Notes_examples.pdf

Complex varibles:Contour integration examples

1

Problem 1

Consider the problem

Z ∞

dx x2 + 1

I= −∞

If we take the substitution Z

2

x = tan θ then dx = sec θdθ, I = which leads to

sec2 θ dθ tan2 θ + 1

Z

I=

dθ = π

Next we consider the same problem using complex variables and a contour integration in the complex plane. We can recall from our earlier introduction to complex variables that if we have the given integral on a closed contour I

f (z)dz = 2πi

X

C

Res(f (zn ))

n

where zn are the poles of f(z). Now, suppose we have, dz +1 C where the contour C is shown in figure 1 with the curved portion being a semi-circle with an infinite radius. The integrand has poles at Z = i. Then I

I2

I

I2 = C

dz = z2 + 1

Z ∞ −∞

dz + z2 + 1

z2

dz i = 2πiRes(f (z = i)) = 2π |z=i = π z2 + 1 z+i

Z C

and I2 = I provided dz =0 +1

Z

z2

C

which we have to prove. Considering the integral Z C

at R = ∞ we substitute z =

Reiθ

and get

iReiθ dθ | ≤ lim R→∞ 0 R2 e2iθ + 1 R→∞ where we have used the inequality | lim

Z π

dz +1

z2

Z π 0

|

iReiθ dθ | ≤ lim R→∞ R2 e2iθ + 1

Z π 0

|

Rdθ πR | = lim 2 →0 2 R→∞ R − 1 R −1

|Z1 + Z2| ≥ ||Z1 | − |Z2 || And so we find that using contour integration and the residue theorem, Z ∞

I= −∞

dx =π +1

x2

1

Im

R=

8

C x Re x Figure 1: Contour for problem 1.

2

Problem 2

Consider

x2 dx x4 + 1

Z ∞

I= −∞

Again we replace this with I

I2 = C

z2 dz = 4 z +1

Z ∞ −∞

x2 dx + 4 x +1

Z C

X z2 dz = 2πi Res(f (z = zn ) +1 n

z4

π

π

π

π

Here the integrand has the following poles, Z = ei 4 , e3i 4 and Z = e−i 4 , e−3i 4 . The contour remains the same, except that now there are two poles within the contour as given above. Thus, the solution is π

I2 = 2πi

π

(ei 4

π

(ei 4 )2 (e3i 4 )2 + 2πi = π π π π π π π π π π π − e3i 4 )(ei 4 − e−i 4 )(ei 4 − e−3i 4 ) (e3i 4 − ei 4 )(e3i 4 − e−i 4 )(e3i 4 − e−3i 4 )

2πi

π

(ei 4

i −i √ √ + 3i π π π π 3i π4 i π4 −3i π4 i 3i − e )(i 2)(e − e ) (e 4 − e 4 )(e 4 − e−i 4 )(i 2)

i 1 1 1 2πi √ + 3i π π i π4 3i π4 i π4 −3i π4 4 i 2 (e − e ) (e − e ) (e − e−i 4 ) √ π π π π ! 2πi (e3i 4 − e−i 4 ) + (ei 4 − e−3i 4 ) = π π −1 − 1 − 1 − 1 (ei 4 − e3i 4 ) √ √   2πi 2 2isin(3π/4) + 2isin(π/4) = 1 + i − (−1 + i) −4 2

!!

=

!

=

√ ! π 2 2i =√ −4 2

iπ

3

Problem 3: sinx/x

Integration of sin x/x from −∞ to ∞ is an interesting problem

3.1

Method 1

In the first method let us consider Z ∞ iax e −∞

x

dx =

Z ∞ cos(ax)

x

−∞

dx + i

Z ∞ sin(ax) −∞

x

dx

So when Z is real the integral is the imaginary part of the above integral. So if we select a contour as shown below, then on the real axis, Z = x and if we can find the values of the other integrals we can find the answer as the imaginary part of the real line integral. Thus, we replace the integral with Z − iaz Z Z ∞ iaz Z I e eiaz e eiaz eiaz dz = dz + dz + dz + dz z −∞ z C z  CR z C z As  tends to zero we find that the first and the third part give us the real line integral Z ∞ iax e

x

−∞

dx

As for CR integral, using Jordan’s lemma, it goes to zero for a > 0. With regard to C we have for Z eiaz dz C z z = 0 is a simple pole. Thus, one of the theorems says Z C

eiaz dz = iφC−1 = −iπ z

where C−1 is the residue at the pole. φ = −π because we go in the CW direction. Thus we have I C

eiaz dz = z

Z ∞ iax e

x

−∞

dx − iπ = 0

since no pole is enclosed. Thus, Z ∞ iax e −∞

Matching real to real etc., we have

x

dx = iπ

Z ∞ sinax −∞

x

3

dx = π

CR

C

3

-R

R z=0

Figure 2: Contour for problem 3.

3.2

Method 2

In the second method, we will include the pole within the contour. Thus, I C

eiaz dz = z

Or I C

Z − iaz e −∞

eiaz dz = z

z

Z

dz + CCC

Z ∞ iaz e −∞

Or

z

Z

dz + C CC

Z ∞ iaz e −∞

Or

eiaz dz + z

z

z



Z

dz + CR

eiaz dz z

eiaz dz = 2πi(ResF (z = 0)) z

dz + iπ = 2πi

Z ∞ iaz e −∞

Z ∞ iaz e

z

dz = πi

And the result follows.

3.3

Method 3 Z ∞ sin(ax) −∞

x

dx =

Z ∞ iax e − e−iax −∞

2ix

dx →

Z ∞ iaz e − e−iaz −∞

2iz

dz

The logic here is this: The first integral (what we want) is exactly equal to the second representation. We make it into a complex integral (with real limits) such that the complex integral will equal the two prior integrals when z = x. Hence the complex integral (with real limits) must be a branch of any contour we choose. Next, since the complex integral is a sum then it has to be split

4

into its components and then each integral has a singularity at z = 0. We must then calculate the penalty for integrating through the singularity. The first term is Z ∞ iaz e

dz

2iz

−∞

If this is integrated over the same contour eiaz dz = 2iz

I C

Z ∞ iaz e −∞

Z

dz +

2iz

C Cw

eiaz dz + 2iz

Z CR

The CR integral is zero. Or eiaz dz = 2iz

I C

Or I C

Z ∞ iaz e

2iz

−∞

eiaz dz = 2iz

Or

Z

dz + C Cw

Z ∞ iaz e

2iz

−∞

Z ∞ iaz e −∞

The second term is −

2iz

dz +

2iz

−∞

−iπ =0 2i

π 2

dz =

Z ∞ −iaz e

eiaz dz = 0 2iz

dz

If this is integrated over the same contour I C

e−iaz dz = 2iz

Z ∞ −iaz e

2iz

−∞

Z

dz + C Cw

eiaz dz + 2iz

Z

→∞

CR

The CR integral is infinity. So we take the bottom contour and include the pole. I C

e−iaz dz = 2iz

Or I C

Z ∞ −iaz e −∞

e−iaz dz = 2iz

Or

2iz

Z

dz + C Cw

Z ∞ −iaz e −∞

2iz

dz +

Z ∞ −iaz e −∞

Thus −

2iz

So

Z ∞ iaz e − e−iaz −∞

2iz

−π = (−1)2πiRes(F (z = 0)) 2

dz = +

Z ∞ −iaz e −∞

eiaz dz = −2πiRes(F (z = 0)) 2iz

2iz

π −π 2

dz = +

dz = π =

π 2

Z ∞ sin(ax) −∞

5

x

dx

3.4

Method 4

It should be mentioned here that if an integral has to go through a singularity then there is no choice but to go through it using an  contour and see if this is a integrable singularity. There is no avoiding it by deforming the contour. Avoiding the ’singularity’ by starting with an indented contour can be done only when the function is analytic in the domain. With a singularity it is not. However, in this case, there is no singularity in sin(ax)/x to begin with at x = 0. So when we break it into Z ∞ sin(ax)

x

−∞

dx =

Z ∞ iax e − e−iax

2ix

−∞

Z ∞ iaz e − e−iaz

dx →

−∞

2iz

dz

the complex integral is analytic all the way along the real axis. Thus, we can deform the contour as we want. Hence, we choose an indented contour to begin with. The value of the integral must be the same. First let us observe that around the origin I

lim

→0 

eiaz − e−iaz dz, 2iz

withz = iθ ,

dz = iiθ dθ

we get I

lim

→0 

eia(cosθ+isinθ) − e−ia(cosθ+isinθ) iθ i dθ = 0 2iiθ

Thus, the function is not singular and has no integral contributions from z = 0. So nothing new can come integrating through the z = 0 point. We now choose to deform the already indented path. If we continuously deform the path now to +∞ then one of the exponents becomes non-analytic at z = ∞. (If the value of a function becomes infinity then it has a singularity at that point). The singularity can be avoided by deforming the contour, as long as the resultant integrands acquire finite values else they are considered to hit a singularity. We will now be forced to break the integrals into two parts. The first chosen path is indented not going through the z = 0 point since there is no reason to. So one of the exponents goes to zero as we go to infinity in the upper half. The other exponent is not analytic and so has to go to −∞. Here it will go through a singularity. So we deform it the other way. So for the first integral we have Z ∞ iaz e dz −∞ 2iz If this is integrated over the same contour I C

eiaz dz = 2iz

Z ∞ iaz e −∞

2iz

Z

dz + C Cw

eiaz dz + 2iz

Z CR

We will hold onto the end points, i.e., +∞ and −∞, and continuously deform the contour upwards, into CR at infinity. The value of this integral by Jordan’s Lemma is zero.

4

Problem 4

Consider I=

Z ∞ cosx − cosa −∞

x2 − a2

6

dx, a real

CR

C 2

C 1

3

3

-R

R

Figure 3: Contour for problem 4. This integral is convergent and well defined at x = ±a because the lHospitals rule shows cosx − cosa −sinx −sina = lim = 2 2 x→±a x→±a x −a 2x 2a lim

To convert the problem to complex domain we consider I

J= C

eiz − cosa dZ, a real z 2 − a2

where C is the contour in figure 3. Here the original function has no singularity. However, once we define the complex function it has a singularity and finally we must take the real part of the answer. So the real argument carries throughout the derivation. Thus, we must stick with the original real line integral for the complex integrand as one branch of the contour. Because, only then after taking the real argument, the original integral is available on the real axis (z = x). Next, no poles are enclosed by C and so I

0 = C

eiz − cosa dZ z 2 − a2

Z −a−1

Z a−2

+

= −R

Z R

+ −a+1

C1

!

+

+

+ a+2

Z

Z

Z

CR

C2

eiz − cosa dZ z 2 − a2

(1)

From theorems 4.2.1 and 4.2.2 (Jordans lemma), the integral along CR is zero . From theorem 4.3.1, the integrals around the small arcs become Z

lim

1 →0 C 1

and Z

lim

2 →0 C 2

eiz − cosa dz = −iπ z 2 − a2

eiz−cosa 2z

eiz − cosa dz = −iπ z 2 − a2

!

|z=−a =

eiz−cosa 2z

!

|z=a =

Thus as R → ∞ (in the CPV sense) I=

Z ∞ cosx − cosa −∞

x2 − a2

dx = −

and hence by taking the real part, I=−

πsina a

7

πsina a

πsina 2a πsina 2a

Discussion: The original integral is not singular at the two denominator zeros. Thus, we could choose the indented contour for the complex integral provided we choose the complex integrand as Z cos(z) − cosa J= dZ, a real z 2 − a2 C which also has no singularity at (z = ±a). In this case we can start with an already indented contour and proceed like we did for the earlier problem. However, the function could also be eiz − cosa dZ, a real z 2 − a2

Z

J= C

where we have to take the real value at every step. However, there is now a singularity on the real axis and if we must obtain the original integral we must stick to the real axis limits. Thus, the singularity is unavoidable. We cannot choose an indented contour.

5

Problem 5

Consider I=

Z ∞ xdx

x3 + 1

0

We write it as

dx

∞ xdx zdz zdz = + + 3+1 3+1 3+1 z x z C 0 CR For finding the singularities of the integrand we have

I

Z

Z

Z

J=

C

zdz z3 + 1

z3 + 1 = 0 ⇒ z 3 = e(iπ+i2πk)

(2)

z = e

(iπ/3+i2πk/3)

z = e

(iπ/3)

(iπ)

,e

, k = 0, 1, 2 (i5π/3)

,e

(3) (4)

Below we take a closed contour which encloses only the first pole. One can see that the CR integral goes to zero. Thus, J=

Z ∞ xdx 0

x3

+1

Z

+ C

The residue is

p e(iπ/3) | = (iπ/3) q 0 z=e 3e(2iπ/3)

Res(e(iπ/3) ) = Thus

Z ∞ xdx

zdz = 2πi Res(e(iπ/3) ) +1

z3

zdz 2πi (−iπ/3) = e +1 +1 3 0 C Let us look at the integral along the straight line contour C. J=

x3

Z

+

Z C

z3

zdz +1

z3

Set z = reiθ , dz = dreiθ z = 0, r = 0

(5)

z = ∞, r = ∞

(6)

8

CR x

C

Figure 4: Contour for problem 5.

Z C

zdz = 3 z +1

Z 0 reiθ dreiθ ∞

r3 e3iθ + 1

=−

Z ∞ rdre2iθ 0

r3 + 1

= −Iei4π/3

since θ = 2π/3. So 2πi (−iπ/3) e 3 2π I= √ 3 3

I(1 − ei4π/3 ) =

6

Problem 5a

Let us do the same problem using the method of contour deformation. We consider the original integral I as an integral of a complex integrand. I=

Z ∞ zdz

z3 + 1

0

where the integral is still on the real axis. Now we deform this straight line ’contour’ as shown in the figure below. Since we cannot cut across a pole, the contour encircles the pole with two

CR x

C

Figure 5: Contour for problem 5a. opposite going segments. The function is well defined on these segments. Thus we get I = Ic + ICR + Itwo−opposites + Iround−the−pole ICR and Itwo−opposites are zero so we are left with I = Ic + Iround−the−pole

9

Ic from the previous method is Iei4π/3 (mind the sign). Thus, I(1 − ei4π/3 ) = Iround−the−pole = 2πi Res(e(iπ/3) ) This gives the same answer as before.

7

Problem 6

Consider I=

Z ∞ 2 x dx

x6 + 1

0

We replace it with I

J= C

z 2 dz z6 + 1

As before the poles are z6 + 1 = 0 ⇒ z 6 = e(iπ+i2πk)

(7)

z = e

(iπ/6+i2πk/6)

, k = 0, 1, 2

z = e

(iπ/6)

....

(iπ/2)

,e

(8) (9)

We can do this in more than one way. We take the following contour where ’x’ is the location of

C

CR x

I

Figure 6: Contour for problem 6. the first pole. Only one pole is enclosed. We get I

J= C

z 2 dz = I + ICR + Ic z6 + 1

We can see that ICR is zero. Thus, J = I + Ic = 2πi Res(eiπ/6 ) For Ic , we do the following z = reiπ/3 ⇒ dz = dreiπ/3 Z 0 2 r dreiπ eiπ/3 J =I+ = 2πi = π/3 6 6e5iπ/6 ∞ r +1 Or 2I = π/3 or I = π/6 10

8

Problem 7

Consider I=

Z ∞ ax e dx

−∞ az e dz

I

J=

1 + ez

C

1 + ex

,

= I + [extra terms]

Z −R+i2π

Z −R+i2π

Z R+i2π

+

+

J =I+

a<1

−R

R

R+2iπ

J = I + term2 + term3 + term4 The poles are at z = e−iπ+i2πk , k = 0, 1, 2.. See the contour below X

X i 3π -R+i 2π

R+i 2π

i 2π Xiπ

R

-R

Figure 7: Contour for problem 7.

8.1

term 2 lim

Z −R+i2π az e dz

R→∞ R+2iπ

1 + ez

z = t + i2π dz = dt

(10)

z = R + i2π ⇒ t = R

(11)

z = −R + i2π ⇒ t = −R

(12)

The integral becomes lim

Z −R+i2π az e dz

R→∞ R+2iπ

1 + ez

= lim

Z −R ai2π at e e dt

R→∞ R

1 + et+i2π

11

= lim

Z −R ai2π at e e dt

R→∞ R

1 + et

= −Ieia2π

8.2

term 3 lim

Z R+i2π az e dz

1 + ez

R→∞ R

z = R + ip dz = idp

(13)

z = R⇒p=0

(14)

z = R + i2π ⇒ p = 2π

(15)

The integral becomes lim

Z 2π aR aip e e idp

R→∞ 0

1 + eR eip

Z 2π

e(a−1)R

= lim

R→∞ 0

eaip idp =0 e−R + eip

since a < 1. Similarly term 4 is also zero. We have the residue at z = iπ left. The residue is eiaπ eiπ Finally considering I and term 2 along with the residue we get I(1 − eia2π ) = −2iπeiaπ I(−2i)sin(aπ)eiaπ = −2iπeiaπ Or I=

9

π sin(aπ)

Problem 7a

We will do the same problem by the method of deformation of contours. Consider I=

Z ∞ ax e dx −∞

1 + ex

Z

= C

eaz dz 1 + ez

Now it is an integral on the complex domain. If we continuously deform the contour the value of the integral remains the same as long as we do not cut across a pole. Thus, we continuously deform the contour upward and as we hit a pole we deform the contour round it. The length of the contour keeps stretching. There will be vertical parts of the contour going in opposite directions and the poles get encircled in the anti-clockwise direction. We go till infinity upwards and in R. Thus the CR integral goes to 0. Thus, we are left with integrations around the poles, i.e., residues only !

e3iaπ e5iaπ eiaπ + + + ... J = 2πi eiπ eiπ eiπ Or





J = −2πi eiaπ + e3iaπ + e5iaπ + ... Using the geometric series formula with r = eia2π 



J = −2πieiaπ 1 + e2iaπ + e4iaπ + ... 12

X i 5π CR X i 3π

X iπ R

-R

Figure 8: Contour for problem 7a. Or

eiaπ 1 − e2iaπ eiaπ J = −2πi iaπ −iaπ e (e − eiaπ ) π J= sin(aπ) J = −2πi

10

Problem 8

Consider

Z ∞

I=

√

0

dx x(1 + x2 )

on a key hole contour given below

CR

Cε

Real axis

Figure 9: Contour for problem 8. We replace the integral by dz z(1 + z 2 ) C The square root function has a branch cut along the positive x axis. So we get I

J=

Z ∞

J=

√

+ 

Z 

Z

+ CR

Z

+ ∞

C

The CR integral is zero. For the C we have z = eiθ dz = ieiθ dθ Z 0

Z

= lim C

→0 2π

√

eiθ idθ eiθ/2 (1 + 2 ...) 13

=0

(16) (17)

Thus, we are left with

Z ∞

Z 

= upper + Lower Branch

+

J=

∞



z = reiθ dz = dreiθ Z r

J = lim

r→∞

0

√

dreiθ reiθ/2 (1 + r2 e2iθ ) Z r

J = lim

r→∞

0

|θ=0 +

Z 0 r

√ iθ/2 |θ=2π re (1 + r2 e2iθ )

dr √ |θ=0 + r(1 + r2 ) Z r

J = lim 2 r→∞

√

0

Z 0

√

r

dr iπ re (1 + r2 )

(18)

!

dreiθ

(19)



dr = 2πiRes(i, −i) r(1 + r2 )

The residues at i, −i. √ √

1 1 |z=i + √ |z=−i z(z + i) z(z − i)

1 1 |z=eiπ/2 + √ | i3π/2 z(z + i) z(z − i) z=e 1 eiπ/4 2i 2I =

1 2i

1

+

ei3π/4 (−2i)

1



+

1



eiπ/4 e−iπ/4 1 2I = 2cos(π/4)2πi 2i π I = cos(π/4)π = √ 2

11

2πi

Problem 9

Consider the integral I=

Z 1 √ 1 − x2 −1

1 + x2

We replace this with the contour integral √ I

J= C

z2 − 1 1 + z2

Using polar coordinates (z 2 − 1)1/2 =

√ ρ1 ρ2 ei(φ1 +φ2 )/2 , 0 ≤ φ1 , φ2 < 2π

where (z − 1)1/2 = (z + 1)1/2 =

√ √

ρ1 eiφ1 /2 , ρ1 = |z − 1| ρ2 eiφ2 /2 , ρ2 = |z + 1| 14

(20)

with this choice of the branch we find  √ x2 − 1     −√x2 − 1

√

(z 2 − 1)1/2 =

 i √ 1 − x2    −i 1 − x2

1<x<∞ − ∞ < x < −1 − 1 < x < 1, y → 0+ − 1 < x < 1, y → 0−

Using the expressions in the contour integral J, it follows that √ Z Z −1+1 Z Z  2 Z 1−2 √ −i 1 − x2 i 1 − x2 (z − 1)1/2 dx + dx + + + dz J= 2 1 + x2 1 + z2 1−2 −1+1 1 + x C1 C2 CR "

(z 2 − 1)1/2 2z

= i2π

!

|eiπ/2 +

(z 2 − 1)1/2 2z

!

#

|ei3π/2

(21)

We note that the crosscut integrals vanish Z 

Z

+ L0

Li

Ce1

(z 2 − 1)1/2 dz = 0 1 + z2

z=i

CR

Ce2

z=1 z=-i

z=-1

Figure 10: Contour for problem 9.

r2=|z+1| r1=|z-1| t1

t2

z=-1

z=1

Figure 11: Square root function of problem 9. because L0 and Li are chosen in a region where (z 2 − 1)1/2 is continuous Rand single valued, and L0 and Li are arbitrarily close to each other. Theorem 4.3.1a shows that Ci → 0 as  → 0, i.e., |

Z Ci

(z 2 − 1)1/2 dz| ≤ 1 + z2

Z 2π (|2i e2iθ − 1|1/2 )

1 + 2i e2iθ

0

Z 2π

< 0

q

2i + 1

1 − 2i

i dθ using the inequalities at the end q

i dθ = 15

2i + 1 Z 2π i dθ → 0 as i → 0 1 − 2i 0

(22)

The contribution from CR is calculated as follows Z CR

(z 2 − 1)1/2 dz = 1 + z2

Z 2π (R2 e2iθ − 1)1/2

1 + R2 e2iθ

0

iReiθ dθ

We note that (R2 e2iθ − 1)1/2 ≈ Reiθ as R → ∞ because the chosen branch implies limz→∞ (z 2 − 1)1/2 = z. Hence, Z (z 2 − 1)1/2 lim dz = 2πi R→∞ CR 1 + z2 Calculations of the residues requires computing the correct branch of (z 2 − 1)1/2 . √ √ i(3π/4+π/4)/2 z2 − 1 2e 1 |z=eiπ/2 = =√ 2z 2i 2 √ √ i(5π/4+7π/4)/2 2 z −1 2e 1 |z=ei3π/2 = =√ (23) 2z −2i 2 Taking i → 0, R → ∞ and substituting the above results in the expression for J, we find Z 1 √ √ 1 − x2 2i dx = 2iπ( 2 − 1) 2 −1 1 + x Thus √ I = π( 2 − 1)

12

Some useful theorems

Theorem 4.2.1 Let f (z) = N (z)/D(z) be a rational function such that the degree of D(z) exceeds the degree of N (z) by at least two. Then Z lim

R→∞ CR

f (z)dz = 0

Theorem 4.2.2: Jordan Lemma Suppose on a circular arc CR that is in the upper half of the complex plane, f (z) → 0 uniformly as R → ∞ then Z lim f (z)eikz dz = 0, k > 0 R→∞ CR

Alternatively Suppose on a circular arc CR that is in the lower half of the complex plane, f (z) → 0 uniformly as R → ∞ then Z lim f (z)e−ikz dz = 0, k > 0 R→∞ CR

Theorem 4.3.1a Suppose on a contour C , as shown in the figure below, (z − z0 )f (z) → 0 unif ormly as  → 0, then lim

Z

→0 C

f (z)dz = 0

Theorem 4.3.1b Suppose f (z) has a simple pole at z = z0 with a residue= C−1 then Z

lim

→0 C

f (z)dz = iφC−1

where φ is the angle subtended by the arc and is taken positive in the anti-clockwise sense. 16

Ce f

Z0 Figure 12: Figure for a theorem.

13

Some useful inequalities

If z1 and z2 are any two complex numbers |z1 + z2 | ≤ |z1 | + |z2 | |z1 + z2 | ≥ ||z1 | − |z2 || |z1 − z2 | ≤ |z1 | + |z2 | |z1 − z2 | ≥ ||z1 | − |z2 || (24)

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Finding residues at singularities

If we have a function that is a ratio of two polynomials, for example, z , singular at z = 1, and z = 2 (z − 1)(z − 2) • then the residue at z = 1 is found as follows: z z Res(z = 1) = (z − 1)|z=1 = |z=1 = −1 (z − 1)(z − 2) (z − 2) • the residue at z = 2 is found as follows: z z Res(z = 2) = (z − 2)|z=2 = |z=2 = 2 (z − 1)(z − 2) (z − 1) • If f (z) is a ratio of two polynomials p(z) and q(z) then the residue at z = zo is also found as p(z) p(zo ) Resz=zo = 0 q(z) q (zo ) 0 provided q (zo ) 6= 0. If q(zo ) = 0, then this is the only way. • If f (z) =

φ(z) (z − zo )m

If m = 1 Resz=zo f (z) = φ(z = zo ) and if m ≥ 2 then Resz=zo f (z) =

1 dm−1 φ(z = zo ) (m − 1)! dz m−1

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