

x

y

1

2

o

Sec. 13.2 Prob. 5. Graph ofzD 12

in the complex plane

7. Polar form.For the givenz we have

jzj Dq

12 C 12

2 D q1 C 1

4 2;

tanD yxD

12

1 D 1

2I D arctan

1

2

:

The desired polar form ofz is

zD jzj .cosC isin / Dq

1 C 14

2

cos

arctan 1

2

C isin

arctan 1

2

:

9. Principal argument.The first and second quadrants correspond to0 Arg z . The third andfourth quadrants correspond to < Argz 0. Note that Arg z is continuous on the positive realsemiaxis and has a jump of2 on the negative real semiaxis. This is a convenient convention. Pointson the negative real semiaxis, e.g., 4:7, have the principal argument Arg zD :

To find the principal argument ofzD 1 C i , we convertz to polar form:

jzj Dp

.1/2 C 12 Dp

2;

tanD yxD 11D 1:

Hence

D 34

D 135:

Hencez, in polar form, iszD

p2cos 3

4C isin 3

4

:

As explained near the end of p. 613,is called the argument ofz and denoted byarg z: Thus is

D arg zD 34

2n; n D 0;1;2; :


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The reason is that sine and cosine are periodic with2 , so135 looks the same as135 C 360, etc.To avoid this concern, we define the principal argument Argz [see (5), p. 614]. We have

ArgzD 34

:

You should sketch the principal argument.

13. Principal argument.The complex number 1 C i in polar form is

1 C iDp

2

cos

4C isin

4

byProb. 1.

Then, using DeMoivres formula (13), p. 616, with rDp

2and n D 20,

.1 C i /20 Dp

220

cos

20 4

C isin

20

4

byProb. 1:

D 210 .cos5C isin 5/

D210 .cos

Cisin / :

Hence

arg zD 2n; n D 0;1;2; I Arg zD :Furthermore, note that

.1 C i/20 D 210 .cosC isin / D 210.1 C i 0/ D 210 D 1024:

Graph the prinicipal argument.

17. Conversion tox + i y.To convert from polar form to the form x C iy, we have to evaluatesin andcos for the given :Here

p8cos 14 C i sin 14 D p8p

2

2 C ip

2

2! D

p16

2 Cp

16

2 iD 2 C 2i:

21. Roots.From Prob. 1 and Example 1, p. 614 in this section, we know that1 C i in polar form is

1 C iDp

2cos 1

4C icos 1

4

:

Hence by (15), p. 617,

3p

1 C iD .1 C i /1=3 Dp

21=3

cos14

C 2k3

C icos14

C 2k3

:

Now we can simplify

p21=3 D 21=21=3 D 21=6and

14

C 2k3

D =43

C 2k 3

D 12

C 8k 12

D .1 C 8k/12

:


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Hence

3p

1 C iD 21=6cos

.1 C 8k/12

C isin .1 C 8k/12

;

where kD 0;1;2 (3 roots; thus 3 values ofk). Written out we get

ForkD 0 z0D 21=6cos

12C isin

12

:

ForkD 1 z1D 21=6cos

9

12C isin 9

12

:

ForkD 2 z2D 21=6cos

17

12 C isin 17

12

:

The three roots are regularly spaced around a circle of radius21=6 D 1:122 5with center0.

120

~1.12

~1.12

~1.12

15

x

y

o

120o

o

Sec. 13.2. Prob. 21. The three rootsz0; z1; z2 ofzD 3p1 C i in the complex plane

29. Equations involving roots of complex numbers.Applying the usual formula for the solutionsof a quadratic equation

zDb p

b2 4ac2a

to

(Eq) z2 z C 1 iD 0;we first have

(A) zD1

p12

4

1

.1

i/

2 1 D1

p

3

C4i

2 :

Now, in (A), we have to simplifyp3 C 4i :Let zD p C qi be a complex number wherep; qare

real. Then

z2 D .p C qi /2 D p2 q2 C 2pqiD 3 C 4i :


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We know that for two complex numbers to be equal, their real parts and imaginary parts must beequal, respectively. Hence, from the imaginary part

2pqD 4;pqD 2;(B)

pD 2q

:

This can then be used, in the real part,

p2 q2 D 3;

p2 C 4p2

D 3;

p4 C 4 D 3p2;p4 3p2 C 4 D 0:

To solve this quartic equation, we set h D p2 and get the quadratic equationh2

C3h

4

D0;

which factors into

.h 1/.h C 4/ D 0 so that h D 1 and h D 4:Hence

p2 D 1 and p2 D 4:Sincep must be real, p2 D 4is of no interest. We are left withp2 D 1so(C) (a) pD 1; (b) pD 1:Substituting [C(a)] into (B) gives

pqD 1 qD 2 so qD 2:

Similarly, substituting [C(b)] into (B) givespqD .1/ qD 2 so qD 2:

We havepD 1; qD 2 andpD 1,qD 2. Thus, forzD p C qi (see above), we get1 C 2i and 1 2iD .1 C 2i/:

Hence (A) simplifies to

zD1 p3 C 4i

2 D1

p.1 C 2i /22

D1 .1 C 2i /2

:

This gives us the desired solutions to (Eq), that is,

z1D1 C .1 C 2i /

2 D 2i

2D i

and

z2D1 .1 C 2i /

2 D2 2i

2 D 1 i:

Verify the result by plugging the two values into equation (Eq) and see that you get zero.


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Sec. 13.3 Derivative. Analytic Function

The material follows the calculus you are used to withcertain differencesdue to working in the complexplane with complex functions f.z/. In particular,the concept of limitis differentas z may approachz0from any direction (see pp. 621622 andExample 4). This also means that thederivative, which looks

the same as in calculus, is different in complex analysis. Open the textbook on p. 623 and take a look atExample 4. We show from the definition ofderivative(4), p. 622, which uses the concept of limit, thatf.z/ D Nz is not differentiable. The essence of the example is that approachingz along path I in Fig. 334,p. 623, gives a value different from that along path II. This is not allowed with limits (see pp. 621622).

We call those functions that are differentiable in some domain analytic(p. 623). You can think of themas the good functions, and they will form the preferred functions of complex analysis and itsapplications. Note thatf .z/ D Nz is not analytic. (You may want to build a small list of nonanalyticfunctions, as you encounter them. In Sec. 13.4 we shall learn a famous method for testing analyticity.)

The differentiation rules are the same as in real calculus (see Example 3, pp. 622623 and Prob. 19).Here are two examples

f.z/ D .1 z/16;

f0.z/ D 16.1 z/15.1/ D 16.1 z/15;where the factor.

1/comes from the chain rule;

f.z/ D i; f0.z/ D 0sincei is a constant.

Go over the material to see that many concepts from calculus carry over to complex analysis. Use thissection as a reference section for many of the concepts needed for Part D.


1. Regions of practical interest. Closed circular disk.We want to write

jz C 1 5i j 32

in the form

jz aj pas suggested on p. 619. We can write

jz C 1 5i jD jz C .1 5i /jDjz ..1 5i//jDjz .1 C 5i /j :

Hence the desired region

jz .1 C 5i /j 32

is a closed circular disk with center 1 C 5i (not1 5i !) and radius 32

.

7. Regions. Half-plane.Let z

Dx

Cyi . Then Re z

Dx as defined on p. 609. We are required to

determine whatRe z 1

means. By our reasoning we have Re zD x 1so that the region of interest isx 1:


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1 5i

x

y

1

5

3

2

Sec. 13.3. Prob. 1. Sketch of closed circular disk

jz

C1

5i

j 3

2

This is a closed right half-plane bounded byxD 1, that is, a half-plane to the right ofxD 1thatincludes the boundary.

y

x1

Sec. 13.3. Prob. 7. Sketch of half-plane Re z

1

11. Function valuesare obtained, as in calculus, by substitution of the given value into the givenfunction. Perhaps a quicker solution than the one shown on p. A35 of the textbook, and following theapproach of p. 621, is as follows. The function

f.z/ D 11 z evaluated atzD 1 i

is

f .1 i/ D 11 .1 i/D

1

1 1 C iD 1

iD i;

with the last equality by (I5) in Prob. 1 of Sec. 13.1 on p. 258 of this Manual. HenceRe fD Re.i/ D 0, Im fD Im.i/ D 1:

17. Continuity.Let us use polar coordinates (Sec. 13.2) to see whether the function defined by

f.z/ D8 0but the product in (C) must equal zero requires that

sin yD 0 which means that (D) yD 0; ; 2; 3 ; : : : :Since the product in (B) is positive,cos y has to be positive. If we look at (D), we know thatcos y is1for yD ; 3; 5 ;: : : but C1for yD 0; 2; 4 ; : : : :Hence (B) and (D) give(E) yD 0; 2; 4 ; : : : :Since (B) requires that the product be equal to1 and the cosine for the values ofy in (E) is 1, wehaveex D 1:Hence(F) xD 0:Then (E) and (F) together yield

xD 0 yD 0; 2; 4 ;: : : ;

and the desired solution to (A) is

zD x C yiD 2n i; n D 0 ; 1 ; 2 ; : : : :Note that (A), being complex, has infinitely many solutions in contrast to the same equation in real,which has only one solution.


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Sec. 13.6 Trigonometric and Hyperbolic Functions. Eulers Formula

In complex, the exponential, trigonometric, and hyperbolic functions are related by the definitions (1),p. 633, and (11), p. 635, and by the Euler formula (5), p. 634, as well as by (14) and (15), p. 635. Thus wecan convert them back and forth. Formulas (6) and (7) are needed for computing values. Problem 9uses

such a formula to compute function values.


1. Formulas for hyperbolic functions.To show that

cosh zD cosh xcosy C isinh xsin y

we do the following. We start with the definition ofcosh z. Since we want to avoid carrying a factor12

along, we multiply both sides of (11), p. 635, by2and get

2 cosh zD ez C ez

D exCiy C exiy (settingzD x C iy)

D ex

.cosy C isin y/ C ex

.cosy isin y/ (by (1), p. 630)D cosy.e x C ex/ C isin y.e x ex/D cosy.2 cosh x/ C isin y.2 sinhx/ (by (17), p. A65 of Sec. A3.1 of App. 3)D 2 cosh xcosy C 2i sinh xsin y:

Division by2 on both sides yields the desired result. Note that the formula just proven is usefulbecause it expressescosh z in terms of its real and imaginary parts.

The related formula forsinh z follows the same proof pattern, this time start with2 sinh zD ez ez. Fill in the details.

9. Function values.The strategy forProbs. 612is to find formulas in this section or in the problemset that allow us to get, as an answer, a real number or complex number. For example, the formulasin Prob. 1 are of the type we want for this kind of problem.

In the present case, by Prob. 1 (just proved before!), we denote the first given complex numberbyz1D 1 C 2i so thatx1D 1and y1D 2and use

cosh z1D cosh x1cosy1 C isinh x1sin y1:

Then

cosh z1D cosh .1 C 2i / D cosh.1/ cos2 C isinh.1/ sin 2:Now by (11), p. 635,

cosh x1D cosh.1/ D 1

2.e1 C e1/ D 1 C e

2

2e I sinh x1D sinh.1/ D

1 e22e

:

Using a calculator (or CAS) to get the actual values we have

cosh .1 C 2i / D 1 C e2

2e cos2 C i1 e

2

2e sin 2

D 1:543081 .0:4161468/ C i.1:752011/ .0:9092974/D 0:642148 1:068607i;


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which corresponds to the rounded answer on p. A36.For the second function valuez2D 2 i we notice that, by (1), p. 633,

coszD 12

eiz C eiz

and, by (11), p. 635,

cosh zD 12

.e z C ez/ :Now

(A) iz2D i.2 i/ D 2i i 2 D 1 2iD z1:Hence

cosz2D 12

eiz2 C eiz2D 1

2.ez1 C ez1/ [by (A)]

D cosh z1D cosh.1 2i /

so we get the same value as before!

13. Equations.We want to show that the complex cosine function is even.First solution directly from definition (1), p. 633.We start with

cos.z/ D 12

ei.z/ C ei.z/

:

We see that for any complex number zD x C iy:i.z/ D i .x C iy/ D i.x iy/ D ix i 2yD y ix :

Similarly,

izD i.x C iy/ D ix i 2yD y ixD i.z/:So we have

izD i.z/:

Similarly,

i.z/ D i.x yi / D y C xiand

izD i.x C iy/ D ix C i 2yD i.z/so that

izD i.z/:

Putting these two boxed equations to good use, we have

cos.z/ D 12

ei.z/ C ei.z/

D 1

2

eiz C eiz D 1

2

eiz C eiz D cosz:

Thuscos.z/ D cosz, which means that the complex cosine function (like its real counterpart) iseven.


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Second solution by using(6a), p. 634. From that formula we know

coszD cosxcosh y isin xsinh y:

We consider

cos.z/ D cos.x iy/D cos.x C i.y//D cos.x/ cosh.y/ isin.x/ sinh.y/D cosxcosh y i. sin x/. sinhy/D cosh xcosh y isin xsinh yD cosz:

The fourth equality used that, for realx and y, bothcosand coshare even andsin and sinhare odd,that is,

cos.x/ D cosxI cosh.y/ D cosh yI

sin.x/ D sinx sinh.x/ D sinhx:Similary, show that the complex sine function is odd, that is, sin.z/ D sin z:

17. Equations.To solve the given complex equation,cosh zD 0, we use that, by the first equality inProb. 1, p. 636, of Sec. 13.6, the given equation is equivalent to a pair of real equations:

Re .cosh z/ D cosh xcosyD 0;Im .cosh z/ D sinhxsin yD 0:

Sincecosh x 0for allx, we must havecos yD 0, henceyD .2n C 1/=2wheren D 0 ; 1 ; 2 ; : : : : For thesey we havesin y 0, noting that the realcosand sinhave no commonzeros! Hencesinh xD 0so thatxD 0. Thus our reasoning gives the solution

zD .x;y/ D .0;.2n C 1/=2/; that is, zD .2n C 1/i=2 where n D 0 ; 1 ; 2 ; : : : :

Sec. 13.7 Logarithm. General Power. Principal Value

Work this section with extra care, so that you understand:

1. The meaning of formulas (1), (2), (3), p. 637.

2. The difference between the real logarithmln x, which is a function defined forx > 0, and thecomplex logarithmln z, which is an infinitely many-valued relation, which, by formula (3),p. 637, decomposes into infinitely many functions.

Example 1, p. 637, andProbs. 5,15, and21illustrate these formulas.General powerszc are defined by (7), p. 639, and illustrated in Example 3 at the bottom of that page.


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5. Principal value.Note that the real logarithm of a negative number is undefined. The principalvalue Lnz ofln z is defined by (2), p. 637, that is,

LnzD

lnjzj C

iArgz

where Arg z is the principal value ofarg z . Now recall from (5), p. 614 of Sec. 13.2, that theprincipal value of the argument ofz is defined by


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Next we note that

zD elnz D elnjzjCiarg z D e0:6e0:4i :

We consider

e0:4i D e0C0:4i D e0.cos0:4 C isin 0:4/ [by (1), p. 630, Sec. 13.5]D cos0:4 C isin 0:4:

Putting it together, we get

zD e0:6e0:4iD e0:6.cos0:4 C isin 0:4/D 1:822119 .0:921061 C 0:389418i/D 1:6783 C 0:70957i:

23. General powers. Principal value.We start with the given equation and use (8), p. 640, and thedefinition of principal value to get

.1 C i /11 D e.1i/ Ln.1Ci/ :

Now the principal value

Ln.1 C i / D ln j1 C i j C i Arg.1 C i/ [by (2), p. 637].

Also

j1 C i j D p12 C 12 D

p2

and

Arg.1 C i / D 4

[see (5) and Example 1, both on p. 614].

Hence

Ln.1 C i/ D lnp

2 C i4

so that

.1 i/Ln.1 C i / D .1 i/ln

p2 C i

4 D lnp2 C i4 iln p2 i 2

4

D lnp

2 C 4C i

4 ln

p2

:


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Thus

.1 C i/11 D exphln

p2 C

4C i

4 ln

p2i

D expln

p2

C

4

exphi

4ln

p2i

D expln

p2

exp

4

hcos

4 ln

p2

C isin

4 ln

p2i

[by (1), p. 630]

Dp

2e=4hcos

4 ln

p2

C isin

4 ln

p2i

:

Numerical values are

4 ln

p2 D 0:4388246;

cos

4 ln

p2D cos.0:4388246/ D 0:9052517;

sin

4 ln

p2

D sin.0:4388246/ D 0:4248757;

p2e=4 D 3:1017664:Hence.1 C i/11 evaluates to

.1 C i /11 D 2:8079 C 1:3179i:


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Chap. 14 Complex Integration

The first method of integration (indefinite integration and substitution of limits) is a direct analog of

regular calculus and thus a good starting point for studying complex integration. The focal point ofChap. 14 is the very important Cauchy integral theorem(p. 653) in Sec. 14.2. This leads toCauchysintegral formula(1), p. 660 in Sec. 14.3, allowing us to evaluate certain complex integrals whoseintegrand is of the formf .z/=.z z0/withfbeing analytic. The chapter concludes with the surprisingresult that all analytic functions have derivatives of all orders. Complex integration has a very distinctflavor of its own and should therefore make an interesting study. The amount of theory in this chapter isvery manageable but powerful in that it allows us to solve many different integrals.

General orientation.Chapter 13 provides the background material for Chap. 14. We can broadlyclassify the material in Chap. 14 as a first approachto complex integration based on Cauchys integraltheorem and his related integral formula. The groundwork to asecond approachto complex integration isgiven in Chap. 15 with the actual method of integration (residue integration) given in Chap. 16.

Prerequisite. You should remember the material of Chap. 13, including the concept ofanalyticfunctions(Sec. 13.3), the important CauchyRiemann equationsof Sec. 13.4, and Eulers formula (5),p. 634 in Sec. 13.6. We make use of some of the properties of elementary complex functions when solving

problemsso, if you forgot,consult Chap. 13 in your textbook. You should recall how to solve basicreal integrals (see inside cover of textbook if needed). You should also have some knowledge of roots ofcomplex polynomials.

Sec. 14.1 Line Integral in the Complex Plane

The indefinite complex integrals are obtained from inverting, just as in regular calculus. Thus the startingpoint for the theory of complex integration is the consideration of definite complex integrals, which aredefined ascomplex line integralsand explored on pp. 643646. As an aside, the reader familiar with realline integrals (Sec. 10.1, pp. 413419 in the text, pp. 169172 in Vol. 1 of this Manual) will notice asimilarity between the two. Indeed (8), p. 646, can be used to make the relationship between complex lineintegrals and real line integrals explicit, that is,

ZC

f .z/ dzDZC

u dx ZC

v dy C iZ

C

u dy CZC

v dx

DZC

.udx v dy/ C iZ

C

u dy C v dx

;

whereC is the curve of integration and the resulting integrals are real.(However, having not studied real line integrals is not a hindrance to learning and enjoying complex

analysis as we go in a systematic fashion with the only prerequisite for Part D being elementary calculus.)The first practical method of complex integration involves indefinite integration and substitution of

limitsand is directly inspired from elementary calculus. It requires that the function be analytic. Thedetails are given in Theorem 1, formula (9), p. 647, and illustrated below by Examples 14and Probs. 23

and27.A prerequisite to understanding the second practical method of integration (use of a representation of apath) is to understandparametrization of complex curves(Examples 14, p. 647,Probs. 1, 7,and 19).Indeed, (10), p. 647, ofTheorem 2is a more general approach than (9) of Theorem 1, because Theorem 2applies toanycontinuous complex function not just analytic functions. However, the price of generality isa slight increase in difficulty.


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1. Path.We have to determine the path of

z.t/ D .1 C 12

i /t .2 t 5/:

Since the parametric representation

z.t/ D x.t/ C iy.t/ D 1 C 12

i

t

D tC i 12

t

is linear in the parametert , the representation is that of a straight line in the complexz-plane. Itsslope is positive, that is

y.t/

x.t/D

12

t

t D 1

2:

The straight-line segment starts attD 2, corresponding to

z0D z.2/ D 2 C i 12 2 D 2 C iand ends attD 5:

z1D z.5/ D 5 C 52 i:

Sketch it.

7. Path.To identify what path is represented by

z.t/ D 2 C 4eit=2 with 0 t 2

it is best to derive the solution stepwise.From Example 5, p. 648, we know that

z.t/ D eit with 0 t 2

represents a unit circle (i.e., radius 1, center0) traveled in the counterclockwise direction. Hence

z.t/ D eit=2 with 0 t 4

also represents that unit circle. Then

z.t/ D 4eit=2 with 0 t 2

represents a semicircle (half circle) of radius 4with center0traversed in the counterclockwisedirection.

Finally

z.t/ D 2 C 4eit=2 with 0 t 2

is a shift of that semicircle to center2, corresponding to the answer on p. A36 in App. 2 of thetextbook.


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Chap. 14 Complex Integration 285

x

y

2

Sec. 14.1 Prob. 7. Semicircle

Remark.Our solution demonstrates a way of doing mathematics by going from a simpler problem,whose answer we know, to more difficult problems whose answers we infer from the simple problem.

19. Parametric representation. Parabola. We are given that

yD 1 14

x2 where 2 x 2:

Hence we may set

xD t so that yD 1 14

x2 D 1 14

t2:

Now, forxD tD 2, we get

yD 1 14

t2 D 1 14

.2/2 D 0

so that

z0D 2 C 0i:

Similarly, z1D 2 C 0i and corresponds totD 2. Hence

z.t/ D x.t/ C iy.t/D tC i 1 1

4t2

; ( 2 t 2).

21. Integration.Before we solve the problem we should use the CauchyRiemann equations todetermine if the integrand Re z is analytic. The integrand

wD u C ivD f.z/ D Re zD x

is not analytic. Indeed, the first CauchyRiemann equation

uvD vy [by (1), p. 625 in Sec. 13.4]

is not satisfied because

uxD 1 but vD 0 so that vyD 0:

(The second CauchyRiemann equation is satisfied, but, of course, that is not enough for analyticity.)Hence wecannotapply the first method (9), p. 647, which would be more convenient, but we mustuse the second method (10), p. 647.

The shortest path fromz0D 1 C i toz1D 3 C 3i is a straight-line segment with these points asendpoints. Sketch the path. The difference of these points is

(A) z1 z0D .3 C 3i / .1 C i/ D 2 C 2i:


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We set

(B) z.t/ D z0 C .z1 z0/t:

Then, by taking the valuestD 0and tD 1, we have

z.0/ D z0 and z.1/ D z1becausez0cancels whentD 1:Hence (B) is a general representation of a segment with givenendpointsz0and z1, andt ranging from0to 1.

Now we start with Equation (B) and substitute (A) into (B) and, by use ofz0D z.0/ D 1 C i ,we obtain

z.t/ D x.t/ C iy.t/D z0 C .z1 z0/tD 1 C iC .2 C 2i/t(C)D 1 C 2tC i.1 C 2t/:

We integrate by using (10), p. 647. In (10) we need

f.z.t// D x.t/ D 1 C 2t ;

as well as the derivative ofz.t/with respect tot , that is,

Pz.t/ D dzdt

D 2 C 2i:

Both of these expressions are obtained from (C).We are now ready to integrate. From (10), p. 647, we obtain

ZC

f .z/ dzDZ ba

f z.t/Pz.t/dt

DZ 10

.1 C 2t/.2 C 2i/ dt

D .2 C 2i /Z 10

.1 C 2t/ dt:

Now

Z .1 C 2t/ dtD

Z dtC 2

Z t dtD tC 2 t

2

2D tC t2;

so that

ZC f .z/ dzD .2 C 2i / tC t

210

D .2 C 2i/.1 C t2/D 2.2 C 2i /D 4 C 4i ;


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which is the final answer on p. A37 in App. 2 of the textbook (with a somewhat differentparametrization).

23. Integration by the first method (Theorem 1, p. 647). From (3), p. 630 of Sec. 13.5 of the text,we know thatez is analytic. Hence we use indefinite integration and substitution of upper and lower

limits. We have Z ez dzD ez C const [by (2), p. 630].

(I1)

Z 2ii

ez dzDh

ezi2ii

D e2i ei :

Eulers formula (5), p. 634, states that

eiz D cosz C i sin z:

Hence

e2i

Dcos2

Ci sin 2

D1

Ci

0

D1;

ei D cosC i sin D 1 C 0 D 1:

Hence the integral (I1) evaluates to1 .1/ D 2:

27. Integration by the first method (Theorem 1, p. 647).The integrand sec2 z is analytic exceptat the points where cos z is 0[see Example 2(b), pp. 634635 of the textbook]. Since

.tanz/0D sec2 z [by (4), p. 634],

(I2)

Z i=4=4

sec2 z dzD

htan z

ii=4=4

D tan 14

i tan 14

:

This can be simplified because

tan1

4D sin

14

cos14

D

1p21p2

D 1:

Also

tan1

4 iD sin

14

i

cos 14

iD i sinh

14

cosh 14

D i tanh 1

4;

since, by (15), p. 635 of Sec. 13.6 of textbook,

sin izD i sinh z

and

cos izD cosh z

withzD 14

.


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A numeric value to six significant digits of the desiredrealhyperbolic tangent is0:655794:Hence(I2) evaluates to

i tanh 14

1 D 0:655794i 1:

Remember that the real hyperbolic tangent varies between 1and 1, as can be inferred from thebehavior of the curves ofsinh x and cosh x in Fig. 551 and confirmed in Fig. 552, p. A65 (in PartA3.1 of App. 3 of the textbook).

Sec. 14.2 Cauchys Integral Theorem

Cauchys integral theorem, p. 653, is the most important theorem in the whole chapter. It states that theintegral around a simple closed path (acontour integral) is zero, provided the integrand is an analyticfunction. Expressing this in a formula

(1)

IC

f .z/ dzD 0 whereCis a simple closed path

andClives in a complex domain D that is simply connected. The little circle on the integral signH

marksa contour integral.

Take a look atFig. 345, p. 652, for the meaning of simple closed path and Fig. 346, p. 653, for a simplyconnected domain. In its basic form, Theorem 1 (Cauchys integral theorem) requires that the path nottouch itself (a circle, an ellipse, a rectangle, etc., but not a figure 8) and lies inside a domainD that has noholes (see Fig. 347, p. 653).

You have to memorize Cauchys integral theorem. Not only is this theorem important by itself, as amain instrument of complex integration, it also has important implications explored further in this sectionas well as in Secs. 14.3 and 14.4.

Other highlights in Sec. 14.2 are path independence (Theorem 2, p. 655),deformation of path (p. 656,Example 6, Prob. 11), and extending Cauchys theorem to multiply connected domains (pp. 658659). Weshow where wecanuse Cauchys integral theorem (Examples 1and 2, p. 653,Probs. 9and 13) andwhere wecannotuse the theorem (Examples 3and 5, pp. 653654,Probs. 11and 23). Often the decisionhinges on the location of the points at which the integrandf .z/is not analytic. If the points lie insideC

(Prob. 23) then we cannot use Theorem 1 but use integration methods of Sec. 14.1. If the points lie outsideC (Prob. 13) we can use Theorem 1.


3. Deformation of path.In Example 4, p. 654, the integrand is not analytic atzD 0, but it iseverywhere else. Hence we can deform the contour (the unit circle) into any contour that containszD 0in its interior. The contour (the square) in Prob. 1 is of this type. Hence the answer is yes.

9. Cauchys integral theorem is applicablesincef.z/ D ez2 is analytic for allz, and thus entire(see p. 630 in Sec. 13.5 of the textbook). Hence, by Cauchys theorem (Theorem 1, p. 653),

IC

ez2

dzD 0 withCunit circle, counterclockwise.

More generally, the integral is0aroundanyclosed path of integration.

11. Cauchys integral theorem (Theorem 1, p. 653) is not applicable. Deformation of path.We see that2z 1 D 0at zD 1

2. Hence, at this point, the function

f.z/ D 12z 1


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is not analytic. SincezD 12

lies inside the contour of integration (the unit circle), Cauchys theoremis not applicable. Hence we have to integrate by the use of path. However, we can choose a mostconvenient path by applying the principle of deformation of path, described on p. 656 of thetextbook. This allows us to move the given unit circle eit by 1

2. We obtain the path Cgiven by

z.t/ D 12C e

it

where 0 t 2:Note thatt is traversed counterclockwise ast increases from0 to 2 , as required in the problem.Then

f.z/ D f.z.t// D 12z.t/ 1D

1

2 12C eit 1D

1

2e it:

Differentiation gives

Pz.t/ D ieit ; (chain rule!).

Using the second evaluation method (Theorem 2, p. 647, of Sec. 14.1) we get

ZC

f .x/ dzDZ ba

f z.t/ Pz.t/ dt [by (10), p. 647]

DZ 20

1

2e it ieit dt

D iZ 20

eit

2eit dt

D iZ 20

1

2dt

D i

t

2

20

D i:Note that the answer also follows directly from (3), p. 656, with m D 1and z0D 12 .

13. Nonanalytic outside the contour.To solve the problem, we consider z4 1:1 D 0, so thatz4 D 1:1:By (15), p. 617 of Sec. 13.2,

4p

zD rcos

C 2k4

C i sin C 2k4

; kD 0;1;2;3;

whererD 4p

1:1 D 1:0241and the four roots are

z0D 4p

1:1.cos0 C i sin 0/ D 4p

1:1;

z1D 4

p1:1cos

2C i sin

2 D 4p1:1 i;

z2D 4p

1:1 .cosC i sin / D 4p

1:1;

z3D 4p

1:1

cos

3

2C i sin 3

2

D 4

p1:1 i:


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C

x

y

1 1

i

1.0241

i

Sec. 14.2 Prob. 13. Area of integrationCversus location of rootsz0,z1,z2,z3 of denominator of integrand

Sincez0,z1; z2; z3 all lie on the circle with center .0; 0/and radiusrD

4p

1:1D

1:0241 > 1, theyareoutsidethe given unit circleC. Hencef .z/is analytic on and inside the unit circleC. HenceCauchys integral theorem applies and gives us

IC

f .z/ dzDIC

1

z4 1:1dzD 0:

23. Contour integration.We want to evaluate the contour integral

IC

2z 1z2 zd z whereC as given in the accompanying figure on p. 659.

We use partial fractions (given hint) on the integrand. We note that the denominator of the integrandfactors intoz2 zD z.z 1/so that we write

2z 1z2 zD

A

zC B

z 1 :

Multiplying the expression byz and then substitutingzD 0gives the value for A:2z 1z 1 D A C

Bz

z 1 ; 1

1D A C 0; A D 1 :

Similarly, multiplyingz 1and then substitutingzD 1, gives the value forB :

2z 1z

D A.z 1/z

C B; 11D 0 C B; BD 1 :

Hence

2z 1z.z 1/D

1

zC 1

z 1 :

The integrand is not analytic atzD 0and zD 1, which clearly lie insideC. Hence Cauchysintegral theorem, p. 653, does not apply. Instead we use (3), p. 656, withm D 1for the two


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integrands obtained by partial fractions. Note that z0D 0, in the first integral, and thenz0D 1in thesecond. Hence we get

IC

2z 1z2 z dz D

IC

1

z dz C

IC

1

z 1d z D 2 iC 2 iD 4i:

Sec. 14.3 Cauchys Integral Formula

Cauchys integral theorem leads to Cauchys integral formula (p. 660):

(1)

IC

f.z/

z z0dzD 2if .z0/:

Formula (1) evaluates contour integrals

(A)

IC

g.z/dz

with an integrand

g.z/ D f.z/z z0

withf .z/analytic:

Hence one must first find

f.z/ D .z z0/g.z/:

For instance, inExample 1, p. 661 of the text,

g.z/ D ez

z 2 hence f.z/ D .z 2/g.z/ D ez:

The next task consists of identifying where the pointz0lies with respect to the contourC of

integration. Ifz0 lies insideC (and the conditions of Theorem 1 are satisfied), then (1) is applied directly(Examples 1 and 2, p. 661). Ifz0lies outsideC, then we use Cauchys integral theorem of Sec. 14.3(Prob. 3). We extend our discussion to several points at which g.z/is not analytic.

Example 3, pp. 661662, and Probs.1 and 11illustrate that the evaluation of (A) depends on thelocation of the points at whichg.z/is not analytic, relative to the contour of the integration. The sectionends with multiply connected domains (3), p. 662 (Prob. 19).


1. Contour integration by Cauchys integral formula (1), p. 660.The contour jz C 1j D 1canbe written as jz .1/j D 1. Thus, it is a circle of radius 1with center 1. The given function to beintegrated is

g.z/

D

z2

z2

1:

Our first task is to find out where g.z/is not analytic. We consider

z2 1 D 0 so that z2 D 1:

Hence the points at whichg.z/is not analytic are

zD 1 and zD 1:


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Our next task is to find out which of these two values lies inside the contour and make sure thatneither of them lies on the contour (a case we would not yet be able to handle). The value zD 1liesoutside the circle (contour) and zD 1lies inside the contour. We have

g.z/

D z2

z2 1D z2

.z C 1/.z 1/:

Also

g.z/ D z2

z2 1D f.z/

z z0D f.z/

z .1/ :

Together

f.z/

z C 1D z2

.z C 1/.z 1/ :

Multiplying both sides byz C 1gives

f.z/ D z2

z 1 ;

which we use for (1), p. 660. Hence

IC

z2

z2 1dzDIC

f.z/

z z0dz [in the form (1), p. 660]

DIC

z2=.z 1/z .1/ dz [Notez0D 1

D 2i f.z0/D 2i f .1/

D 2 i 1

2D i:

x

y

C

1 1

Sec. 14.3 Prob. 1. ContourCof integration


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3. Contour integration. Cauchys integral theorem, p. 653.The contourC3W j z C i j D 1:4isa circle of radius1:4and centerz0D i . Just as in Prob. 1, we have to see whether the pointsz1D 1andz2D 1lie inside the contourC3. The distance between the pointsz0D i andz1D 1is, by (3)and Fig. 324, p. 614 in Sec. 13.2, as follows.

jz0 z1j D ji 1j D j1 C i j D px2 C y2 D p.1/2 C 12 Dp

2 > 1:4:

Hencez1lies outside the circleC3.By symmetryz2D 1also lies outside the contour.Henceg.z/ D z2=.z 1/is analytic on and insideC3:We apply Cauchys integral theorem and

get, by (1) on p. 653 in Sec. 14.2,

IC3

z2

z2 1 dzD 0

by settingf .z/ D z2

z2 1 in (1)

:

11. Contour integral.The contourC is an ellipse with focal points0 and 2i . The given integrand is

g.z/ D 1z2

C4

:

We considerz2 C 4 D 0so thatzD 2i . Hence the points at whichg.z/is not analytic arezD 2iandzD 2i .

To see whether these points lie inside the contour Cwe calculate forzD 2iD x C yi so thatxD 0and yD 2and

4x2 C .y 2/2 D 4 02 C .2 2/2 D 0 < 4;

so thatzD 2i lies inside the contour. Similarly,zD 2i corresponds toxD 0; yD 2and

4x2 C .y 2/2 D .2 2/2 D 16 > 4;

so thatzD 2i lies outside the ellipse.We have

g.z/ D 1z2 C 4D

f.z/

z z0D f.z/

z 2i:

Together

f.z/

z 2i D 1

z2 C 4D 1

.z C 2i/.z 2i /where

f.z/ D 1z C 2i:


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Cauchys integral formula gives us

IC

dz

z2 C 4 dzDIC

f.z/

z C z0dz [by (1), p. 660]

D IC

1=.z

C2i /

z 2i dzD 2if .z0/D 2if.2i/

D 2 i 12iC 2i

D 2 i 14i

D 12

.

13. Contour integral.We use Cauchys integral formula. The integral is of the form (1), p. 660, with

z z0D z 2, hencez0D 2. Also,f .z/ D z C 2is analytic, so that we can use (1) and calculate2if.2/ D 8 i .

19. Annulus.We have to find the points in the annulus1 < j z j < 3at which

g.z/ D ez2

z2.z 1 i / D ez

2

z2z .1 C i /is not analytic. We see thatzD 1 C i is such a point in the annulus. Another point iszD 0, but thisis not in the annulus, that is, not between the circles, but in the hole. Hence we calculate

f.z/ D z .1 C i/g.z/ D ez2

z2 :

We evaluate it atzD 1 C i and also note that

(C) z2 D .1 C i /2 D 2i:

We obtain by Cauchys integral formula, p. 660,

2if .1 C i/ D 2 ie.1Ci/2

2i

D e.1Ci/2

D e2i [by (C)]D .cos2 C i sin 2/ [by Eulers formula]:

A numeric value is

.0:416147 C 0:909297i/ D 1:30736 C 2:85664i:


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Sec. 14.4 Derivatives of Analytic Functions

The main formula is (1), p. 664. It shows the surprising fact that complex analytic functions havederivatives of all orders. Be aware that, in the formula, the power in the denominator is one degree higher(n C 1) than the order of differentiation (n).


1. Contour integration. Use of a third derivative.Using (1), p. 664, we see that the givenfunction is

sin z

z4 D f.z/

.z z0/nC1 with f.z/ D sin zI z0D 0 and n C 1 D 4:

Thusn D 3:By Theorem 1, p. 664, we have

(A)

IC

f.z/

.z z0/4dzD 2 i

3 f.3/.z0/:

Sincef .z/D

sin z,f0

.z/D

cosz,f00

.z/D

sin z, so that

f.3/ D . sin z/0D cosz:

Furthermorez0D 0and

f.3/.z0/ D cos.0/ D 1:

Hence, by (A), we get the answer that

IC

sin z

z4 dzD 2 i

3 .1/

D 23

2

1

i

D 13

i:

5. Contour integration.This is similar toProb. 1. Here the denominator of the function to be

integrated is

z 12

4; and

z 1

2

4 D 0givesz0D 12 which lies inside the unit circle. To useTheorem 1, p. 664, we need the third derivative ofcosh 2z. We have, by the chain rule,

f.z/ D cosh 2zf0.z/ D 2 sinh 2z

f00.z/ D 4 sinh 2zf.3/.z/ D 8 sinh 2z:


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41/222


Hence (1), p. 664, gives you the value of the integral in thecounterclockwise direction, that is,

(B) 2if0.0/ D 2 i 1

D 22i:

Since the contour is to be traversed in theclockwisedirection, we obtain a minus sign in result (B)and get the final answer 22i:

13. First derivative. Logarithm.The question asks us to evaluate

IC

Lnz

.z 2/2 dz; C W jz 3j D 2traversed counterclockwise.

We see that the given integrand is Ln.z/=.z 2/2 and the contour of integration is a circle ofradius 2 with center3: At0and on the ray of the real axis, the function Ln z is not analytic, and it isessential that these points lie outside the contour. Otherwise, that is, if that ray intersected or touchedthe contour, we would not be able to integrate. Fortunately, in our problem, the circle is always to theright of these points.

In view of the fact that the integrand is not analytic at zD z0D 2, which lies inside the contour,then, according to (1), p. 664, withn C 1 D 2, hencen D 1, andz0D 2; the integral equals2 itimes the value of the first derivative of Lnz evaluated at at z0D 2. We have the derivative of Ln z is

.Lnz/0D 1z

which, evaluated atzD z0D 2, is 12 :This gives a factor 12 to the result, so that the final answer is12 2 iD i:


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Chap. 15 Power Series, Taylor Series

We shift our studies from complex functions to power seriesof complex functions, marking the beginning

of anotherdistinctapproach to complex integration. It is called residue integration and relies ongeneralizedTaylor seriestopics to be covered in Chap. 16. However, to properly understand these topics,we have to start with the topics of power series and Taylor series, which are the themes of Chap. 15.

Thesecond approachto complex integration based on residues owes gratitude to Weierstrass (seefootnote 5, p. 703 in the textbook), Riemann (see footnote 4, p. 625 in Sec. 13.4), and others. Weierstrass,in particular, championed the use of power series in complex analysis and left a distinct mark on the fieldthrough teaching it to his students (who took good lecture notes for posterity; indeed we own such ahandwritten copy) and his relatively few but important publications during his lifetime. (His collectedwork is much larger as it also contains unpublished material.)

The two approaches of complex integration coexist and should not be a source of confusion. (For moreon this topic turn to p. x of the Preface of the textbook and read the first paragraph.)

We start with convergence tests for complex series, which are quite similar to those for real series.Indeed, if you have a good understanding of real series, Sec. 15.1 may be a review and you could move onto the next section on power series and theirradius of convergence. We learn that complex power series

represent analytic functions (Sec. 15.3) and that, conversely, every analytic function can be represented bya power series in terms of a (complex)Taylor series(Sec. 15.4). Moreover, we can generate new powerseries from old power series (of analytic functions) by termwise differentiation and termwise integration.We conclude our study with uniform convergence.

From calculus, you want to review sequences and series and their convergence tests. You shouldrememberanalytic functionsand Cauchys integral formula(1), p. 660 in Sec. 14.3. A knowledge ofhow to calculate real Taylor series is helpful for Sec. 15.4. The material is quite hands-on in that you willconstruct power series and calculate their radii of convergence.

Sec. 15.1 Sequences, Series, Convergence Tests

This is similar to sequences and series in real calculus. Before you go ontest your knowledge of realseries and answer the following questions: What is the harmonic series? Does it converge or diverge? Canyou show that your answer is correct? Close the book and work on the problem. Compare your answerwith the answer on p. 314 at the end of this chapter in this Manual. If you got a correct answer, great! Ifnot, then you should definitely study Sec. 15.1 in the textbook.

Most important, from a practical point of view, is the ratio test(see Theorem 7, p. 676 and Theorem 8,p. 677).

The harmonic series is used in the proof of Theorem 8 (p. 677) and in the Caution after Theorem 3,p. 674. One difference between calculus and complex analysis is Theorem 2, p. 674, which treats theconvergence of a complex series as the convergence of its real part and its complex part, respectively.


3. Sequence.The sequence to be characterized is

znD n

4 C 2ni:


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Chap. 15 Power Series, Taylor Series 299

First solution method:

znD n

4 C 2ni

D n

4 C 2ni4 2ni4 2ni

[by (7), p. 610 of Sec. 13.1]

D n.4 2ni/42 C .2n/2

D 4n42 C 4n2C i

n

2

8 C 2n2

:

We have just writtenznin the form

znD xn C iyn:By Theorem 1, p. 672, we treat each of the sequences fxng and fyng separately when characterizingthe behavior offzng :Thus

limn!1

xnD limn!1

4n4 C 4n2

D limn!1

nn2

1Cn2

n2

(divide numerator and denominator by4n2/

D limn!1

n

1n2

C 1

Dlimn!1

n

limn!1

1n2

C limn!1

1

D 00

C1D 0:

Furthermore,

limn!1

ynD limn!1

n

2

8 C 2n2

D limn!1

n2n2

8C2n2

n2

D limn!1

8n2

C 2

D 0

C2D

2:

Hence the sequence converges to

0 C i

2

D 1

2 i:


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Second solution method(as given on p. A38):

znD n

4 C 2ni

Dn2ni

4C2ni2ni (division of numerator and denominator by2n i )

D2i

2ni

C 1D2 1

i2ni

C 1D2

.i/2ni

C 1D1

2 i

1 C 2ni

:

Now

limn!1

znD limn!1

12

i

1 C 2ni

Dlimn!1

12

i

limn!1

1 C limn!1

2ni

D12

i

1 C 0D 1

2 i:

Since the sequence converges it is also bounded.

5. Sequence.The termsznD .1/n C 10i; n D 1;2;3; , arez1D 1 C 10i; z2D 1 C 10i; z3D 1 C 10i; z4D 1 C 10i; :

The sequence is bounded because

j zn jDj.1/n C 10i jD

p.1/n2 C 102

Dp

1 C 100D

p101

< 11:

For odd subscripts the terms are 1 C 10i and for even subscripts1 C 10i . The sequence has twolimit points 1 C 10i and1 C 10i , but, by definition of convergence (p. 672), it can only have one.Hence the sequence fzng diverges.

9. Sequence.Calculate

j zn j D j0:9 C 0:1i j2n

D .j0:9 C 0:1i j2/nD .0:81 C 0:01/nD 0:82n ! 0 as n ! 0:

Conclude that the sequence converges absolutely to 0.

13. Bounded complex sequence. To verify the claim of this problem, we first have to show that:

(i) If a complex sequence is bounded, then the two corresponding sequences of real parts and

imaginary parts are also bounded.Proof of (i).Let fzng be an arbitrary complex sequence that is bounded. This means that there is aconstantKsuch that

j zn j < K for alln (i.e., all terms of the sequence).


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Set

znD xn C i ynas on p. 672 of the text. Then

jznj D qx2n C y2n [by (3), p. 613 of Sec. 13.2]and

jznj2 D x2n C y2n:Now

x2n x2n C y2nD jznj2 sincex2n 0; y2n 0:Furthermore,

x2nD jxnj2 sincex2n 0:Thus

jxnj2 jznj2

jxnjjznjso that

jxnj < K:Similarly,

jynj2 y2n jznj2 < K2so that

jynj < K:Sincen was arbitrary, we have shown that fxng and fyng are bounded by some constantK:

Next we have to show that:

(ii) If the two sequences of real parts and imaginary parts are bounded, then the complex sequence is

also bounded.

Proof of (ii). Let fxng and fyng be bounded sequences of the real parts and imaginary parts,respectively. This means that there is a constant Lsuch that

j xn j < Lp

2; jyn j 0:Furthermore, (D) also gives us

rD R :Together,

jz C i j R Dp

3 ( > 0).This is the form in which the answer is given on p. A39 in App. 2 of the textbook.

7. Power series. No uniform convergence.We have to calculate the radius of convergence for

1XnD1

n

n2

z C 1

2i

n:

We want to use the CauchyHadamard formula (6), p. 683 of Sec. 15.2. We start with

an

anC1D n

n2 .n C 1/

2

.n C 1/ ;

which is written out

D n 1n2

.n C 1/.n C 1/.n C 1/.n 1/

and, with cancellations, becomes

D n C 1n2

:

Thus

limn>1

ananC1

D lim

n>1

n C 1n2

D limn>1

1nC 1

n2

D lim

n>1

1

n 0

C limn>1

1

n2 0

D 0:

HenceRD 0, which means that the given series converges only at the center:z0D 12 i:

Hence it does not converge uniformly anywhere. Indeed, the result is not surprising since

n >> n2;

and thus the coefficients of the series

1;

2

4 ;

6

9 ;

24

16 ;

120

25;

720

36; n

n! 1 as n ! 1:13. Uniform convergence. Weierstrass M-test.We want to show that

1XnD1

sinn jzj

n2

converges uniformly for allz:


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59/222


s32D 1 C 12C C 132 31terms

D 1 C 12C 1

3C 1

4

C 15C C 1

8

4terms

C 19C C 1

16

8terms

C 117

C C 132

16terms

> 1 C 12C 1

4C 1

4

2terms

C 18C C 1

8

4terms

C 116

C C 116

8terms

C 132

C C 132

16termsD 1 C 1

2C 2

4C 4 1

8C 8 1

16C 16 1

32D 1 C 1

2C 2

4C 4

8C 8

16C 16

32 5fractions of value

12

D 1 C 5 12

:

Thus in general

s2n

> 1 C n 12

:

Asn ! 1, then1 C n 1

2! 1 and hence s

2n! 1:

This shows that the sequence of partial sumss2n

is unbounded, and hence the sequence ofall partial

sums of the series is unbounded. Hence, the harmonic series diverges.Another way to show that the hamonic series diverges is by the integral testfrom calculus

(which we can use sincef .x/is continuous, positive, and decreasing on the real interval 1; 1)

(A)

Z 1

1

1

xdxD lim

t!1

Z t1

1

xdxD lim

t!1lnxtxD1D lim

t!1ln t ln 1

0

D limt!1

ln t! 1:

Since the integral in (A) does not exist (diverges), the related harmonic series (HS) [whosenth termequalsf.n/] diverges.

Remark. The nameharmoniccomes from overtones in music (harmony!). The harmonic series is soimportant because, although its terms go to zero asm ! 1, it still diverges. Go back to p. 298.


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Chap. 16 Laurent Series. Residue Integration

In Chap. 16, we solve complex integrals over simple closed paths Cwhere the integrandf .z/is analytic

exceptat a pointz0 (or at several such points) inside C: In this scenario we cannot use Cauchys integraltheorem (1), p. 653, but need to continue our study of complex series, which we began in Chap. 15. Wegeneralize Taylor series toLaurent serieswhich allow such singularities atz0. Laurent series have bothpositive andnegativeinteger powers and have no significant counterpart in calculus. Their study providesthe background theory (Sec. 16.2) needed for these complex integrals with singularities. We shall useresidue integration, in Sec. 16.3, to solve them. Perhaps most amazing is that we can use residueintegration to even solve certain types ofrealdefinite integrals (Sec. 16.4) that would be difficult to solvewith regular calculus. This completes our study of thesecond approach to complex integration based onresiduesthat we began in Chap. 15.Before you study this chapter you should know analytic functions (p. 625, in Sec. 13.4), Cauchys integraltheorem (p. 653, in Sec. 14.2), power series (Sec. 15.2, pp. 680685), and Taylor series (1), p. 690. Fromcalculus, you should know how to integrate functions in the complex several times as well as know how tofactor quadratic polynomials and check whether their roots lie inside a circle or other simple closed paths.

Sec. 16.1 Laurent Series

Laurent seriesgeneralize Taylor series by allowing the development of a function f.z/in powers ofz z0whenf .z/is singular atz0(for singular, see p. 693 of Sec. 15.4 in the textbook). A Laurent series (1),p. 709, consists of positive as well asnegativeinteger powers ofz z0and a constant. The Laurentseries converges in an annulus, a circular ring with centerz0 as shown in Fig. 370, p. 709 of the textbook.

The details are given in the important Theorem 1, p. 709, and expressed by (1) and (2), which can bewritten in shortened form (10) and (20), p. 710.

Take a look atExample 4, p. 713, andExample 5, pp. 713714. A function may have different Laurentseries in different annuli with the same centerz0. Of these series, the most important Laurent series is theone that converges directly near the center z0, at which the given function has a singularity. Its negativepowers form the so-called principal partof the singularity off .z/at z0 (Example 4withz0D 0andProbs. 1and 8).


Hint. To obtain the Laurent series for probs. 18 use either a familiar Maclaurin series of Chap. 15 or aseries in powers of 1=z:

1. Laurent series near a singularity at 0.To solve this problem we start with the Maclaurin seriesfor cos z, that is,

coszD 1 z2

2C z

4

4 z

6

6C [by (14), p. 695].(A)

Next, since we want

cosz

z4

;

we divide (A) byz4, that is,

cosz

z4 D 1

z4

1 z

2

2C z

4

4 z

6

6C

D 1z4

12z2

C 124

z2

720C :


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Chap. 16 Laurent Series. Residue Integration 317

The principal part consists of

1

z4 1

2z2:

Furthermore, the series converges for allz

0.

7. Laurent series near a singularity at 0.We start with

coshwD 1 C w2

2C w

4

4C [by (15), p. 695].

We setwD 1=z:Then

cosh1

zD 1 C 1

2

1

z2C 1

4

1

z4C .

Multiplication byz3 yields

z3 cosh1

zDz3

C

1

2

1

z

2z3

C

1

4

1

z

4z3

C D z3 C 1

2z C 1

4

1

zC 1

6

1

z2

D z3 C 12

z C 124

z1 C 1720

z2 C :

We see that the principal part is

1

24z1 C 1

720z2 C :

Furthermore, the series converges for allz 0;or equivalently the region of convergence is0 < jzj < 1:

15. Laurent series. Singularity atz0D

.We use (6), p. A64, of Sec. A3.1 in App. 3 of thetextbook and simplify by noting that cos D 1and sin D 0:

coszD cos..z / C /D cos.z / cosC sin.z / sin(B)D cos.z /:

Now

coswD 1 w2

2C w

4

4 w

6

6C :

We set

w

Dz

and get

cos.z / D 1 .z /2

2 C .z /

4

4 .z /

6

6 C :

Then

cos.z / D 1 C .z /2

2 .z /

4

4 C .z /

6

6 C .


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We multiply by

1

.z /2

and get

cos.z /.z /2 D

1

.z /2C 1

2 .z /

2

4 C .z /

4

6 C .

Hence by (B)

cosz

.z /2D .z /2 C 1

2 1

24.z /2 C 1

720.z /4 C .

The principal part is .z /2 and the radius of convergence is 0 < jz j < 1 (converges forallz ).

19. Taylor and Laurent Series.The geometric series is

1

1 wD1X

nD0wn j w j < 1 [by (11), p. 694].

We need

1

1 z2 so we set wD z2:

Then we get the Taylor series

1

1 z2D1X

nD0.z2/n

z2< 1

D1X

nD0z2n or

z2 D jzj2 < 1 so thatj z j < 1

D 1 C z2 C z4 C z6 C :

Similarly, we obtain the Laurent series converging forj z j > 1by the following trick, which youshould remember:

1

1 z2D 1

z2

1 1z2

D 1z2 1

1

1

z

2

D 1z21X

nD0

1

z

2n

D 1z2

1 C z2 C z4 C z6 C

D 1z2

1z4

1z6

1z8

D 1X

nD0

1

z2nC2 j z j > 1:


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23. Taylor and Laurent series.We want all Taylor and Laurent series for

z8

1 z4 with z0D 0:

We start with

1

1 wD1X

nD0wn j w j < 1 [by (11), p. 694].

We setwD z4 and get

1

1 z4D1X

nD0z4n j z j < 1:

We multiply this byz8 to obtain the desiredTaylor series:

z8

1

z4

D z81

XnD0z4n D

1

XnD0z4nC8 j z j < 1

D z8 C z12 C z16 C :From Prob. 19 we know that the Laurent series for

1

1 w2D 1X

nD0

1

w2nC2 j w j > 1:

We setwD z2

1

1 z4D 1X

nD0

1

.z2/2nC2D

1XnD0

1

z4nC4

z2> 1 so thatj z j > 1:

Multiply the result byz8:

z8

1 z4D z8

1XnD0

1

z4nC4D

1XnD0

z8

z4nC4:

Now

z8

z4nC4D z8.4nC4/ D z44n:

Hence the desiredLaurent seriesfor

z8

1 z4 with center z0D 0

is

z8

1 z4D 1X

nD0z44n D z4 1 z4 z8

so that the principal part is

z4 z8

and j z j > 1:


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Note that we could have developed the Laurent series without using the result by Prob. 19 (but inthe same vein as Prob. 19) by starting with

1

1 z4D 1

z4 1

1

z4

; etc.

Sec. 16.2 Singularities and Zeros. Infinity

Major points of this section are as follows. We have to distinguish between the concepts of singularity andpole. A functionf .z/has asingularityat z0 iff .z/is not analytic atzD z0, but every neighborhood ofzD z0 contains points at whichf.z/is analytic.

Furthermore, if there is at least one such neighborhood that does not contain any other singularity, thenzD z0 is called anisolated singularity. For isolated singularities we can develop a Laurent seriesthatconverges in the immediate neighborhood ofzD z0. We look at the principal part of that series. If it is ofthe form

b1

z z0(withb1 0),

then the isolated singularity atzD z0is asimple pole (Example 1, pp. 715716).However, if theprincipal part is of the form

b1

z z0C b2

.z z0/2C bm

.z z0/m;

then we havea pole of orderm. It can also happen that the principal part has infinitely many terms; thenf.z/has anisolated essential singularityat zD z0 (seeExample 1, pp. 715716,Prob. 17).

A third concept is that of a zero, which follows our intuition. A functionf .z/has azeroat zD z0 iff .z0/ D 0:

Just as poles have orders so do zeros. Iff .z0/ D 0but the derivativef0.z/ 0, then the zero is asimple zero(i.e., a first-order zero). Iff .z0/ D 0; f0.z0/ D 0, butf00.z/ 0, then we have asecond-order zero, and so on (see Prob. 3for fourth-order zero). This relates toTaylor seriesbecause,when developing Taylor series, we calculate f .z0/; f

0.z0/; f00.z0/; :f.n/ .z0/by (4), p. 691, inSec. 15.4. In the case of a second-order zero, the first two coefficients of the Taylor series are zero. Thuszeros can be classified by Taylor series as shown by (3), p. 717.

Make sure that you understand the material of this section, in particular the concepts of pole and orderof pole as you will need these for residue integration.Theorem 4, p. 717, relates poles and zeros and willbe frequently used in Sec. 16.3.


3. Zeros.We claim thatf.z/ D .z C 81i/4 has a fourth-order zero atzD 81i . We show this directly:f.z/ D .z C 81i/4 D 0 gives zD z0D 81i:

To determine the order of that zero we differentiate until f.n/ .z0/ 0:We have

f.z/

D .z

C81i/4; f .

81i/

D f .z0/

D0

If0.z/ D 4.z C 81i/3; f 0.81i/D 0If00 .z/ D 12.z C 81i/2; f 00.81i/D 0I

f000 .z/ D 24.z C 81i/; f 000.81i/D 0If iv .z/ D 24; f iv .81i/ 0:

Hence, by definition of order of a zero, p. 717, we conclude that the order at z0 is 4. Note that wedemonstrated a special case of the theorem that states that ifg has a zero of first order (simple zero)atz0, theng

n (na positive integer) has a zero ofnth order atz0.


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5. Zeros. Cancellation.The point of this, and similar problems, is that we have to be cautious. In thepresent case,zD 0is not a zero of the given function because

z2 sin2 zD z2 .z/2 C D 2 C :

11. Zeros.Show that the assumption, in terms of a formula, is(A) f.z/ D .z z0/ng.z/ with g.z0/ 0;

so that

f .z0/ D 0; f0.z0/ D 0; ; f.n1/.z0/ D 0;as it should be for annth-order zero.

Show that (A) implies

h.z/ D f2 .z/ D .z z0/2ng2.z/;

so that, by successive product differentiation, the derivatives ofh.z/will be zero at z0as long as afactor ofz z0is present in each term. Ifn D 1, this happens forhand h0, giving a second-orderzeroz0 ofh. Ifn

D2, we have.z

z0/

4 and obtainf, f0 ,f00 ,f000 equal to zero atz0, giving afourth-order zeroz0ofh. And so on.

17. Singularities.We start with cot z: By definition,

cotzD 1tanz

D coszsin z

:

By definition on p. 715,cot z is singular where cot zis not analytic. This occurs wheresin zD 0,hence for

zD 0; ; 2 ;: : : D n, where n D 0 ; 1 ; 2 ; : : : :(B)Since cos z and sin z share no common zeros, we conclude that cot z is singular where sin z is 0; asgiven in (B). The zeros are simple poles.

Next we consider

cot4 zD cos

4 z

sin4 z

:

Nowsin4 zD 0for z as given in (B). But, since sin4 z is the sine function to the fourth power andsin z has simple zeros, the zeros ofsin4 zare of order4: Hence, by Theorem 4, p. 717, cot4 zhaspoles of order4 at (B).

But we are not finished yet. Inspired by Example 5, p. 718, we see that cos zalso has an essentialsingularity at 1. We claim that cos4 z also has an essential singularity at 1. To show this we wouldhave to develop the Maclaurin series ofcos4 z. One way to do this is to develop the first few terms ofthat series by (1), p. 690, of Sec. 15.4. We get (using calculus: product rule, chain rule)

cos4 wD 1 1

24w2 C 1

440w4 C :(C)

The odd powers are zero because in the derivation of (C) these terms contain sine terms (chain rule!)that are zero atw0D 0.

We setwD 1=z and multiply out the coefficients in (C):

cos4 1

zD 1 2z2 C 5

3z4 C :(D)

We see that the principal part of the Laurent series (D) is (D) without the constant term1. It isinfinite and thus cot4 z has an essential singularity at 1 by p. 718. Since multiplication of the series


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by1= sin4 z does not change the type of singularity, we conclude that cot4 zalso has an essentialsingularity at 1:

Sec. 16.3 Residue Integration Method

This section deals with evaluating complex integrals (1), p. 720, taken over a simple closed path C. Theimportant concept is that of a residue, which is the coefficient b1of a Laurent series that converges for allpoints near a singularityzD z0insideC, as explained on p. 720.Examples 1and 2 show how to evaluateintegrals that have only one singularity withinC.

A systematic study of residue integration requires us to consider simple poles (i.e., of order1) and polesof higher order. For simple poles, we use (3) or (4), on p. 721, to compute residues. This is shown inExample 3, p. 722, andProb. 5. The discussion extends to higher order poles (of orderm) and leads to(5), p. 722, andExample 4, p. 722.It is critical that you determine theorderof the poles insideCcorrectly. In many cases we can use Theorem 4, on p. 717 of Sec. 16.2, to determine m. However, whenh.z/in Theorem 4 is also zero at z0;the theorem cannot be applied. This is illustrated inProb. 3.

Having determined the residues correctly, it is fairly straightforward to use the residue theorem(Theorem 1, p. 723) to evaluate integrals (1), p. 720, as shown inExamples 5and 6, p. 724, andProb. 17.


3. Use of the Laurent series.The function

f.z/ D sin 2zz6

has a singularity atzD z0D 0:

However, since both sin 2z and z6 are0for z0D 0, we cannot use Theorem 4 of Sec. 16.2, p. 717, todetermine the order of that zero. Hence we cannot apply (5), p. 722, directly as we do not know thevalue ofm.

We develop the first few terms of the Laurent series for f .z/:From (14) in Sec. 15.4, p. 695, weknow that

sinwD w w3

3C w

5

5 w

7

7C :

We setwD 2z and getsin 2zD 2z .2z/

3

3 C .2z/

5

5 .2z/

7

7 C :(A)

Since we need the Laurent series ofsin 2z=z6 we multiply (A) byz6 and get

z6 sin 2zD z6

2z .2z/3

3 C .2z/

5

5 .2z/

7

7 C

(B)

D 2z5

83

1

z3C 32

5

1

z 128

7 z C :

The principal part of (B) is (see definition on p. 709)

2z5

83

1z3

C 325

1z

:

We see that

f.z/ D sin 2zz6

has a pole of fifth order at zD z0D 0 [by (2), p. 715].

Note that the pole off is only of fifth order and not of sixth order because sin 2z has a simple zeroatzD 0:


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Using the first line in the proof of (5), p. 722, we see that the coefficient ofz1 in the Laurentseries (C) is

b1D 32

5D 32

5 4 3 2 1D 4

15:

Hence the desired residue at0 is 415

:

Checking our result by(5), p. 722.Having determined that the order of the singularity atzD z0D 0is 5and that we have a pole of order5at z0D 0, we can use (5), p. 722, with m D 5. Wehave

ReszDz0D0

sin 2z

z6 D 1

.6 1/ limz!0

d61

dz 61

.z 0/6f.z/

D 15

limz!0

d5

dz 5

z6

sin 2z

z6

D 15

limz!0

d5

dz 5sin 2z

:

We needg.z/ D sin 2zI

g0.z/ D 2 cos2zIg00.z/ D 4 sin 2zI

g000.z/ D 8 cos2zIg.4/.z/ D 16 sin 2zIg.5/.z/ D 32 cos2z:

Then

limz

!0f32 cos2zg D 32 lim

z

!0fcos2zg D 32 1:

Hence

Resz D z0 D 0

sin 2z

z6 D 1

5 32 1 D 4

15; as before.

Remark.In certain problems, developing a few terms of the Laurent series may be easier than using(5), p. 722, if the differentiation is labor intensive such as requiring several applications of thequotient rule of calculus (see p. 623, of Sec. 13.3).

5. Residues. Use of formulas (3) and (4), p. 721.Step 1. Find the singularities off.z/. From

f.z/ D 81

Cz2

we see that 1 C z2 D 0 implies z2 D 1, hencezD i andzD i:

Hence we have singularities atz0D i andz0D i:Step 2. Determine the order of the singularities and determine whether they are poles.Since thenumerator off is8 D h.z/ 0(in Theorem 4), we see that the singularities in step 1 are simple,i.e., of order1. Furthermore, by Theorem 4, p. 717, we have two poles of order1 at i and i ,respectively.


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Step 3. Compute the value of the residues.We can do this in two ways.

Solution 1. By(3), p. 721, we have

Resz

Di

f.z/ D limz

!i

.z i / 8

1

Cz2

D limz!i

.z i / 8

.z i/.z C i /

D limz!i

8

z i

D 82i

D 4iD 4i :

Also

ReszDi

f.z/ D limz!i

.z .i// 8

.z i/.z C i/

D limz!i

.z C i/ 8

.z i/.z C i/

D 82i D 4i :

Hence the two residues are

ReszDi

f.z/ D 4i and Resz Di

f.z/ D 4i :

Solution 2. By(4), p. 721, we have

Resz = z0

f.z/ D Resz= z0

p.z/

q.z/D p.z0/

q0.z0/D 8

.1 C z2/0

z= z0

D 82z

z= z0

D 82z0

:

Forz0D i we have

Resz0Di

f.z/ D 82i

D 4i ;

and forz0D i

Resz0Di

f.z/ D 82i D 4i ;

as before.

15. Residue theorem.We note that

f.z/ Dtan

2 zD sin 2z

cos2z

is singular where cos 2zD 0:This occurs at

2zD 2

; 32

; 52 ;

and hence at

(A) zD 14

; 34

; 54

:


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Since sin 2z 0at these points, we can use Theorem 4, p. 717, to conclude that we have infinitelymany poles at (A).

Consider the path of integrationCW jz 0:2j D 0:2. It is a circle in the complex plane withcenter0:2and radius0:2:We need to be only concerned with those poles that lie insideC: There isonly one pole of interest, that is,

zD 14D 0:25 (i.e.,j0:25 0:2j D 0:05 < 0:2):

We use (4) p. 721, to evaluate the residue off atz0D 14 :We have

p.z/ D sin 2z; p

1

4

D sin

2D 1;

q.z/ D cos2z; q0 .z/ D 2 sin 2z (chain rule!); q0 14

D 2 sin 2 14D 2:

Hence

Resz0 =

14

f.z/ D p.14

/

q0. 14

/D 12D

1

2:

Thus, by (6) of Theorem 1, p. 723,

IC

f.z/dzDI

CWjz0:2jD0:2Wtan2z dz

D 2 i Resz0 =

14

f.z/

D 2 i 1

2

D i:

17. Residue integration.We use the same approach as in Prob. 15. We note that

coszD 0 at zD 2

; 32

; 52 :

Alsoez is entire, see p. 631 of Sec. 13.5.From Theorem 4, p. 717, we conclude that we have infinitely many simple poles at

zD 2

; 32

; 52 .

Here the closed path is a circle:

CWz

i

2

D 4:5 and only zD

2 and zD

2 lie withinC:

This can be seen because for

zD 2W

2 i

2

D

2 i

2

D

r2

2C

2

2Dr

2 2

4 D

p2

2 D 2:2214 < 4:5:

Same for zD =2:Hence, by (4) p. 721,

Resz = =2

f.z/ D e=2

sin=2D e=2;


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and

Resz=

2

f.z/ D e=2

sin .=2/D e=2

sin=2D e=2:

Using (6), p. 723,IC

f .z/ dzDI

CWjz i=2jD4:5

ez

coszd z

D 2 i

Resz = =2

f.z/ C Resz = =2

f.z/

D 2 ie=2 C e=2

D 2 i

2 sinh

2

[by (17), p. A65, App. 3]

D 2 i2 sinh

2

[since sinh is an odd function]

D 4 i sinh 2

D 28:919i:

Sec. 16.4 Residue Integration of Real Integrals

It is surprising that residue integration, a method ofcomplexanalysis, can also be used to evaluate certainkinds of complicatedrealintegrals. The key ideas in this section are as follows. To apply residueintegration, we need a closed path, that is, a contour. Take a look at the different real integrals in thetextbook, pp. 725732. For real integrals (1), p. 726, we obtain a contour by the transformation (2), p. 726.This is illustrated in Example 1 and Prob. 7.

For real integrals (4), p. 726, and (10), p. 729 (real Fourier integrals), we start from a finite intervalfrom Rto R on the real axis (the x-axis) and close it in complex by a semicircle S as shown in Fig. 374,p. 727. Then we blow up this contour and make an assumption (degree of the denominator

degree of

the numerator C2) under which the integral over the blown-up semicircle will be 0. Note thatwe only taketho

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