complex chapter 7 final[1]

8/14/2019 Complex Chapter 7 Final[1]

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In the last chapter we introduced the notion of complex integration. An important part of

our development was the statement of Cauchys integral formula. In this chapter were

going to extend this technique using reside theory. This is an elegant formulation that notonly allows you to calculate many complex integrals, but also gives you a trick you can

use to calculate many real integrals. We begin by stating some theorems related toCauchys integral formula.

Theorems Related To Cauchys Integral Formula

We begin the chapter by writing down another form of Cauchys integral formula. First

lets write (6.16) in the following way:

1

2

f z f a dz

i z a Now lets take the derivative of this expression, with respect to a. This gives:

2

1 1 1

2 2 2

f z f z f z d d f a dz dz dz

da i z a i da z a i z a

We can repeat this process multiple times. That is, take the derivative again. Each time

the exponent, which is negative, cancels out the minus sign we pick up by computing the

derivative with respect to a ofz a

. For example the second derivative is:

2 3

1 1

2

f z f z d f a dz dz

da i i z a z a

This process can be continued. For an arbitrary n, we obtain a second Cauchy Integral

formula for the nth derivative of f a :

1

!

2

n

n

f zn f a dz

i z a

for 1,2,3,...n . There are two facts you should come away with from the CauchyIntegral formulas:

If a function f z is known on a simple closed curve , then that function isknown at all points inside . Moreover, all of the functions derivatives can befound inside .


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If a function is analytic in a simply connected region of the complex plane, and

hence has a first derivative, all of its higher derivatives exist in that simply

connected region.

Now we turn to a statement known as Cauchys inequality. This statement is related to ,

which gives us an expression we can use to calculate the derivative of an analyticfunction in a simply connected region. Consider a circle of radius rwhich has the point

z a at its center, and suppose that f z is analytic on the circle and inside the circle.

Let Mbe a positive constant such that f z M in the region z a r . Then:

!n

n

Mnf a

r

The next theorem, which is due to Liouville, tells us that an entire function cannot be

bounded unless it is a constant. This statement is calledLiouvilles theorembut it was first

proved by Cauchy. So maybe we should call it the Cauchy-Liouville theorem. In anycase, it simply says that if f z is analytic and bounded in the entire complex plane, i.e. t

f z M for some constant M, then f z is a constant.

Liouvilles theorem implies theFundamental Theorem of Algebra. Consider a polynomial

with degree 1n and coefficient 0na :

20 1 2n

nP z a a z a z a z L

The fundamental theorem of algebra tells us that every polynomial P z has at least oneroot. The proof follows from Liouvilles theorem and a proof by contradiction. Suppose

that instead 0P z for allz. Then

1f z

P z

is analytic throughout the complex plane and is bounded outside some circle z r .

Moreover, the assumption that 0P z implies that 1/f P is also bounded for z r .

Hence 1/ P z is bounded in the entire complex plane. Using Liouvilles theorem,

1/ P z must be a constant. This is a contradiction, since

20 1 2n

nP z a a z a z a z L is clearly not constant. Therefore P z must have at

least one root such that 20 1 2 0n

nP z a a z a z a z L is satisfied.


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Next we state the Maximum modulus theorem and the Minimum modulus theorem. The

maximum modulus theorem tells us the following. Let f z be a complex valued

function which is analytic inside and on a simple closed curve . If f z is not a

constant, then the maximum value of f z is found on the curve .

Now we state the minimum modulus theorem. Assume once again that f z is a complex

valued function which is analytic inside and on a simple closed curve . If 0f z

inside , then f z assumes its minimum value on the curve .

The next theorem is theDeformation of path theorem. Consider a domainD in the

complex plane, and two curves inD we call 1 2and . We suppose that

1 2is larger than or lies outside of , and that 1can be deformed into 2 without leaving

the domainD (that is, we can shrink the first curve down to the second one without

crossing any holes or discontinuities in the domain-see Figure 7-1). If f z is analytic in

D then

1 2

f z dz f z dz

i i

Figure 7-1: A graphic illustration of the deformation of path theorem.

Next, we state Gauss mean value theorem. Consider a circle of radius rcentered at the

point a. Let f z be a function which is analytic on and inside . The mean value of

f z on is given by f a :

1

2

Can have hold insidesecond curve.

If first curve has to

cross hole to deform

into second curve,theorem does not

work.


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2

0

1

2

i f a f a re d

Once again, let f z be a function which is analytic on and inside a simple, closed curve

. Now assume that f z has a finite number of poles inside . IfMis the number of

zeros of f z inside andNis the number of poles inside , the argument theorem

states that:

1

2

f zdz M N

i f z

Next is a statement ofRouches theorem. Let and f z g z be two functions which are

analytic inside and on a simple closed curve

. If ong z f z , then andf z g z f z have the same number of zeros inside .

Finally, we end our whirlwind tour of theorems and results related to the Cauchy integral

formula with a statement of Poissons integral formula for a circle. This expresses thevalue of a harmonic function inside of a circle in terms of its values on the boundary. Let

f z be analytic inside and on the circle , centered at the origin with radiusR. Suppose

that i z re is any point inside . Then

2 2

2

2 20

Re1

2 2 cos

i R r f

f z d R Rr r

This example illustrates the solution of Laplaces equation on a disk. First, show that

01

, cos sinn nn nn

u r a a r n b r n

is the solution of Laplaces equation on the disc 0 1r with Dirichlet boundaryconditions:

2

2 2

1 10, 0 1,0 2

1, , , bounded as 0

u ur r

r r r r r

u f u r r


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Show the coefficients in the series expansion are given by:

2 2 2

00 0 0

1 1 1, cos , sin

2n na f d a f n d b f n d

Use the result to deduce Poissons integral formula for a circle of radius 1:

2

2

20

1 1,

2 1 2 cos

ru r f d

r r

We try separation of variables. Let ,u r R r . Then it follows that:

2 2 2 2

2 2 2 2, ,

u R u R uR r

r r r r

The statement of the problem tells us that:

2 2

2 2 2

1 10

u u u

r r r r

Hence:

2 2

2 2 2

1 10

R RR r

r r r r

We divide every term in this expression by ,u r R r . This allows us to write:

2 2 2

2 2

1r R r R

R r R r

The left hand side and the right hand side are functions ofronly and only respectively.Therefore they can be equal only if they are both equal to a constant. We call this constant

2n . Then we have the equation in :

2 2

2 2

2 2

1

, 0

d d

n nd d

Note that partial derivatives can be replaced by ordinary derivatives at this point, since

each equation involves one variable only. This familiar differential equation has solution

given by:

cos sinn na n b n


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Now, turning to the equation in r, we have:

2 2 2

2 2 2

2 2, 0

r d R r dR d R dRn r r n R

R dr R dr dr dr

You should also be familiar with this equation from the study of ordinary differential

equations. It has solution:

n nn nR r c r c r

The total solution, by assumption is the product of both solutions, i.e.

,u r R r . So we have:

, cos sinn nn n n nu r c r c r a n b n

The condition that ,u r is bounded as 0r imposes a requirement that the constant0nc since:

as 0nn

cr

r

Therefore, we take , cos sinnn n nu r c r a n b n . We can just absorb the

constant nc into the other constants, and still designate them by the same letters. Then:

, cos sinn n nu r r a n b n

The most general solution is a superposition of such solutions which ranges over all

possible values ofn. Therefore we write:

00 1

, cos sin cos sinn nn n n nn n

u r r a n b n a r a n b n

To proceed, the following orthogonality integrals are useful:

2

0

for 0sin sin

0 for 0

mn nm n d

n

2

0

for 0cos cos

2 for 0

mn

mn

nm n d

n


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2

0

sin cos 0m n d

Here, 1 for ,0mn m n otherwise is theKronecker delta function.Now we apply the

boundary condition 1,u f for 0 2 :

01

cos sinn n nn

f a r a n b n

Multiply through this expression by sin m and integrate. We obtain:

2 2 2 2

00 0 0 0

1

2

01 1

sin sin cos sin sin sin

sin sin

n n

n

n n mn m

n n

f m d a m d a n m d b n m d

b n m d b b

where - were used. We conclude that:

2

0

1sinnb f n d

Now we return to , and multiply by cos m and integrate. This time:

2 2 2 2

00 0 0 0

1

2

01 1

cos cos cos cos sin cos

sin cos

n n

n

n n mn m

n n

f m d a m d a n m d b n m d

a n m d a a

Hence:

2

0

1cosma f m d

To obtain the constant 0a , we integrate without first multiplying by any trig functions, i.e.


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2 2 2 2

00 0 0 0

1

0

2

00

cos sin

2

1

2

n n

n

f d a d a n d b n d

a

a f d

This should be obvious since:

2

0

2

0

21cos sin 0

0

21sin cos 00

n d nn

n d nn

Now we are in a position to derive Poissons formula. We have:

2 2 2

0 0 01

1 1 1, cos cos sin sin

2

n

n

u r f d r f n d n f n d n

We can move the summation inside the integrals:

2 2 2

0 0 01 1

2

01 1

2

01

1 1 1, cos cos sin sin

2

11 2 cos cos 2 sin sin

2

11 2 cos cos sin sin

2

n n

n n

n n

n n

n

n

u r f d f r n n d f r n n d

d f r n n r n n

d f r n n n n

Now recall that:

cos cos sin sin cosn n n n n

Its also true that:


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2

2

1

11 2 cos

1 2 cos

n

n

rr n

r r

So we arrive at the Poisson formula for a disc of radius 1:

2

2

20

1 1,

2 1 2 cos

ru r f d

r r

This tells us that the value of a harmonic function at a point inside the circle is the

average of the boundary values of the circle.

The Cauchy Integral Formula as a Sampling Function

TheDirac delta function has two important properties. First if we integrate over theentire real line then the result is unity:

1 x dx

Second, it acts as asamplingfunction-that is, it picks out the value of a real function

f x at a point:

f x x a dx f a

In complex analysis, the function 1/ zplays an analogous role. It has a singularity at0z , and:

0 if 0 is not in the interior of1

1 if 0 is in the interior of2

dz

i z

It also acts as a sampling functionfor analytic functions f z in that:

1

2

f z dz f a

i z a

Some Properties of Analytic Functions

Now we are going to lay some more groundwork before we state the residue theorem. In

this section, we consider some properties of analytic functions.


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Suppose that a function f z is analytic inside a disc centered at a point a of radius r:

z a r . Then f z has a power series expansion given by:

0

n

nn

f z a z a

The coefficients of the expansion can be calculated using the Cauchy integral formula :

!

n

n

f aa

n

Note the following result:

0 if 1ln if 1

m m z a dz z a m

Hence:

0

n

n

n

f z dz a z a dz

since n is never equal to -1.

f(z)Consider the punctured disc of radius rcentered at the point a. We denote this by writing

0 z a r . If f z is analytic in this region, it is analytic inside the disc but not at thepoint a. In this case, the function has a Laurent expansion:

n

n

n

f z a z a

As stated in chapter 5, we can classify the points at which the function blows up or goes

to zero. A removable singularity is a point a at which the function appears to be

undefined, but it can be shown by writing down the Laurent expansion that in fact thefunction is analytic at a. In this case the Laurent expansion assumes the form

n

n

n k

f z a z a

where 0k . Then it turns out the point z a is azero of orderk.


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On the other hand, suppose that the series expansion retains terms with 0n :

n

n

n k

f z a z a

Then we say that the point z a is apole of orderk. Simply put, a pole is a point that

behaves like the point 0z for 1

g zz

. That is, as z a , then f z . A function

might have multiple poles. For example, in Figure 7-2 we illustrate the poles of the

modulus of the gamma function, z , which are points where the function blows up.

Figure 7-2: When the real part ofzis negative, the modulus of the gamma function blowsto infinity at several points. These points are the poles of the function.

A Laurent series expansion of this type can be split into two parts:

1

0

n n n

n n n

n k n k n

f z a z a a z a a z a F G

The second series, which we have denoted by G, looks like a plain old Taylor expansion.

The other series, which we have denoted byF, is called theprincipal partand it includesthe singularities (the real ones-the poles) of the function.

Is the point 0z a removable singularity of sin z

f zz

?


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At first glance, the behavior of the function at 0z cant really be determined. To seewhats going on we expand the sin function in Taylor:

3 5

2 4

sin 1 1 1

3! 5!

1 11

3! 5!

z f z z z z

z z

z z

L

L

From this expression, its easy to see that:

2 40 0 0

sin 1 1lim lim lim1 1

3! 5! z z z

z f z z z

z L

Therefore, the point 0z is a zero of order 1.

Describe the nature of the singularities of ze

f zz

.

We follow the same procedure used in Example 7-2. First expand in Taylor:

2 3 21 1

1 12! 3! 2 6

ze z z z z f z z

z z z

L L

The principal part of this series expansion is given by:

1

z

It follows that the point 0z is a pole of order 1.

Is the point 0z a removable singularity of 4

sin zf z

z ?

Contrast this solution with that found in Example 7-2. Expanding in Taylor we find:


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3 5 74 4

3

3

sin 1 1 1 1

3! 5! 7!

1 1 1 1

6 5! 7!

z f z z z z z

z z

z z

z z

L

L

This time, the singularity cannot be removed. So the point 0z is a pole. The principalpart in this series expansion is:

3

1 1

6z z

The leading power (most negative power) in the expansion gives the order of the pole.Hence 0z is a pole of order 3.

Next we consider the essential singularity. In this case the Laurent series expansion of the

function includes a principal part that is non-terminating. That is, all terms out to minusinfinity are included in the Laurent expansion with negative n, i.e. there are no non-zero

terms in the expansion for 0n :

n

n

n

f z a z a

Describe the nature of the singularity at 0z for 1

z f z e .

This function is the classic example used to illustrate an essential singularity. We justwrite down the series expansion:

1

2 3 4 5

1 2 3 4 5

1 1 1 1 1 1 1 1 1

1 2 6 4! 5!

1 1 1 11

2 6 4! 5!

z f z e

z z z z z

z z z z z

L

L

This series expansion has a non-terminating principal part. Therefore 0z is an essentialsingularity.


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The Residue Theorem

Now were in a position where we can describe one of the central results of complex

analysis, the residue theorem. We consider a function f z in a region enclosed by a

curve that includes isolated singularities at the points 1 2, ,..., k z z z . The function is

analytic everywhere on the curve and inside it except at the singularities. This is

illustrated in Figure 7-3.

Figure 7-3: A function f z is analytic in a certain region enclosed by a curve, except at

a set of isolated singularities.

We can use the deformation of path theorem to shrink the curve down. In fact, we canshrink it down into isolated curves enclosing each singularity. This is shown in figure 7-4.

0z 1

z

kz

0z 1

z

kz

1

2

k


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Figure 7-4: If the region is simply connected, we can apply the deformation of path

theorem to shrink the curve down, until we have circles around each isolated singularpoint.

After application of the deformation of path theorem, the integral is broken up into a sumof integrals about each singular point:

1

j

k

j

f z dz f z dz

This expression can be written in terms of the Laurent expansion. Note that there will bea series expansion (which is local) about each singular point:

1 2 j j j

n n j j j

n j n j

n n

f z dz a z z dz a z z dz a i

We call the coefficient in the expansion 1ja the residue. Summing over all of the integrals

for each singular point, we get the residue theorem. This states that the integral is

proportional to the sum of the residues:

1

2k

j

f z dz i residues

Residues are computed by finding the limit of the function f z as zapproaches each

singularity. This is done for a singularity at z a as follows:

1

1

1lim

1 !

kk

kz a

dresidue z a f z

k dz

where kis the order of the singularity.

Compute the integral

5 2

2

z

dzz z

where is a circle of radius 3r centered at the origin.

The singularities of this function are readily identified to be located at 0,2z . Bothsingularities are enclosed by the curve, since 3z in both cases. To find each residue,


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we compute the limit of the function for each singularity. The residue corresponding to

is:

0 05 2 5 2 2

lim lim 12 2 2z z

z zz

z z z

The residue corresponding to the singularity at 2z is:

2 25 2 5 2 8

lim 2 lim 42 2z z

z zz

z z z

Therefore using the integral evaluates to:

5 22 2 1 4 10

2

zdz i residues i i

z z

Compute the integral of 3cosh z

dzz

where is the unit circle centered about the origin.

The function has a singularity at 0z of 3d order. Using:

1

!

2

n

n

f zn f a dz

i z a

We have:

2

0 03 2

cosh 2cosh cosh

2!z z

z i d dz z i z i

z dz

Evaluation of Real, Definite Integrals

One of the most powerful applications of the residue theorem is in the evaluation of

definite integrals of functions of a real variable. We start by considering:

2

0

cos , sinf d

Now write the complex variablezin polar form on the unit circle, that is let iz e .Notice that:


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1,i

idz ie d d dz dz

iz z

As increases from 0 to 2 , one sees that the complex variablezmoves around the unit

circle in a counter clockwise direction. Using Eulers formula, we can also rewrite cosand sin in terms of complex variables. In the first case:

2 2 21 1 1cos

2 2 2 2

i i i ii

i

e e e e z e

e z

Similarly, we find that:

2 1sin

2

z

iz

Taking these facts together, we see that 2

0

cos , sinf d

can be rewritten as acontour integral in the complex plane. We only need to include residues that are inside theunit circle.

Compute2

0 24 8cos

d

.

Usingi

d dzz

together with2 1

cos2

z

z

we have:

2

20

2

2

2

24 8 cos 124 8

2

24 4 4

4 24 4

4 6 1

d dzi

zz

z

dzi

z z

dzi

z z

i dz

z z

i

i

i

i


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We will choose the unit circle for our contour. To find the singularities, we find the roots

of the denominator. Some algebra shows that they are located at:

3 2 2z

The first residue is given by:

3 2 2

1 1lim 3 2 2

4 23 2 2 3 2 2zz

z z

The residue corresponding to 3 2 2z is given by:

3 2 2

1 1lim 3 2 2

4 23 2 2 3 2 2zz

z z

You should always check that your singularities lie inside the curve you are using to

integrate. If they do not, they do not contribute to the integral. In this case both residues

do not contribute. This is because

3 2 2 1z

lies outside the unit circle. So we will only include the second residue, because the

singularity it corresponds to, 3 2 2 1z and so is inside the unit circle.

Using we have:

2

12 2

6 1 4 2 2 2

dz ii residues i

z z

i

Hence:

2

20 24 8cos 4 6 1 4 2 2 8 2

d i dz i i

z z

i

The next type of definite integral we consider is one of the form:

cos

sin

mx f x dx

mx

This type of integral can be converted into a contour integral of the form:


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imz f z e dz

To obtain the desired result, we take the real or imaginary part of depending on whetheror not a cos orsin function is found in the original integral. A useful tool when evaluating

integrals of the form is calledJordans Lemma. Imagine that we choose to be a semi-circle located at the origin and in the upper half plane, as illustrated in Figure 7-3.

Figure 7-3: A semi-circle in the upper half plane, of radiusR.

Jordans lemma states that:

1

lim 0mzR

C

f z e dz

Jordans lemma does not hold in all cases. To use , if 0m then it must be the case that 0f z as R . We can also apply it in the following case:

1

lim 0R

C

f z dz

provided that 0f z faster than 1/zas R .

Compute2

cos kxdx

x

.

We can compute this integral by computing:
http://upload.wikimedia.org/wikipedia/commons/f/fb/Jordan%27s_Lemma.svg


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2

2

cosRe ,

Where:

z

ikz

z

kxdx I

x

e I P dz

z

ThePstands forprincipal part. To do the integral, we will take a circular contour in the

upper half plane which omits the origin. This is illustrated in Figure 7-4.

Figure 7-4: We use a semicircular contour in the upper half plane, omitting the origin

using a small semicircle or radius rthat gives us a curve that omits the origin.

Now we can write out the integral piecewise, taking little chunks along the curves

1 2andC C . Note that when directly on the real axis, we set z x . This gives:

2 1

2 2 2 2 20

ikz ikx ikz ikx ikz r R

R rC C

e e e e edz dx dz dx dz

z x z x z

The entire sum of these integrals equals zero because the contour encloses no singularpoints. However, individual integrals in this expression are not all zero. By Jordans

lemma:

1

20

ikz

C

edz

z

So, we only need to calculate the residues for the curve 2C . This curve is in the clockwise

direction, so we need to add a minus sign when we do our calculation. Also, up to this


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point, we have been using full circles in our calculations. The curve in this case is a semi-

circle, so is written as:

1

k

j

f z dz i residues

The singularity at 0z is inside the curve 2C , of order 2. The residue corresponding tothis singularity is:

2

02

0

ikzikz

z

z

d e z ike ik

dz z

Therefore:

2

2

ikz

C

edz i ik k

z

Now,

1

1

2 2 2 2

2 2 2

2 2

0,

, as 0,

0

ikz ikx ikx ikz r R

R rC

ikx ikx ikxr R

R r

ikx ikz

C

e e e edz dx dx dz

z x x z

e e e P dx dx dx r R

x x x

e e P dx dz

x z

Therefore we find that:

1

2 2 2

cosRe

ikx ikz

z

C

kx e edx I P dx dz k

x x z

Integral of a Rational Function

The integral of a rational function f x :

f x dx


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can be calculated by computing:

f z dz

using the contour shown in Figure 7-3, which consists of a line along thex axis from RtoR and a semicircle above thex axis the same radius. Then we take the limit R .

Consider the Poisson kernel:

2 2

1y

yp x

x y

Treatingy as a constant, use the residue theorem to show that its Fourier transform is

given by 12

k ye

.

The Fourier transform of a function f x is given by the integral:

1

2

ikx F k f x e dx

So, we are being asked to evaluate the integral:

2 21 12

ikxy I e dxx y

We do this by considering the contour integral:

2 2 2

1

2

ikzy e dzz y

First, note that:

2 2 2 2

1 1

2 2

y y

x y z iy z iy

Therefore, there are two simple poles located at z iy . These lie directly on they axis,one in the upper half plane and one in the lower half plane. To get the right answer for the

integral we seek, we need to compute using both cases. First we consider the pole in the

upper half plane. The residue corresponding to z iy is:


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1 2 2 2

1

2 2 2 4

ikz kyky

z iy

ye yea z iy e

z iy z iy iy i

Applying the residue theorem, we find that:

2 2 2

1 1 12

2 4 2

kyikx kyy e I e dx i e

x y i

However, now lets consider enclosing the other singularity, which would be an equally

valid approach. The singularity is located at z iy , which is below thex axis, so wewould need to use a semicircle in the lower half plane to enclose it. This time the residue

is:

1 2 2

1

2 4

ikzky

z iy

yea z iy e

z iy z iy i

Using this result, we obtain:

2 2 2

1 1 12

2 4 2

kyikx kyy e I e dx i e

x y i

Combining both results gives the correct answer, which is:

2

k ye

I

Compute the integral given by:

2

41

x I dx

x

This integral is given by:

2 residues in upper half planeI i

We find the residues by considering the complex function:

2

41

zf z

z

The singularities are found by solving the equation:


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4 1 0z

This equation is solved by 1/4

1z . But, remember that 1 ie .That is, there are four roots given by:

/4

3 /4

1/ 4 2 1

5 /4

7 /4

i

ii n

i

i

e

ez e

e

e

These are shown in Figure 7-5. Notice that two of the roots are in the upper half plane,

while two of the roots are in the lower half plane. We rejectthe roots in the lower halfplane because we are choosing a closed semicircle in the upper half plane (as in Figure 7-

3) as our contour. We only consider the singularities that are inside the contour, the othersdo not contribute to the integral.

Figure 7-5: An illustration of the roots in Example 7-11. For our contour, we will enclose

the upper half plane-so we ignore the roots that lie in the lower half plane.

We proceed to compute the two residues. They are all simple so in the first case we have:

/ 4ie 3 / 4ie

5 / 4ie 7 / 4ie


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/4

/4

/4 2

/ 4 3 / 4 5 / 4 7 / 4

2

3 / 4 5 / 4 7 / 4

/ 4 3 / 4 / 4 5 / 4 / 4 7 / 4

3 /4

/ 4 / 2 3 / 2

lim

lim

1

42

i

i

i

i i i iz e

i i iz e

i i i i i i

i

i i i

z e z

z e z e z e z e

z

z e z e z e

i

e e e e e e

ie

e e e

And so:

A similar calculation shows that the residue corresponding to the pole at

3 / 4 / 41

4

i i z e e . Hence:

/ 4 3 / 41 1

4 4

1cos / 4 sin / 4 cos3 / 4 sin 3 / 4

4

1 1 11 1 2

4 2 4 2 2 2

i iresidues e e

i i

ii i i

Therefore the integral evaluates to:

2 residues in upper half plane 22 2 2

i I i i

Compute2

cos

2 2

xdx

x x

We can compute this integral by considering:


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2 2 2 1 1

iz iz e e dz dz

z z z i z i

The root 1z i lies in the upper half plane, while the root 1z i lies in the lower halfplane. We choose a contour which is a semicircle in the upper half plane, enclosing the

first root. This is illustrated in Figure 7-6.

Figure 7-6: The contour used in Example 7-12.

The residue is given by:

1

1

1

lim 1 1 1

lim1

2

iz

z i

iz

z i

i

e

z i z i z i

e

z i

e e

i

Therefore we have:

1

22

2 2 2

iz iie dz e ei e

z z i e

But, using Eulers identity, we have:


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cos1 sin1ie i

And so:

2 cos 1 sin 12 2

ize dz

i z z e e

Now, we have:

1

1

2 2

2 2 2

2 2 2 2

cos sin

2 2 2 2 2 2

cos 1 sin 1

ix iz R

RC

izR R

R RC

e dx e dz

x x z z

xdx xdx e dz i

x x x x z z

ie e

Now we let R . By Jordans lemma:

1

20

2 2

iz

C

e dz

z z

So we have:

2 2

cos sincos 1 sin 1

2 2 2 2

xdx xdxi i

x x x x e e

Equating real and imaginary parts gives the result we are looking for:

2

coscos 1

2 2

xdx

x x e

Chapter SummaryBy computing the Laurent expansion of an analytic function in a region containing one or

more singularities, we were able to arrive at the residue theorem which can be used tocalculate a wide variety of integrals. This includes integrals of complex functions, but the

residue theorem can also be used to calculate certain classes of integrals involving

functions of a real variable.


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Chapter Quiz

1. Compute 3sinh z

dzz

2. Compute 4sinh z

dzz

3. Find the principal part of

23

1

1f z

z

4. What are the singular points and residues of

sin

5

2

z

z z

?

5. What are the singularities and residues of 2sin z

z z ?

6. Evaluate2

0 24 6sin

d

7. Using the technique outlined in Example 7-9, compute2

20

sin xdx

x

8. Use the residue theorem to compute2

lim1

R

RR

dx

x

9. Compute

2

2 2

3

1 4

xdx

x x

10. Compute2

cos

1

xdx

x

complex chapter 7 final[1]

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