The Residue Theorem · 2010-10-04 · 5.We will prove the requisite theorem (the Residue Theorem)...

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Singularities Residues Residue Theorem Residue at Infinity Types of Isolated Singularities Residues at Poles

The Residue Theorem

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Residue Theorem

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Introduction

1. The Cauchy-Goursat Theorem says that if a function is analyticon and in a closed contour C, then the integral over the closedcontour is zero.

2. But what if the function is not analytic?3. We will avoid situations where the function “blows up” (goes to

infinity) on the contour. So we will not need to generalizecontour integrals to “improper contour integrals”.

4. But the situation in which the function is not analytic inside thecontour turns out to be quite interesting.

5. We will prove the requisite theorem (the Residue Theorem) inthis presentation and we will also lay the abstract groundwork.

6. We will then spend an extensive amount of time with examplesthat show how widely applicable the Residue Theorem is.


The Residue Theorem

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Introduction1. The Cauchy-Goursat Theorem says that if a function is analytic

on and in a closed contour C, then the integral over the closedcontour is zero.







The Residue Theorem

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2. But what if the function is not analytic?

3. We will avoid situations where the function “blows up” (goes toinfinity) on the contour. So we will not need to generalizecontour integrals to “improper contour integrals”.





The Residue Theorem

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infinity) on the contour.

So we will not need to generalizecontour integrals to “improper contour integrals”.





The Residue Theorem

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The Residue Theorem

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5. We will prove the requisite theorem

(the Residue Theorem) inthis presentation and we will also lay the abstract groundwork.



The Residue Theorem

logo1







5. We will prove the requisite theorem (the Residue Theorem)

inthis presentation and we will also lay the abstract groundwork.



The Residue Theorem

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The Residue Theorem

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Example.

The function f (z) =1

(z−1)(1+ z2)is not analytic at

z = 1, i,−i.

-

6ℑ(z)

ℜ(z)1−1

i

−i

rr

r


The Residue Theorem

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Example. The function f (z) =1


z = 1, i,−i.

-

6ℑ(z)

ℜ(z)1−1

i

−i

rr

r


The Residue Theorem

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z = 1, i,−i.

-

6ℑ(z)

ℜ(z)

1−1

i

−i

rr

r


The Residue Theorem

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z = 1, i,−i.

-

6ℑ(z)

ℜ(z)1

−1

i

−i

rr

r


The Residue Theorem

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z = 1, i,−i.

-

6ℑ(z)

ℜ(z)1−1

i

−i

rr

r


The Residue Theorem

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z = 1, i,−i.

-

6ℑ(z)

ℜ(z)1−1

i

−i

r

r

r


The Residue Theorem

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z = 1, i,−i.

-

6ℑ(z)

ℜ(z)1−1

i

−i

rr

r


The Residue Theorem

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Example.


sin(

π

z

) is not analytic at

z = 1,12,13,14, . . . and at 0 and at z =−1,−1

2,−1

3,−1

4, . . ..

-

6ℑ(z)

ℜ(z)1−1

i

−i

rrr· · ·rr r r· · ·


The Residue Theorem

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sin(

π

z


z = 1,12,13,14, . . . and at 0

and at z =−1,−12,−1

3,−1

4, . . ..

-

6ℑ(z)

ℜ(z)1−1

i

−i

rrr· · ·rr r r· · ·


The Residue Theorem

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sin(

π

z


z = 1,12,13,14, . . . and at 0 and at z =−1,−1

2,−1

3,−1

4, . . ..

-

6ℑ(z)

ℜ(z)1−1

i

−i

rrr· · ·rr r r· · ·


The Residue Theorem

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sin(

π

z


z = 1,12,13,14, . . . and at 0 and at z =−1,−1

2,−1

3,−1

4, . . ..

-

6ℑ(z)

ℜ(z)1−1

i

−i

r

rr· · ·rr r r· · ·


The Residue Theorem

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sin(

π

z


z = 1,12,13,14, . . . and at 0 and at z =−1,−1

2,−1

3,−1

4, . . ..

-

6ℑ(z)

ℜ(z)1−1

i

−i

rr

r· · ·rr r r· · ·


The Residue Theorem

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sin(

π

z


z = 1,12,13,14, . . . and at 0 and at z =−1,−1

2,−1

3,−1

4, . . ..

-

6ℑ(z)

ℜ(z)1−1

i

−i

rrr

· · ·rr r r· · ·


The Residue Theorem

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sin(

π

z


z = 1,12,13,14, . . . and at 0 and at z =−1,−1

2,−1

3,−1

4, . . ..

-

6ℑ(z)

ℜ(z)1−1

i

−i

rrr· · ·

rr r r· · ·


The Residue Theorem

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sin(

π

z


z = 1,12,13,14, . . . and at 0 and at z =−1,−1

2,−1

3,−1

4, . . ..

-

6ℑ(z)

ℜ(z)1−1

i

−i

rrr· · ·r

r r r· · ·


The Residue Theorem

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sin(

π

z


z = 1,12,13,14, . . . and at 0 and at z =−1,−1

2,−1

3,−1

4, . . ..

-

6ℑ(z)

ℜ(z)1−1

i

−i

rrr· · ·rr

r r· · ·


The Residue Theorem

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sin(

π

z


z = 1,12,13,14, . . . and at 0 and at z =−1,−1

2,−1

3,−1

4, . . ..

-

6ℑ(z)

ℜ(z)1−1

i

−i

rrr· · ·rr r

r· · ·


The Residue Theorem

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sin(

π

z


z = 1,12,13,14, . . . and at 0 and at z =−1,−1

2,−1

3,−1

4, . . ..

-

6ℑ(z)

ℜ(z)1−1

i

−i

rrr· · ·rr r r

· · ·


The Residue Theorem

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sin(

π

z


z = 1,12,13,14, . . . and at 0 and at z =−1,−1

2,−1

3,−1

4, . . ..

-

6ℑ(z)

ℜ(z)1−1

i

−i

rrr· · ·rr r r· · ·


The Residue Theorem

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Definition.

If the function f is analytic for 0 < |z− z0|< R, then z0 iscalled an isolated singularity of f .

Notes.1. z = 1, i,−i are isolated singularities of f (z) =

1(z−1)(1+ z2)

.

2. z = 1,12,13,14, . . . are isolated singularities of f (z) =

1sin(

π

z

) .

But 0 is not an isolated singularity of f (z) =1

sin(

π

z

) .

3. 0 is not an isolated singularity of f (z) = Log(z) or of any rootfunction. (Remember that every branch cut must contain zero, sothese functions will not be analytic on a set 0 < |z|< R.)


The Residue Theorem

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Definition. If the function f is analytic for 0 < |z− z0|< R

, then z0 iscalled an isolated singularity of f .


1(z−1)(1+ z2)

.


1sin(

π

z

) .


sin(

π

z

) .



The Residue Theorem

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Definition. If the function f is analytic for 0 < |z− z0|< R, then z0 iscalled an isolated singularity of f .


1(z−1)(1+ z2)

.


1sin(

π

z

) .


sin(

π

z

) .



The Residue Theorem

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Notes.

1. z = 1, i,−i are isolated singularities of f (z) =1

(z−1)(1+ z2).


1sin(

π

z

) .


sin(

π

z

) .



The Residue Theorem

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1(z−1)(1+ z2)

.


1sin(

π

z

) .


sin(

π

z

) .



The Residue Theorem

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1(z−1)(1+ z2)

.


1sin(

π

z

) .


sin(

π

z

) .

3. 0 is not an isolated singularity of f (z) = Log(z) or of any rootfunction.

(Remember that every branch cut must contain zero, sothese functions will not be analytic on a set 0 < |z|< R.)


The Residue Theorem

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1(z−1)(1+ z2)

.


1sin(

π

z

) .


sin(

π

z

) .

3. 0 is not an isolated singularity of f (z) = Log(z) or of any rootfunction. (Remember that every branch cut must contain zero

, sothese functions will not be analytic on a set 0 < |z|< R.)


The Residue Theorem

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1(z−1)(1+ z2)

.


1sin(

π

z

) .


sin(

π

z

) .



The Residue Theorem

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Definition.

If the function f is analytic for 0 < |z− z0|< R, then it has

a Laurent expansion∞

∑n=−∞

cn(z− z0)n about z0. The coefficient c−1 is

called the residue of f at z0. It is also denoted Resz=z0(f ) := c−1.

The next result will show the relevance of residues.


The Residue Theorem

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Definition. If the function f is analytic for 0 < |z− z0|< R

, then it has


∑n=−∞





The Residue Theorem

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Definition. If the function f is analytic for 0 < |z− z0|< R, then it has


∑n=−∞

cn(z− z0)n about z0.

The coefficient c−1 is




The Residue Theorem

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∑n=−∞


called the residue of f at z0.

It is also denoted Resz=z0(f ) := c−1.



The Residue Theorem

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∑n=−∞





The Residue Theorem

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Theorem.

If the function f is analytic for 0 < |z− z0|< R and C is apositively oriented simple closed contour around z0 that is contained

in 0 < |z− z0|< R, then∫

Cf (z) dz = 2πiResz=z0(f ).

Proof. From the theorem on Laurent expansions, we have that

Resz=z0(f ) = a−1 =1

2πi

∮C

f (ξ )(ξ − z0)(−1)+1 dξ =

12πi

∮C

f (ξ ) dξ ,

where C is any circle around z0 with radius < R. Replacement of thecircle with any contour around the origin requires an argument similarto the one that shows that we can use circles of any radius.


The Residue Theorem

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Theorem. If the function f is analytic for 0 < |z− z0|< R

and C is apositively oriented simple closed contour around z0 that is contained

in 0 < |z− z0|< R, then∫



Resz=z0(f ) = a−1 =1

2πi

∮C

f (ξ )(ξ − z0)(−1)+1 dξ =

12πi

∮C

f (ξ ) dξ ,



The Residue Theorem

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Theorem. If the function f is analytic for 0 < |z− z0|< R and C is apositively oriented simple closed contour around z0 that is contained

in 0 < |z− z0|< R

, then∫



Resz=z0(f ) = a−1 =1

2πi

∮C

f (ξ )(ξ − z0)(−1)+1 dξ =

12πi

∮C

f (ξ ) dξ ,



The Residue Theorem

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in 0 < |z− z0|< R, then∫



Resz=z0(f ) = a−1 =1

2πi

∮C

f (ξ )(ξ − z0)(−1)+1 dξ =

12πi

∮C

f (ξ ) dξ ,



The Residue Theorem

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in 0 < |z− z0|< R, then∫


Proof.

From the theorem on Laurent expansions, we have that

Resz=z0(f ) = a−1 =1

2πi

∮C

f (ξ )(ξ − z0)(−1)+1 dξ =

12πi

∮C

f (ξ ) dξ ,



The Residue Theorem

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in 0 < |z− z0|< R, then∫



Resz=z0(f )

= a−1 =1

2πi

∮C

f (ξ )(ξ − z0)(−1)+1 dξ =

12πi

∮C

f (ξ ) dξ ,



The Residue Theorem

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in 0 < |z− z0|< R, then∫



Resz=z0(f ) = a−1

=1

2πi

∮C

f (ξ )(ξ − z0)(−1)+1 dξ =

12πi

∮C

f (ξ ) dξ ,



The Residue Theorem

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in 0 < |z− z0|< R, then∫



Resz=z0(f ) = a−1 =1

2πi

∮C

f (ξ )(ξ − z0)(−1)+1 dξ

=1

2πi

∮C

f (ξ ) dξ ,



The Residue Theorem

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in 0 < |z− z0|< R, then∫



Resz=z0(f ) = a−1 =1

2πi

∮C

f (ξ )(ξ − z0)(−1)+1 dξ =

12πi

∮C

f (ξ ) dξ

,



The Residue Theorem

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in 0 < |z− z0|< R, then∫



Resz=z0(f ) = a−1 =1

2πi

∮C

f (ξ )(ξ − z0)(−1)+1 dξ =

12πi

∮C

f (ξ ) dξ ,

where C is any circle around z0 with radius < R.

Replacement of thecircle with any contour around the origin requires an argument similarto the one that shows that we can use circles of any radius.


The Residue Theorem

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in 0 < |z− z0|< R, then∫



Resz=z0(f ) = a−1 =1

2πi

∮C

f (ξ )(ξ − z0)(−1)+1 dξ =

12πi

∮C

f (ξ ) dξ ,



The Residue Theorem

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Proof (concl.)

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The Residue Theorem

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Proof (concl.)

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The Residue Theorem

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Proof (concl.)

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The Residue Theorem

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Proof (concl.)

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rrr

rIR


The Residue Theorem

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Proof (concl.)

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rrrr

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The Residue Theorem

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Proof (concl.)

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1

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0

rrrrI

R


The Residue Theorem

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Proof (concl.)

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-

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0

rrrrIR


The Residue Theorem

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Example.

Let C be the unit circle, traversed in the positive

orientation. Then∫

C

ez

zdz = 2πi

The functionez

zhas only the singularity at 0 inside the contour. The

residue ofez

z=

1z

+∞

∑n=0

zn

(n+1)!at z = 0 is 1. Now apply the

preceding theorem.Note that direct computation gives the same result, because the

integral of∞

∑n=0

zn

(n+1)!over any closed contour is 0 and

∫C

1z

dz =∫ 2π

0

1eiθ ieiθ dθ = 2πi.


The Residue Theorem

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Example. Let C be the unit circle, traversed in the positive

orientation.

Then∫

C

ez

zdz = 2πi

The functionez


residue ofez

z=

1z

+∞

∑n=0

zn



integral of∞

∑n=0

zn


∫C

1z

dz =∫ 2π

0



The Residue Theorem

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C

ez

zdz = 2πi

The functionez


residue ofez

z=

1z

+∞

∑n=0

zn



integral of∞

∑n=0

zn


∫C

1z

dz =∫ 2π

0



The Residue Theorem

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C

ez

zdz = 2πi

The functionez

zhas only the singularity at 0 inside the contour.

The

residue ofez

z=

1z

+∞

∑n=0

zn



integral of∞

∑n=0

zn


∫C

1z

dz =∫ 2π

0



The Residue Theorem

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C

ez

zdz = 2πi

The functionez


residue ofez

z

=1z

+∞

∑n=0

zn



integral of∞

∑n=0

zn


∫C

1z

dz =∫ 2π

0



The Residue Theorem

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C

ez

zdz = 2πi

The functionez


residue ofez

z=

1z

+∞

∑n=0

zn

(n+1)!

at z = 0 is 1. Now apply the


integral of∞

∑n=0

zn


∫C

1z

dz =∫ 2π

0



The Residue Theorem

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C

ez

zdz = 2πi

The functionez


residue ofez

z=

1z

+∞

∑n=0

zn

(n+1)!at z = 0 is 1.

Now apply the


integral of∞

∑n=0

zn


∫C

1z

dz =∫ 2π

0



The Residue Theorem

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C

ez

zdz = 2πi

The functionez


residue ofez

z=

1z

+∞

∑n=0

zn


preceding theorem.

Note that direct computation gives the same result, because the

integral of∞

∑n=0

zn


∫C

1z

dz =∫ 2π

0



The Residue Theorem

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C

ez

zdz = 2πi

The functionez


residue ofez

z=

1z

+∞

∑n=0

zn


preceding theorem.Note that direct computation gives the same result

, because the

integral of∞

∑n=0

zn


∫C

1z

dz =∫ 2π

0



The Residue Theorem

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C

ez

zdz = 2πi

The functionez


residue ofez

z=

1z

+∞

∑n=0

zn



integral of∞

∑n=0

zn

(n+1)!over any closed contour is 0

and

∫C

1z

dz =∫ 2π

0



The Residue Theorem

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C

ez

zdz = 2πi

The functionez


residue ofez

z=

1z

+∞

∑n=0

zn



integral of∞

∑n=0

zn


∫C

1z

dz

=∫ 2π

0



The Residue Theorem

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C

ez

zdz = 2πi

The functionez


residue ofez

z=

1z

+∞

∑n=0

zn



integral of∞

∑n=0

zn


∫C

1z

dz =∫ 2π

0

1eiθ ieiθ dθ

= 2πi.


The Residue Theorem

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C

ez

zdz = 2πi

The functionez


residue ofez

z=

1z

+∞

∑n=0

zn



integral of∞

∑n=0

zn


∫C

1z

dz =∫ 2π

0



The Residue Theorem

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Example.

Let C be the unit circle, traversed in the positive


C

ez2

z2 dz = 0

The functionez2

z2 has only the singularity at 0 inside the contour. The

residue ofez2

z2 =1z2 +

∞

∑n=0

z2n


preceding theorem.


The Residue Theorem

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orientation.

Then∫

C

ez2

z2 dz = 0

The functionez2


residue ofez2

z2 =1z2 +

∞

∑n=0

z2n


preceding theorem.


The Residue Theorem

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C

ez2

z2 dz = 0

The functionez2


residue ofez2

z2 =1z2 +

∞

∑n=0

z2n


preceding theorem.


The Residue Theorem

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C

ez2

z2 dz = 0

The functionez2

z2 has only the singularity at 0 inside the contour.

The

residue ofez2

z2 =1z2 +

∞

∑n=0

z2n


preceding theorem.


The Residue Theorem

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C

ez2

z2 dz = 0

The functionez2


residue ofez2

z2

=1z2 +

∞

∑n=0

z2n


preceding theorem.


The Residue Theorem

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C

ez2

z2 dz = 0

The functionez2


residue ofez2

z2 =1z2 +

∞

∑n=0

z2n

(n+1)!

at z = 0 is 0. Now apply the

preceding theorem.


The Residue Theorem

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C

ez2

z2 dz = 0

The functionez2


residue ofez2

z2 =1z2 +

∞

∑n=0

z2n

(n+1)!at z = 0 is 0.

Now apply the

preceding theorem.


The Residue Theorem

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C

ez2

z2 dz = 0

The functionez2


residue ofez2

z2 =1z2 +

∞

∑n=0

z2n


preceding theorem.


The Residue Theorem

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Theorem.

Residue Theorem. Let C be a simple closed positivelyoriented contour, let z1, . . . ,zn be points in the interior of C, and letthe function f be analytic on C and in its interior, except possibly atthe zj. Then

12πi

∫C

f (ξ ) dξ =n

∑j=1

Resz=zj(f ).

Proof. Let Cj be positively oriented circle around zj so that no two ofC1, . . . ,Cn intersect and so that all are contained in the interior of C.Then by extension of Cauchy-Goursat theorem∫

Cf (z) dz =

n

∑j=1

∫Cj

f (z) dz = 2πin

∑j=1

Resz=zj(f ).


The Residue Theorem

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Theorem. Residue Theorem. Let C be a simple closed positivelyoriented contour, let z1, . . . ,zn be points in the interior of C, and letthe function f be analytic on C and in its interior, except possibly atthe zj. Then

12πi

∫C

f (ξ ) dξ =n

∑j=1

Resz=zj(f ).


Cf (z) dz =

n

∑j=1

∫Cj

f (z) dz = 2πin

∑j=1

Resz=zj(f ).


The Residue Theorem

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12πi

∫C

f (ξ ) dξ =n

∑j=1

Resz=zj(f ).

Proof.

Let Cj be positively oriented circle around zj so that no two ofC1, . . . ,Cn intersect and so that all are contained in the interior of C.Then by extension of Cauchy-Goursat theorem∫

Cf (z) dz =

n

∑j=1

∫Cj

f (z) dz = 2πin

∑j=1

Resz=zj(f ).


The Residue Theorem

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12πi

∫C

f (ξ ) dξ =n

∑j=1

Resz=zj(f ).

Proof. Let Cj be positively oriented circle around zj

so that no two ofC1, . . . ,Cn intersect and so that all are contained in the interior of C.Then by extension of Cauchy-Goursat theorem∫

Cf (z) dz =

n

∑j=1

∫Cj

f (z) dz = 2πin

∑j=1

Resz=zj(f ).


The Residue Theorem

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12πi

∫C

f (ξ ) dξ =n

∑j=1

Resz=zj(f ).

Proof. Let Cj be positively oriented circle around zj so that no two ofC1, . . . ,Cn intersect and so that all are contained in the interior of C.

Then by extension of Cauchy-Goursat theorem∫C

f (z) dz =n

∑j=1

∫Cj

f (z) dz = 2πin

∑j=1

Resz=zj(f ).


The Residue Theorem

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12πi

∫C

f (ξ ) dξ =n

∑j=1

Resz=zj(f ).


Cf (z) dz =

n

∑j=1

∫Cj

f (z) dz

= 2πin

∑j=1

Resz=zj(f ).


The Residue Theorem

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12πi

∫C

f (ξ ) dξ =n

∑j=1

Resz=zj(f ).


Cf (z) dz =

n

∑j=1

∫Cj

f (z) dz = 2πin

∑j=1

Resz=zj(f ).


The Residue Theorem

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Example.

Integrate the function f (z) =z2

z2 +1over the positively

oriented boundary of the square of numbers z = x+ iy with (x,y) in[−3,3]× [−3,3].

-

6ℑ(z)

ℜ(z)6

�

?

-

3

ri

r−i


The Residue Theorem

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Example. Integrate the function f (z) =z2



-

6ℑ(z)

ℜ(z)6

�

?

-

3

ri

r−i


The Residue Theorem

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-

6ℑ(z)

ℜ(z)

6

�

?

-

3

ri

r−i


The Residue Theorem

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-

6ℑ(z)

ℜ(z)6

�

?

-

3

ri

r−i


The Residue Theorem

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-

6ℑ(z)

ℜ(z)6

�

?

-

3

r

i

r−i


The Residue Theorem

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-

6ℑ(z)

ℜ(z)6

�

?

-

3

ri

r−i


The Residue Theorem

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-

6ℑ(z)

ℜ(z)6

�

?

-

3

ri

r

−i


The Residue Theorem

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-

6ℑ(z)

ℜ(z)6

�

?

-

3

ri

r−i


The Residue Theorem

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Residue at i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z+ i

1z− i

Resz=i

(z2

z2 +1

)= − 1

i+ i=

i2

Residue at −i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z− i

1z+ i

Resz=−i

(z2

z2 +1

)= − 1

−i− i=− i

2


The Residue Theorem

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Residue at i:

z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z+ i

1z− i

Resz=i

(z2

z2 +1

)= − 1

i+ i=

i2

Residue at −i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z− i

1z+ i

Resz=−i

(z2

z2 +1

)= − 1

−i− i=− i

2


The Residue Theorem

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Residue at i:z2

z2 +1

=z2 +1−1

z2 +1= 1− 1

z2 +1= 1− 1

z+ i1

z− i

Resz=i

(z2

z2 +1

)= − 1

i+ i=

i2

Residue at −i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z− i

1z+ i

Resz=−i

(z2

z2 +1

)= − 1

−i− i=− i

2


The Residue Theorem

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Residue at i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z+ i

1z− i

Resz=i

(z2

z2 +1

)= − 1

i+ i=

i2

Residue at −i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z− i

1z+ i

Resz=−i

(z2

z2 +1

)= − 1

−i− i=− i

2


The Residue Theorem

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Residue at i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z+ i

1z− i

Resz=i

(z2

z2 +1

)

= − 1i+ i

=i2

Residue at −i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z− i

1z+ i

Resz=−i

(z2

z2 +1

)= − 1

−i− i=− i

2


The Residue Theorem

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Residue at i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z+ i

1z− i

Resz=i

(z2

z2 +1

)= − 1

i+ i

=i2

Residue at −i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z− i

1z+ i

Resz=−i

(z2

z2 +1

)= − 1

−i− i=− i

2


The Residue Theorem

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Residue at i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z+ i

1z− i

Resz=i

(z2

z2 +1

)= − 1

i+ i=

i2

Residue at −i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z− i

1z+ i

Resz=−i

(z2

z2 +1

)= − 1

−i− i=− i

2


The Residue Theorem

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Residue at i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z+ i

1z− i

Resz=i

(z2

z2 +1

)= − 1

i+ i=

i2

Residue at −i:

z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z− i

1z+ i

Resz=−i

(z2

z2 +1

)= − 1

−i− i=− i

2


The Residue Theorem

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Residue at i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z+ i

1z− i

Resz=i

(z2

z2 +1

)= − 1

i+ i=

i2

Residue at −i:z2

z2 +1

=z2 +1−1

z2 +1= 1− 1

z2 +1= 1− 1

z− i1

z+ i

Resz=−i

(z2

z2 +1

)= − 1

−i− i=− i

2


The Residue Theorem

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Residue at i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z+ i

1z− i

Resz=i

(z2

z2 +1

)= − 1

i+ i=

i2

Residue at −i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z− i

1z+ i

Resz=−i

(z2

z2 +1

)= − 1

−i− i=− i

2


The Residue Theorem

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Residue at i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z+ i

1z− i

Resz=i

(z2

z2 +1

)= − 1

i+ i=

i2

Residue at −i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z− i

1z+ i

Resz=−i

(z2

z2 +1

)

= − 1−i− i

=− i2


The Residue Theorem

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Residue at i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z+ i

1z− i

Resz=i

(z2

z2 +1

)= − 1

i+ i=

i2

Residue at −i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z− i

1z+ i

Resz=−i

(z2

z2 +1

)= − 1

−i− i

=− i2


The Residue Theorem

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Residue at i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z+ i

1z− i

Resz=i

(z2

z2 +1

)= − 1

i+ i=

i2

Residue at −i:z2

z2 +1=

z2 +1−1z2 +1

= 1− 1z2 +1

= 1− 1z− i

1z+ i

Resz=−i

(z2

z2 +1

)= − 1

−i− i=− i

2


The Residue Theorem

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∫C

z2

z2 +1dz = 2πi

[Resz=i

(z2

z2 +1

)+Resz=−i

(z2

z2 +1

)]= 2πi

[i2

+(− i

2

)]= 0


The Residue Theorem

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∫C

z2

z2 +1dz

= 2πi[

Resz=i

(z2

z2 +1

)+Resz=−i

(z2

z2 +1

)]= 2πi

[i2

+(− i

2

)]= 0


The Residue Theorem

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∫C

z2

z2 +1dz = 2πi

[Resz=i

(z2

z2 +1

)+Resz=−i

(z2

z2 +1

)]

= 2πi[

i2

+(− i

2

)]= 0


The Residue Theorem

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∫C

z2

z2 +1dz = 2πi

[Resz=i

(z2

z2 +1

)+Resz=−i

(z2

z2 +1

)]= 2πi

[i2

+(− i

2

)]

= 0


The Residue Theorem

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∫C

z2

z2 +1dz = 2πi

[Resz=i

(z2

z2 +1

)+Resz=−i

(z2

z2 +1

)]= 2πi

[i2

+(− i

2

)]= 0


The Residue Theorem

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Definition.

Let f be analytic on C except for a finite number ofsingular points z1, . . . ,zn. Assume that R1 is so that all |zj| ≤ R1. ForR0 > R1 let C−0 be the circle around the origin of radius R0, traversedin the clockwise, that is, the mathematically negative direction. Thenwe define the residue at infinity of f as

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz =− 12πi

∫C+

0

f (z) dz =− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1 =− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz =− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz =−Resz=0

(1z2 f(

1z

))


The Residue Theorem

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Definition. Let f be analytic on C except for a finite number ofsingular points z1, . . . ,zn.

Assume that R1 is so that all |zj| ≤ R1. ForR0 > R1 let C−0 be the circle around the origin of radius R0, traversedin the clockwise, that is, the mathematically negative direction. Thenwe define the residue at infinity of f as

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz =− 12πi

∫C+

0

f (z) dz =− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1 =− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz =− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz =−Resz=0

(1z2 f(

1z

))


The Residue Theorem

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Definition. Let f be analytic on C except for a finite number ofsingular points z1, . . . ,zn. Assume that R1 is so that all |zj| ≤ R1.

ForR0 > R1 let C−0 be the circle around the origin of radius R0, traversedin the clockwise, that is, the mathematically negative direction. Thenwe define the residue at infinity of f as

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz =− 12πi

∫C+

0

f (z) dz =− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1 =− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz =− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz =−Resz=0

(1z2 f(

1z

))


The Residue Theorem

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Definition. Let f be analytic on C except for a finite number ofsingular points z1, . . . ,zn. Assume that R1 is so that all |zj| ≤ R1. ForR0 > R1 let C−0 be the circle around the origin of radius R0, traversedin the clockwise

, that is, the mathematically negative direction. Thenwe define the residue at infinity of f as

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz =− 12πi

∫C+

0

f (z) dz =− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1 =− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz =− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz =−Resz=0

(1z2 f(

1z

))


The Residue Theorem

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Definition. Let f be analytic on C except for a finite number ofsingular points z1, . . . ,zn. Assume that R1 is so that all |zj| ≤ R1. ForR0 > R1 let C−0 be the circle around the origin of radius R0, traversedin the clockwise, that is, the mathematically negative

direction. Thenwe define the residue at infinity of f as

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz =− 12πi

∫C+

0

f (z) dz =− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1 =− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz =− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz =−Resz=0

(1z2 f(

1z

))


The Residue Theorem

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Definition. Let f be analytic on C except for a finite number ofsingular points z1, . . . ,zn. Assume that R1 is so that all |zj| ≤ R1. ForR0 > R1 let C−0 be the circle around the origin of radius R0, traversedin the clockwise, that is, the mathematically negative direction.

Thenwe define the residue at infinity of f as

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz =− 12πi

∫C+

0

f (z) dz =− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1 =− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz =− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz =−Resz=0

(1z2 f(

1z

))


The Residue Theorem

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Definition. Let f be analytic on C except for a finite number ofsingular points z1, . . . ,zn. Assume that R1 is so that all |zj| ≤ R1. ForR0 > R1 let C−0 be the circle around the origin of radius R0, traversedin the clockwise, that is, the mathematically negative direction. Thenwe define the residue at infinity of f as

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz =− 12πi

∫C+

0

f (z) dz =− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1 =− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz =− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz =−Resz=0

(1z2 f(

1z

))


The Residue Theorem

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Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz

=− 12πi

∫C+

0

f (z) dz =− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1 =− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz =− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz =−Resz=0

(1z2 f(

1z

))


The Residue Theorem

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Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz =− 12πi

∫C+

0

f (z) dz

=− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1 =− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz =− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz =−Resz=0

(1z2 f(

1z

))


The Residue Theorem

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Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz =− 12πi

∫C+

0

f (z) dz =− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1 =− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz =− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz =−Resz=0

(1z2 f(

1z

))


The Residue Theorem

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Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz =− 12πi

∫C+

0

f (z) dz =− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1

=− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz =− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz =−Resz=0

(1z2 f(

1z

))


The Residue Theorem

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Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz =− 12πi

∫C+

0

f (z) dz =− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1 =− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz

=− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz =−Resz=0

(1z2 f(

1z

))


The Residue Theorem

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Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz =− 12πi

∫C+

0

f (z) dz =− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1 =− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz =− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz =−Resz=0

(1z2 f(

1z

))


The Residue Theorem

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Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz =− 12πi

∫C+

0

f (z) dz =− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1 =− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz =− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz

=−Resz=0

(1z2 f(

1z

))


The Residue Theorem

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Resz=∞(f ) =1

2πi

∫C−0

f (z) dz.

Resz=∞(f ) =1

2πi

∫C−0

f (z) dz =− 12πi

∫C+

0

f (z) dz =− 12πi

∫C+

0

∞

∑n=−∞

cnzn dz

= −c−1 =− 12πi

∫C+

0

∞

∑n=−∞

cn−2

zn dz =− 12πi

∫C+

0

1z2

∞

∑n=−∞

cn−2

zn−2 dz

= − 12πi

∫C+

0

1z2 f(

1z

)dz =−Resz=0

(1z2 f(

1z

))Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Residue Theorem

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Theorem.

If f is analytic on C, except for a finite number of singularpoints that lie in the interior of a positively oriented simple closedcontour C, then ∫

Cf (z) dz = 2πiResz=0

[1z2 f(

1z

)].

Proof. See previous panel.


The Residue Theorem

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Theorem. If f is analytic on C, except for a finite number of singularpoints that lie in the interior of a positively oriented simple closedcontour C, then ∫


[1z2 f(

1z

)].



The Residue Theorem

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[1z2 f(

1z

)].

Proof.

See previous panel.


The Residue Theorem

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[1z2 f(

1z

)].



The Residue Theorem

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Example.

Integrate the function f (z) =z2



-

6ℑ(z)

ℜ(z)6

�

?

-

3

ri

r−i


The Residue Theorem

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-

6ℑ(z)

ℜ(z)6

�

?

-

3

ri

r−i


The Residue Theorem

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-

6ℑ(z)

ℜ(z)

6

�

?

-

3

ri

r−i


The Residue Theorem

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-

6ℑ(z)

ℜ(z)6

�

?

-

3

ri

r−i


The Residue Theorem

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-

6ℑ(z)

ℜ(z)6

�

?

-

3

r

i

r−i


The Residue Theorem

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-

6ℑ(z)

ℜ(z)6

�

?

-

3

ri

r−i


The Residue Theorem

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-

6ℑ(z)

ℜ(z)6

�

?

-

3

ri

r

−i


The Residue Theorem

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-

6ℑ(z)

ℜ(z)6

�

?

-

3

ri

r−i


The Residue Theorem

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1z2 f(

1z

)=

1z2

(1z

)2(1z

)2+1

=1z2

11+ z2

=1z2

11− (−z2)

=1z2

∞

∑n=0

(−1)nz2n

=1z2 +

∞

∑n=1

(−1)nz2n−2

and hence∫


[1z2 f(

1z

)]= 0.


The Residue Theorem

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1z2 f(

1z

)

=1z2

(1z

)2(1z

)2+1

=1z2

11+ z2

=1z2

11− (−z2)

=1z2

∞

∑n=0

(−1)nz2n

=1z2 +

∞

∑n=1

(−1)nz2n−2

and hence∫


[1z2 f(

1z

)]= 0.


The Residue Theorem

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1z2 f(

1z

)=

1z2

(1z

)2(1z

)2+1

=1z2

11+ z2

=1z2

11− (−z2)

=1z2

∞

∑n=0

(−1)nz2n

=1z2 +

∞

∑n=1

(−1)nz2n−2

and hence∫


[1z2 f(

1z

)]= 0.


The Residue Theorem

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1z2 f(

1z

)=

1z2

(1z

)2(1z

)2+1

=1z2

11+ z2

=1z2

11− (−z2)

=1z2

∞

∑n=0

(−1)nz2n

=1z2 +

∞

∑n=1

(−1)nz2n−2

and hence∫

Cf (z) dz

= 2πiResz=0

[1z2 f(

1z

)]= 0.


The Residue Theorem

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1z2 f(

1z

)=

1z2

(1z

)2(1z

)2+1

=1z2

11+ z2

=1z2

11− (−z2)

=1z2

∞

∑n=0

(−1)nz2n

=1z2 +

∞

∑n=1

(−1)nz2n−2

and hence∫


[1z2 f(

1z

)]

= 0.


The Residue Theorem

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1z2 f(

1z

)=

1z2

(1z

)2(1z

)2+1

=1z2

11+ z2

=1z2

11− (−z2)

=1z2

∞

∑n=0

(−1)nz2n

=1z2 +

∞

∑n=1

(−1)nz2n−2

and hence∫


[1z2 f(

1z

)]= 0.


The Residue Theorem

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Definition.

Let z0 be a complex number, let r > 0 and let f be analyticin a deleted neighborhood of z0. In this case z0 is called an isolatedsingularity of f . Consider the Laurent expansion of f around z0

f (z) =∞

∑n=0

an(z− z0)n +b1

z− z0+

b2

(z− z0)2 +b3

(z− z0)3 + · · ·

The sum of the negative powers of z− z0 is also called the principalpart of f .

1. If bn = 0 for all n, then z0 is called a removable singularity.2. If there is a positive number m so that bm 6= 0 and bn = 0 for all

n > m, then z0 is called a pole of order m.3. If the number m in part 2 equals 1, then z0 is also called a

simple pole.4. If z0 is not removable and there is no number as in part 2, then z0

is called an essential singularity.


The Residue Theorem

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Definition. Let z0 be a complex number, let r > 0 and let f be analyticin a deleted neighborhood of z0.

In this case z0 is called an isolatedsingularity of f . Consider the Laurent expansion of f around z0

f (z) =∞

∑n=0

an(z− z0)n +b1

z− z0+

b2

(z− z0)2 +b3

(z− z0)3 + · · ·







The Residue Theorem

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Definition. Let z0 be a complex number, let r > 0 and let f be analyticin a deleted neighborhood of z0. In this case z0 is called an isolatedsingularity of f .

Consider the Laurent expansion of f around z0

f (z) =∞

∑n=0

an(z− z0)n +b1

z− z0+

b2

(z− z0)2 +b3

(z− z0)3 + · · ·







The Residue Theorem

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Definition. Let z0 be a complex number, let r > 0 and let f be analyticin a deleted neighborhood of z0. In this case z0 is called an isolatedsingularity of f . Consider the Laurent expansion of f around z0

f (z) =∞

∑n=0

an(z− z0)n +b1

z− z0+

b2

(z− z0)2 +b3

(z− z0)3 + · · ·







The Residue Theorem

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f (z) =∞

∑n=0

an(z− z0)n +b1

z− z0+

b2

(z− z0)2 +b3

(z− z0)3 + · · ·


1. If bn = 0 for all n, then z0 is called a removable singularity.

2. If there is a positive number m so that bm 6= 0 and bn = 0 for alln > m, then z0 is called a pole of order m.

3. If the number m in part 2 equals 1, then z0 is also called asimple pole.

4. If z0 is not removable and there is no number as in part 2, then z0is called an essential singularity.


The Residue Theorem

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f (z) =∞

∑n=0

an(z− z0)n +b1

z− z0+

b2

(z− z0)2 +b3

(z− z0)3 + · · ·



n > m, then z0 is called a pole of order m.

3. If the number m in part 2 equals 1, then z0 is also called asimple pole.



The Residue Theorem

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f (z) =∞

∑n=0

an(z− z0)n +b1

z− z0+

b2

(z− z0)2 +b3

(z− z0)3 + · · ·




simple pole.



The Residue Theorem

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f (z) =∞

∑n=0

an(z− z0)n +b1

z− z0+

b2

(z− z0)2 +b3

(z− z0)3 + · · ·







The Residue Theorem

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Example.

The function f (z) =sin(z)

zhas a removable singularity at

z = 0.sin(z)

z=

1z

sin(z)

=1z

∞

∑n=0

(−1)n

(2n+1)!z2n+1

=∞

∑n=0

(−1)n

(2n+1)!z2n


The Residue Theorem

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Example. The function f (z) =sin(z)


z = 0.

sin(z)z

=1z

sin(z)

=1z

∞

∑n=0

(−1)n

(2n+1)!z2n+1

=∞

∑n=0

(−1)n

(2n+1)!z2n


The Residue Theorem

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z = 0.sin(z)

z

=1z

sin(z)

=1z

∞

∑n=0

(−1)n

(2n+1)!z2n+1

=∞

∑n=0

(−1)n

(2n+1)!z2n


The Residue Theorem

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z = 0.sin(z)

z=

1z

sin(z)

=1z

∞

∑n=0

(−1)n

(2n+1)!z2n+1

=∞

∑n=0

(−1)n

(2n+1)!z2n


The Residue Theorem

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Example.


1− z2 has simple poles at z = 1 and at

z =−1.1

1− z2 =1

z+11

z−1


The Residue Theorem

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z =−1.

11− z2 =

1z+1

1z−1


The Residue Theorem

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z =−1.1

1− z2

=1

z+11

z−1


The Residue Theorem

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z =−1.1

1− z2 =1

z+11

z−1


The Residue Theorem

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Example.

The function f (z) =1+ z2

z3 has a pole of order 3 at z = 0.

1+ z2

z3 =1z3 +

1z


The Residue Theorem

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Example. The function f (z) =1+ z2


1+ z2

z3 =1z3 +

1z


The Residue Theorem

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1+ z2

z3

=1z3 +

1z


The Residue Theorem

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1+ z2

z3 =1z3 +

1z


The Residue Theorem

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Example.

The function f (z) = e1z has an essential singularity at

z = 0.

e1z =

∞

∑n=0

1n!

1zn


The Residue Theorem

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Example. The function f (z) = e1z has an essential singularity at

z = 0.

e1z =

∞

∑n=0

1n!

1zn


The Residue Theorem

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z = 0.

e1z

=∞

∑n=0

1n!

1zn


The Residue Theorem

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z = 0.

e1z =

∞

∑n=0

1n!

1zn


The Residue Theorem

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Theorem.

Let the function f be analytic in a deleted neighborhood ofthe point z0. Then z0 is a pole of order m if and only if there is functionΦ that is analytic in a neighborhood of z0, Φ(z0) 6= 0 and so that

f (z) =Φ(z)

(z− z0)m

for all z in a deleted neighborhood of z0. Moreover,

Resz=z0(f ) =Φ(m−1)(z0)(m−1)!


The Residue Theorem

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Theorem. Let the function f be analytic in a deleted neighborhood ofthe point z0.

Then z0 is a pole of order m if and only if there is functionΦ that is analytic in a neighborhood of z0, Φ(z0) 6= 0 and so that

f (z) =Φ(z)

(z− z0)m


Resz=z0(f ) =Φ(m−1)(z0)(m−1)!


The Residue Theorem

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Theorem. Let the function f be analytic in a deleted neighborhood ofthe point z0. Then z0 is a pole of order m if and only if there is functionΦ that is analytic in a neighborhood of z0

, Φ(z0) 6= 0 and so that

f (z) =Φ(z)

(z− z0)m


Resz=z0(f ) =Φ(m−1)(z0)(m−1)!


The Residue Theorem

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Theorem. Let the function f be analytic in a deleted neighborhood ofthe point z0. Then z0 is a pole of order m if and only if there is functionΦ that is analytic in a neighborhood of z0, Φ(z0) 6= 0

and so that

f (z) =Φ(z)

(z− z0)m


Resz=z0(f ) =Φ(m−1)(z0)(m−1)!


The Residue Theorem

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Theorem. Let the function f be analytic in a deleted neighborhood ofthe point z0. Then z0 is a pole of order m if and only if there is functionΦ that is analytic in a neighborhood of z0, Φ(z0) 6= 0 and so that

f (z) =Φ(z)

(z− z0)m

for all z in a deleted neighborhood of z0.

Moreover,

Resz=z0(f ) =Φ(m−1)(z0)(m−1)!


The Residue Theorem

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Theorem. Let the function f be analytic in a deleted neighborhood ofthe point z0. Then z0 is a pole of order m if and only if there is functionΦ that is analytic in a neighborhood of z0, Φ(z0) 6= 0 and so that

f (z) =Φ(z)

(z− z0)m


Resz=z0(f ) =Φ(m−1)(z0)(m−1)!


The Residue Theorem

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Proof.

If f has a pole of order m around z0, then the Laurentexpansion of z for 0 < |z− z0|< R (for some R) is (with c−m 6= 0)

f (z) =∞

∑n=−m

cn(z− z0)n =1

(z− z0)m

∞

∑n=−m

cn(z− z0)n+m

=1

(z− z0)m

∞

∑k=0

ck−m(z− z0)k =:Φ(z)

(z− z0)m .

Conversely, if f (z) =Φ(z)

(z− z0)m with Φ(z0) 6= 0, then the above

computation in reverse shows that f has a pole of order m. Moreover,in this situation, the coefficient c−1 = Resz=z0(f ) of the Laurent

expansion of f is the coefficient am−1 =Φ(m−1)(z0)(m−1)!

of the power

series expansion of Φ.


The Residue Theorem

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Proof. If f has a pole of order m around z0, then the Laurentexpansion of z for 0 < |z− z0|< R (for some R) is (with c−m 6= 0)

f (z) =∞

∑n=−m

cn(z− z0)n

=1

(z− z0)m

∞

∑n=−m

cn(z− z0)n+m

=1

(z− z0)m

∞

∑k=0


(z− z0)m .





of the power



The Residue Theorem

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f (z) =∞

∑n=−m

cn(z− z0)n =1

(z− z0)m

∞

∑n=−m

cn(z− z0)n+m

=1

(z− z0)m

∞

∑k=0


(z− z0)m .





of the power



The Residue Theorem

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f (z) =∞

∑n=−m

cn(z− z0)n =1

(z− z0)m

∞

∑n=−m

cn(z− z0)n+m

=1

(z− z0)m

∞

∑k=0

ck−m(z− z0)k

=:Φ(z)

(z− z0)m .





of the power



The Residue Theorem

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f (z) =∞

∑n=−m

cn(z− z0)n =1

(z− z0)m

∞

∑n=−m

cn(z− z0)n+m

=1

(z− z0)m

∞

∑k=0


(z− z0)m .





of the power



The Residue Theorem

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f (z) =∞

∑n=−m

cn(z− z0)n =1

(z− z0)m

∞

∑n=−m

cn(z− z0)n+m

=1

(z− z0)m

∞

∑k=0


(z− z0)m .



computation in reverse shows that f has a pole of order m.

Moreover,in this situation, the coefficient c−1 = Resz=z0(f ) of the Laurent


of the power



The Residue Theorem

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f (z) =∞

∑n=−m

cn(z− z0)n =1

(z− z0)m

∞

∑n=−m

cn(z− z0)n+m

=1

(z− z0)m

∞

∑k=0


(z− z0)m .




expansion of f

is the coefficient am−1 =Φ(m−1)(z0)(m−1)!

of the power



The Residue Theorem

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f (z) =∞

∑n=−m

cn(z− z0)n =1

(z− z0)m

∞

∑n=−m

cn(z− z0)n+m

=1

(z− z0)m

∞

∑k=0


(z− z0)m .





of the power



The Residue Theorem

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Example.

Find all poles, their order and their residues for

f (z) =(

z+1z−1

)3

.

The only singularity of f is z = 1.

f (z) =(

z+1z−1

)3

=1

(z−1)3

(z3 +3z2 +3z+1

)Φ(z) = z3 +3z2 +3z+1, m = 3

Φ′(z) = 3z2 +6z+3

Φ′′(z) = 6z+6

Φ′′(1) = 12

Resz=1(f ) =122!

= 6


The Residue Theorem

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Example. Find all poles, their order and their residues for

f (z) =(

z+1z−1

)3

.


f (z) =(

z+1z−1

)3

=1

(z−1)3

(z3 +3z2 +3z+1

)Φ(z) = z3 +3z2 +3z+1, m = 3

Φ′(z) = 3z2 +6z+3

Φ′′(z) = 6z+6

Φ′′(1) = 12

Resz=1(f ) =122!

= 6


The Residue Theorem

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f (z) =(

z+1z−1

)3

.


f (z) =(

z+1z−1

)3

=1

(z−1)3

(z3 +3z2 +3z+1

)

Φ(z) = z3 +3z2 +3z+1, m = 3

Φ′(z) = 3z2 +6z+3

Φ′′(z) = 6z+6

Φ′′(1) = 12

Resz=1(f ) =122!

= 6


The Residue Theorem

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f (z) =(

z+1z−1

)3

.


f (z) =(

z+1z−1

)3

=1

(z−1)3

(z3 +3z2 +3z+1

)Φ(z) = z3 +3z2 +3z+1

, m = 3

Φ′(z) = 3z2 +6z+3

Φ′′(z) = 6z+6

Φ′′(1) = 12

Resz=1(f ) =122!

= 6


The Residue Theorem

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f (z) =(

z+1z−1

)3

.


f (z) =(

z+1z−1

)3

=1

(z−1)3

(z3 +3z2 +3z+1

)Φ(z) = z3 +3z2 +3z+1, m = 3

Φ′(z) = 3z2 +6z+3

Φ′′(z) = 6z+6

Φ′′(1) = 12

Resz=1(f ) =122!

= 6


The Residue Theorem

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Example.

Show that Resz=iLog(z)

(z2 +1)2 =π +2i

8.

Log(z)

(z2 +1)2 =Log(z)(

(z+ i)(z− i))2 =

1(z− i)2

Log(z)(z+ i)2

Φ(z) =Log(z)(z+ i)2 , m = 2

Φ′(z) =

1z (z+ i)2−2(z+ i)Log(z)

(z+ i)4 =1z (z+ i)−2Log(z)

(z+ i)3

=z+ i−2Log(z)z

z(z+ i)3

Φ′(i) =

i+ i−2Log(i)ii(i+ i)3 =

2i−2i π

2 ii(2i)3 =

2i+π

8

Resz=iLog(z)

(z2 +1)2 =11!

π +2i8

=π +2i

8


The Residue Theorem

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Example. Show that Resz=iLog(z)

(z2 +1)2 =π +2i

8.

Log(z)

(z2 +1)2 =Log(z)(

(z+ i)(z− i))2 =

1(z− i)2

Log(z)(z+ i)2

Φ(z) =Log(z)(z+ i)2 , m = 2

Φ′(z) =

1z (z+ i)2−2(z+ i)Log(z)

(z+ i)4 =1z (z+ i)−2Log(z)

(z+ i)3

=z+ i−2Log(z)z

z(z+ i)3

Φ′(i) =


2i−2i π

2 ii(2i)3 =

2i+π

8

Resz=iLog(z)

(z2 +1)2 =11!

π +2i8

=π +2i

8


The Residue Theorem

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(z2 +1)2 =π +2i

8.

Log(z)

(z2 +1)2 =Log(z)(

(z+ i)(z− i))2

=1

(z− i)2Log(z)(z+ i)2

Φ(z) =Log(z)(z+ i)2 , m = 2

Φ′(z) =

1z (z+ i)2−2(z+ i)Log(z)

(z+ i)4 =1z (z+ i)−2Log(z)

(z+ i)3

=z+ i−2Log(z)z

z(z+ i)3

Φ′(i) =


2i−2i π

2 ii(2i)3 =

2i+π

8

Resz=iLog(z)

(z2 +1)2 =11!

π +2i8

=π +2i

8


The Residue Theorem

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(z2 +1)2 =π +2i

8.

Log(z)

(z2 +1)2 =Log(z)(

(z+ i)(z− i))2 =

1(z− i)2

Log(z)(z+ i)2

Φ(z) =Log(z)(z+ i)2 , m = 2

Φ′(z) =

1z (z+ i)2−2(z+ i)Log(z)

(z+ i)4 =1z (z+ i)−2Log(z)

(z+ i)3

=z+ i−2Log(z)z

z(z+ i)3

Φ′(i) =


2i−2i π

2 ii(2i)3 =

2i+π

8

Resz=iLog(z)

(z2 +1)2 =11!

π +2i8

=π +2i

8


The Residue Theorem

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(z2 +1)2 =π +2i

8.

Log(z)

(z2 +1)2 =Log(z)(

(z+ i)(z− i))2 =

1(z− i)2

Log(z)(z+ i)2

Φ(z) =Log(z)(z+ i)2

, m = 2

Φ′(z) =

1z (z+ i)2−2(z+ i)Log(z)

(z+ i)4 =1z (z+ i)−2Log(z)

(z+ i)3

=z+ i−2Log(z)z

z(z+ i)3

Φ′(i) =


2i−2i π

2 ii(2i)3 =

2i+π

8

Resz=iLog(z)

(z2 +1)2 =11!

π +2i8

=π +2i

8


The Residue Theorem

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(z2 +1)2 =π +2i

8.

Log(z)

(z2 +1)2 =Log(z)(

(z+ i)(z− i))2 =

1(z− i)2

Log(z)(z+ i)2

Φ(z) =Log(z)(z+ i)2 , m = 2

Φ′(z) =

1z (z+ i)2−2(z+ i)Log(z)

(z+ i)4 =1z (z+ i)−2Log(z)

(z+ i)3

=z+ i−2Log(z)z

z(z+ i)3

Φ′(i) =


2i−2i π

2 ii(2i)3 =

2i+π

8

Resz=iLog(z)

(z2 +1)2 =11!

π +2i8

=π +2i

8


The Residue Theorem

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(z2 +1)2 =π +2i

8.

Log(z)

(z2 +1)2 =Log(z)(

(z+ i)(z− i))2 =

1(z− i)2

Log(z)(z+ i)2

Φ(z) =Log(z)(z+ i)2 , m = 2

Φ′(z) =

1z (z+ i)2−2(z+ i)Log(z)

(z+ i)4

=1z (z+ i)−2Log(z)

(z+ i)3

=z+ i−2Log(z)z

z(z+ i)3

Φ′(i) =


2i−2i π

2 ii(2i)3 =

2i+π

8

Resz=iLog(z)

(z2 +1)2 =11!

π +2i8

=π +2i

8


The Residue Theorem

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(z2 +1)2 =π +2i

8.

Log(z)

(z2 +1)2 =Log(z)(

(z+ i)(z− i))2 =

1(z− i)2

Log(z)(z+ i)2

Φ(z) =Log(z)(z+ i)2 , m = 2

Φ′(z) =

1z (z+ i)2−2(z+ i)Log(z)

(z+ i)4 =1z (z+ i)−2Log(z)

(z+ i)3

=z+ i−2Log(z)z

z(z+ i)3

Φ′(i) =


2i−2i π

2 ii(2i)3 =

2i+π

8

Resz=iLog(z)

(z2 +1)2 =11!

π +2i8

=π +2i

8


The Residue Theorem

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(z2 +1)2 =π +2i

8.

Log(z)

(z2 +1)2 =Log(z)(

(z+ i)(z− i))2 =

1(z− i)2

Log(z)(z+ i)2

Φ(z) =Log(z)(z+ i)2 , m = 2

Φ′(z) =

1z (z+ i)2−2(z+ i)Log(z)

(z+ i)4 =1z (z+ i)−2Log(z)

(z+ i)3

=z+ i−2Log(z)z

z(z+ i)3

Φ′(i) =

i+ i−2Log(i)ii(i+ i)3

=2i−2i π

2 ii(2i)3 =

2i+π

8

Resz=iLog(z)

(z2 +1)2 =11!

π +2i8

=π +2i

8


The Residue Theorem

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(z2 +1)2 =π +2i

8.

Log(z)

(z2 +1)2 =Log(z)(

(z+ i)(z− i))2 =

1(z− i)2

Log(z)(z+ i)2

Φ(z) =Log(z)(z+ i)2 , m = 2

Φ′(z) =

1z (z+ i)2−2(z+ i)Log(z)

(z+ i)4 =1z (z+ i)−2Log(z)

(z+ i)3

=z+ i−2Log(z)z

z(z+ i)3

Φ′(i) =


2i−2i π

2 ii(2i)3

=2i+π

8

Resz=iLog(z)

(z2 +1)2 =11!

π +2i8

=π +2i

8


The Residue Theorem

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(z2 +1)2 =π +2i

8.

Log(z)

(z2 +1)2 =Log(z)(

(z+ i)(z− i))2 =

1(z− i)2

Log(z)(z+ i)2

Φ(z) =Log(z)(z+ i)2 , m = 2

Φ′(z) =

1z (z+ i)2−2(z+ i)Log(z)

(z+ i)4 =1z (z+ i)−2Log(z)

(z+ i)3

=z+ i−2Log(z)z

z(z+ i)3

Φ′(i) =


2i−2i π

2 ii(2i)3 =

2i+π

8

Resz=iLog(z)

(z2 +1)2 =11!

π +2i8

=π +2i

8


The Residue Theorem

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Example.

Integrate f (z) =1

z3(z+4)around the positively oriented

circle of radius 5 around the origin.

We will need the residues of f at 0 and at −4, or we need the residue

at 0 of1z2 f(

1z

).

1z2 f(

1z

)=

1z2

1(1z

)3 (1z +4

) =1

1z

1+4zz

=z2

1+4z

As1z2 f(

1z

)does not have a singularity at 0, the integral must be 0.


The Residue Theorem

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Example. Integrate f (z) =1




at 0 of1z2 f(

1z

).

1z2 f(

1z

)=

1z2

1(1z

)3 (1z +4

) =1

1z

1+4zz

=z2

1+4z

As1z2 f(

1z



The Residue Theorem

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We will need the residues of f at 0 and at −4

, or we need the residue

at 0 of1z2 f(

1z

).

1z2 f(

1z

)=

1z2

1(1z

)3 (1z +4

) =1

1z

1+4zz

=z2

1+4z

As1z2 f(

1z



The Residue Theorem

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at 0 of1z2 f(

1z

).

1z2 f(

1z

)=

1z2

1(1z

)3 (1z +4

) =1

1z

1+4zz

=z2

1+4z

As1z2 f(

1z



The Residue Theorem

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at 0 of1z2 f(

1z

).

1z2 f(

1z

)

=1z2

1(1z

)3 (1z +4

) =1

1z

1+4zz

=z2

1+4z

As1z2 f(

1z



The Residue Theorem

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at 0 of1z2 f(

1z

).

1z2 f(

1z

)=

1z2

1(1z

)3 (1z +4

)

=1

1z

1+4zz

=z2

1+4z

As1z2 f(

1z



The Residue Theorem

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at 0 of1z2 f(

1z

).

1z2 f(

1z

)=

1z2

1(1z

)3 (1z +4

) =1

1z

1+4zz

=z2

1+4z

As1z2 f(

1z



The Residue Theorem

The Residue Theorem · 2010-10-04 · 5.We will prove the requisite theorem (the Residue Theorem)...

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Transcript of The Residue Theorem · 2010-10-04 · 5.We will prove the requisite theorem (the Residue Theorem)...