Lecture 1Functions of a Complex Variable

1 Complex Numbers and Functions

The set of real numbers is not a sufficient basis1 for the representation of the complete setof roots of algebraic equations. However, all roots can be expressed as complex numbers.Thus, the location of the singularities of a function, f(z) ( poles), and its zeros, can define thefunction. For example, in electrostatics the position of all the charges and their strengthscompletely specify the electric field. From this we expect a close connection between com-plex functions and the potential equation (Laplace’s equation) of electrostatics. The studyof functions of a complex variable are more important than what one might first suspect,because the properties of these functions can contain information about the geometry ofspace-time.

We introduce complex numbers by defining the operator, i. This operator, when appliedto a 2-D vector, rotates that vector by 90◦ counterclockwise. Then i2 represents an applica-tion of the operator twice in succession, and of course this operator can be extended to allpowers of i. Further, all real numbers are to be plotted on the x-axis, and a real numberoperated on by i is plotted on the y-axis. Thus any 2-D vector (or vector function) is written;

~f = u x̂ + v ŷ

Clearly, there is a connection between complex functions and fields in 2-D space 2

1.1 Complex Numbers and Vector Algebra

From the arguments in the last section, we restrict the discussion to 2 coordinate variables.In Cartesian coordinates define the vector functions, ~A(x, y, xa, ya) and ~B(x, y, xb, yb) at thepoint (x, y) and having lengths√

(xa)2 + (ya)2, etc. Vector algebra provides the definition for the operations of addition,multiplication, and conjugation. In the following notation the position (x, y) of the opera-tion. is suppressed.

Addition at the point (x, y)

1A basis is a collection of elements that includes all objects of a defined set2A vector field is a mathematical construction that defines a function in each coordinate direction at each

point in the 2-D space. Almost always, these functions are considered to be continuous in the coordinatevariables (x,y)

1

~A± ~B = ~W (xa ± xb, ya ± yb)

Multiplication at the point (x, y) (vector length - dot or scalar product)

~A · ~B = W (xaxb + yayb)

Conjugation (Operator C)

C A(xa, yb) = W (xa,−yb)

This latter operation is shown in Figure 1.

C A

A

−ya

ya

x

xa

y

Figure 1: The operation of charge conjugation at the point (x, y)

In polar representation the vector, ~A, is represented by a length, r, and an angle θ, 0 ≤ θ ≤2π. Figure 2 illustrates the vector function, ~A, where the coordinate origin has been placedat the point (x, y). At this point, the vector, written in complex notation is;

A = reiθ = r[cos(θ) + i sin(θ)]

Look briefly at the exponential form, eα. This is represented by the series;

eα = 1 + α/1! + α2/2! + α3/3! + · · ·

As we will later see, the ratio of successive terms, an, is ;

an+1an =

αn+ 1

2

A

r

θ

(x,y)

Figure 2: The representation of a complex number in polar coordinates

which converges for any α as n→ ∞. Then using this notation;

eiα = 1 + (iα)/1! + (iα)2/2! + · · ·

eiα = (1 − α2/2! + · · · ) + i[α− α3/3! + · · · ]

eiα = cos(α) + i sin(α)

In the same way we demonstrate that;

eα = [eα − e−α]/2 + [eα + e−α]/2 = cosh(α) + sinh(α)

Now consider the function eınθ with n an integer and 0 ≤ θ ≤ 2π. We have already the resultfor n = 1. Then for n = 2 we can work out the result by brute force.

ei2θ = [sin(θ) + i sin(θ)]2 = cos(2θ) + i sin(2θ)

In general, this can be obtained by series expansion of the exponential as previously. Thusthe complex number, z in polar form can be written;

z = r ei nθ = r[cos(θ) + i sin(θ)]n

In this case, z, is multi-valued since nθ rotates beyond 2π. Multi-valued maps will bediscussed later.

3

1.2 Mapping

We choose from the description above to let θ range between −π ≤ θ ≤ π in order to havez a single valued complex function. Then for any z = r ei(n+1)θ;

lnz = ln(r) + i[θ + 2nπ];

with n integral. Using the above convention, the negative real axis is defined as a branchcut line. Crossing the cut line selects another value of the function, Figure 3. The choice ofn = 0 provides the principle value.

−θ

+θ

Path on Primary Function

Crossing Branch Cut

Path

Branch Cut x

y

y

+θ

−θBranch Cutx

(x,y) PlaneComplex

Figure 3: The representation of a complex function z(x, y), showing a path in the complexplane lying within the principle value of the function and because it does not cross the branchcut line

Now we look a one-to-one maps to make a connection to single valued functions. Con-sider the transformation equations;

u = ax

v = ay

where a is a real number. The transformation maps the point (x1, y1) into the point (u1, v1);Figure 4. We write the transformation on an ordered pair as (u, v) → a(x, y), where thesymbol, a, represents multiplication of each element of the pair by the real number, a. Thisis a scale transformation.

A more general transformation takes the form;

4

(u,v) Plane(x,y) Plane

x

y

u

v

SlopeConstant

(u ,v )

(u ,v )

(x ,y )

(x ,y )1 1 11

2 2

2 2

Figure 4: A map representing a change of scale

u = ax− by v = bx − ayx = au+ bv

a2 − b2 y = −bu − ava2 + b2

This transformation depends on 2 real numbers, a, b. The transform introduces the map,Figure 5. It can be written in matrix notation as follows;

(u ,v )2 2

1/222[a + b ]

= tan ( )−1b/aα

(u,v) Plane(x,y) Plane

x

y

u

v

(u ,v )

(x ,y )

(x ,y )1 1 11

2 2

α

Scaled by

Rotated by

Figure 5: A map representing a change of scale and a rotation

(u, v) = (x, y)

(

a b−b a

)

This algebra also has an idenity and a zero. Now note that this transformation is equivalentto the operation of complex numbers, since a complex transformation can be defined by theoperator i.

i =

(

0 1−1 0

)

In terms of the i operator with the identity matrix, I, the above map produces the equation;

aI + i b =

(

a b−b a

)

5

The transformation of Figure 5 introduces a scale change of√a2 + b2 and rotates all rays

through and angle tan(α) = b/a. The magnitude,√a2 + b2, is called the modulus, and α is

called the argument.

A point(s) at ∞ are not really defined in Euclidian geometry. However, a working definitionof a point at ∞ is that point at which parallel rays intersect. If we take a bundle of rays anddefine a map, 1/z, all points at ∞ map to the origin. Thus, for a one-to-one map there canbe only one point at ∞.

Consider the function w(z) using complex notation. In this notation, z = x + iy. Thefunction (transformation), w(z), maps the point (x, y) into the (u, v) plane.

w(z) = u(x, y) + i v(x, y)

As an example, let

w(z) = (a + ib)z = (ax - by) + i(bx + iy)

L2 = [(ax)2 + (by)2 − 2abxy)] + [(bx)2 + (ay)2 + 2abxy]

L2 = (a2 + b2)(x2 + y2)

Note that the transformation as described in Figure 5, has a change of scale equal to√a2 + b2 and a phase addition of tan(α) = b/a

1.3 Multiplication of Complex Numbers

The multiplication of complex numbers using complex algebra is;

(a+ ib)(c + id) = (ac− bd) + i(bc + ad)

Using matrix albegra introduced in the last section;

(a+ ib)(c + id) =

(

ac 00 ac

)

+

(

0 ad−ad 0

)

+(

0 bc−bc 0

)

+

(

−bd 00 −bd

)

The above when converted to complex notation, is the expected result. In polar notationwrite;

6

(a+ ib) = r1 eiθ1

(c+ id) = r2 eiθ2

In the above; r1 =√a2 + b2, etc., and tan(θ1) = b/a, etc. Multiply these complex numbers

to get;

(a+ ib)(c + id) = r1r2 ei(θ1+θ2)

Work through the algebra using the series expansions for the sin and cos in powers of theangle to show that the result is the same as the complex number multiplication worked outabove.

2 Cauchy-Riemann Conditions

At function, f(z), has a differential at the point, z0 if the operation;

limz→z0f(z) − f(z0)

z − z0is finite and is the same no matter how z → z0. That is, we assume that the derivative isindependent of direction, if viewed in the complex plane. To be more precise;

|f(z) − f(z0)z − z0 − L| < ǫ ;

for any z sufficiently close to z0. Thus for a function to have a derivative it must have thefollowing properties.

• It must be continuous

• It must be unique

If a function is single valued and differentiable at all but a finite number of points it isanalytic. The exceptional points are called singular points (singularities). If there are nosingularities, the function is regular. Consider the series;

N∑

0

anxn

.By Cauchy’s nth root test 3, this series converges when;

3Series convergence to be discussed later

7

limn→Nmax |anzn|1/n < 1.Then define;

limn→Nmax |an|−1/n = R

The series converges when |z| < R and diverges when |z| > R. Thus R is called the radiusof convergence. Therefore, if a power series has a non-zero radius of convergence, its sumis analytic within this radius. It is also true that an analytic function can be expressed asa power series within the radius of convergence, which means that the function is analyticwithin R. For example,

∑

n

zn for n > 0 converges for z < 1. However,∑

n

(1/z)n has a singi-

larity at z = 0. An analytic function can be studied using its power series representationwithin a circle of convergence.

∑

n

an(z − zz0)n

As another example consider the extension of ex into the complex plane. A representation is;

ez = 1 +∞∑

n=1

zn

n!.This converges for all z, so the radius of convergence is R = ∞. Now z = x + iy, so thefunction can be represented by substitution in the series the term;

zn = u + iv

However;

ez = ex eiy = ex[cos(y) + i sin(y)]

Suppose we have a mapping;

w(z) = u(x, y) + i v(x, y)

For a small, real increment, h, the derivative is written;

f ′x(z) = limh→0u(x+ h, y) − u(x, y)

h−

iv(x+ h, y) − v(x, y)

h

So that;

8

f ′x(z) =∂u∂x

+ i ∂v∂x

= ux + ivx

However, if we take h imaginary then by the same analysis;

f ′y(z) = −iuy + vy

The function f is analytic so f ′x = f′y, which yields;

∂u∂x

= ∂v∂y

−∂u∂y

= ∂v∂x

These are the Cauchy-Riemann conditions, and are Necessary for a function to be analytic,but they are not Sufficient. To be sufficient the derivatives must also be cointinuous, Figure6.

(x, y)

x

y

f(z)

dxdy

The derivative is

independent of direction

Figure 6: Representation of an analytic function showing the directional derivative

3 Analytic Functions

An analytic function has a unique derivative. Thus for the function, f(x, y);

df =∂f∂x

dx +∂f∂y

dy

df = [∂f∂x

x̂ +∂f∂y

ŷ][dx x̂ + dy ŷ]

Suppose the derivative is in the direction, d~r. This is written as;

df = ~∇f · d~r

9

where ~∇f is the 2-D gradient operation on the function, f . Define z = x+ iy and f = u+ iv.the gradient operation is worked out in the following.

dfdz

= [∂u∂x

+ i ∂v∂x

]dxdz

+ [−i∂u∂y

+ ∂v∂y

]dydz

Take the derivatives of z with respect to x and y to obtain;

df = [∂u∂xx̂ + −i ∂u

∂yŷ][dx x̂ + i dy ŷ] =

[∂v∂yx̂ + i ∂v

∂xŷ][dx x̂ + i dy ŷ]

Thus the gradient is independent of the direction of the operation. The above developmentalso shows the equivalent operations of vector calculus and complex arithmatic in 2-D.

Next, take the derivative of the Cauchy-Riemann equations.

∂2u∂x2

= ∂2v

∂x ∂y= −∂

2u∂y2

The above equation shows that in 2-d Laplacian, ∇2u = 0. In a similar way ∇2v = 0. Alsousing the Cauchy-Riemann conditions, one can show that ~∇u · ~∇v = 0. Thus the gradientoperation on u and v finds the change in these functions to be perpendicular to each other,or the lines formed by u equals a constant and v equals a constant are perpendicular (sincethe gradient finds the direction perpendicular to the surface at the point of the derivative).Note, the solutions of Laplace’s equation in 2-D are analytic functions.

4 Examples of Analytic Functions

As an example of a map that is not necessarily one-to-one, define the transformation equa-tions;

ln(z) = w ;

with z = x+ iy and w = u+ iv, Figure 7.

x+ iy = eu+iv = eu cos(v) + i eu sin(v)

The real axis is obtained when y = 0. Therefore;

0 ≤ x ≤ ∞ v = 0

−∞ ≤ x ≤ 0 v = π, −π

10

z Plane w Plane

x

y

u

vπ

π/2

−π/2

−π

3

4

1

3

4

2 1

1

1

(2)

(2)

Figure 7: A representation of the map, ln(z) = w

The upper half plane (x, y) restricts 0 ≤ v ≤ π in order to obtain a single valued map.Perhaps this is more easily seen using the inverse equations;

x = eu cos(v) y = eu sin(v)

v = tan−1(y/x) u = (1/2)ln[x2 + y2]

For a single valued map, −π ≤ v ≤ π. Then points in the complex w plane are mapped intothe right half of the complex z plane. The Cauchy-Riemann conditions are easily shown tobe satisfied.

As another example, consider the simple function, f(z) = z2.

f(z) = (x+ iy)2 = (x2 − y2) + i(2xy)

u = x2 − y2

v = 2xy

Then obtain the partial derivatives to check the Cauchy-Riemann conditions,

∂u∂x

= 2x ∂u∂y

= −2y

∂v∂x

= 2y ∂v∂y

= 2x

Thus the Cauchy-Riemann conditions are satisfied. The gradient operation yields;

~∇u = 2xx̂− 2yŷ

~∇v = 2yx̂+ 2xŷ

11

Then;

~∇u · ~∇v = 0

Suppose we consider lines of constant u and v plotted in the z plane;

Equation 1 x2 − y2 = a2

Equation 2 2xy = b

In the above, U = a2 and v = b are arbitrary constants. These lines are shown in Figure 8and indentified by their equation number.

1

1

2

2

x

y

Figure 8: The map of z2 in the complex z plane for constant lines of u = a2 and v = b, asidentified by equation numbers in the text

From the figure, the slope of line 1 at the point (x0, y0) is

dydx

= x0/y0

For line 2 the slope is;

dydx

= iy0/x0 = − b2x20

12

These lines are mutually perpendicular. We also check the gradients.

~∇f = 2(x+ iy) x̂ + 2(x+ iy) ŷ

For a change in f along a direction d~s = dx x̂+ dy ŷ one obtains;

~∇f · d~s = 2(x+ iy)dx+ 2i(x+ iy)dy = 2(x+ iy)ds

This is independent of the direction d~s, only depending on the magnitude. This demon-strates that the function, f , is analytic.

5 Conformal Transformation

Note that when 2 lines cross in z plane they will also cross in the w plane. This is shown inFigure 9. The map is one-to-one if it is analytic, and if analytic, it is also conformal. Thatis it preserves the angle between the crossing lines. To demonstrate this, the differential lineelements in the z plane are given by;

dz1 = |dz1|eiφ1 dz2 = |dz2|eiφ2

θ θ

x

y

u

v

z Plane w Plane

dz

dz

1

2

dw

dw

1

2

Figure 9: A conformal map preserves the angle between two arcs in the z and w planes.

Transforming to the w plane the line elements are;

dw1 = |dz1dwdz |1 dw2 = |dz2wdz

|2

Since w is analytic, any points, say the point z = x+ iy, will have the same derivative, dwdz

,in all directions. Thus we write;

dw1 = |dz1dwdz |1 ei(α+φ1)

13

dw2 = |dz2dwdz |2 ei(α+φ2)

The difference in these arguments (angle between the lines) remains constant. One can alsoconsider the scale factors of the transformation. These are given by;

h2u = (∂u∂x

)2 + (∂u∂y

)2

h2v = (∂v∂x

)2 + (∂v∂y

)2

These scale factors are equal because of the Cauchy-Riemann conditions. Therefore anyfigure in the z plane is transformed into one of the same shape, albeit of different size, inthe w plane.

Consider the following map, w = z2. Thus w = ρ2ei2φ. The argument is simply doubled.Thus each point within the range 0 ≤ φ ≤ π in the z plane maps into the entire w plane,and the points π ≤ 2π in the z plane map into a second, overlapping w plane. The map is2 to 1. This map yields the equations;

u = x2 − y2

v = 2xy

Then consider the lines u = c1 (constant) and v = c2 (constant). These lines in the w planeare perpendicular, and parallel to the Cartesian axes. They map into hyperbolas in the zplane, Figure 10. Note the corresponding dual points in the w and z planes. Now look atLaplace’s equation under an anaytic map.

90o

90o

x

y

u

v u = c1

v = c2

2xy = c2

x − y = c2 2 1

Planez w Plane

Figure 10: A conformal map illustrating how to use mapping to find the electric potentialsand fields for a particular geometry in 2-dimensions

∇2ψ = |dwdz

|2[∂2ψ∂u2

+∂2ψ∂v2

] = 0

14

From the above, solutions for u and v in one coordinate frame transfom under an analyticmap into solutions in another frame. We will later see that this is tied to a uniquenesstheorem for solutions to eliptic, partial differential equations.

Suppose a constant surface in 2 dimensions. That is, V (x, y) = constant. As an example,this function could be an electric potential, where the electric field lines perpendicular to thissurface are obtained from, ~E = −~∇V . Note that the gradient points points in the directionperpendicular to a surface.

The complex function, F = u(x, y) + iv(x, y) which satisfies Laplace’s equation, yields func-tions which can be identified with the electric potential (say the function u), and electricfield (the corresponding function v) of a particular geometry. An analytic transformationpreserves the angle between any two arcs. Also the resulting real and imaginary functionsremain solutions to Laplace’s equation satisfying the same boundary conditions. Then con-sider the map w = z2 as discussed above, and shown in Figure 10. We can easily findsolutions to Laplace’s equation for the potential and field in the w plane. Then any analyticmap transforms this solution into another geometry, in the above example, the geometrydefined by the transformation into the z plane.

6 Riemann Integration

The operation of integration arises in two ways; 1) an operation inverse to differentiationand 2) a geometric interpretation obtained as the limit of a sum as the discrete elementsbecome continuous. The latter interpretation is most useful in when considering Riemannintegration. Assume a function, f(z), where z is complex. The function is not necessarilyanalytic, but must have a value along all points of an arc given by the paramteric equation,z = x(t) + iy(t). Then suppose the sum;

N∑

n=1

f(ǫn)(zn − zn−1) zn−1 ≤ ǫn ≤ zn

In the above ǫn lies on the arc. If this sum goes to a limit as N → ∞ and (zn − zn−1) → 0,the function f(z) in integrable.

J =∫

P

f(z) dz

Here P represents the path over which the integral is taken. As an example, suppose

f = 1z − a integated over a circular arc. Use the parametric equation;

z = a + Rcos(t) + iR sin(t) 0 ≤ t ≤ 2π

The geometry is shown in Figure 11. Note the direction as well as the path of integration.

15

The integral has the following form.

J =∫

P

1z − a dz =

2π∫

0

1R cos(t) + i R sin(t)

[−Rsin(t) + i R cos(t)]dt

Remove the complex dependence in the denominator to obtain;

J =2π∫

0

−sin(t) cos(t) + cos(t) sin(t) +

i(cos2(t) + i sin2(t)) dt

J = 2πi

x Plane

z = aDirection

Path

R

x

y

Figure 11: An Example of a Riemann integral over a circular arc

We note that the result is independent of the radius of the path R. This integral could alsobe completed in polar coordinates, letting α = z−a and z = a+Reiφ. In polar coordinatesthe integral is;

J =∮

C

dαα = 2πi

16

7 Contours

We obtain a contour (path) for integration by a parameterization, z = z(t), for some limita ≤ t ≤ b. The path is a smooth curve if it has a continuous derivative which does notvanish, and z is a single valued function of t. The contour has a direction given by theordering of points, ti. The contour can be closed so that it contains interior points andexcludes exterior points. The direction is given by the right hand rule which represents apositive rotation when the vector points upward from the plane. Any set of points in thecomplex plane is a domain, and if every contour in a domain can be continuously deformedto a point within the domain, the domain is simply connected. Examples of connected andnot connected domains are given in Figure 12.

Not Simply ConnectedSimply ConnectedSimply Connected

Figure 12: Examples of simply and not simply connected domains

8 Cauchy’s Theorem

Consider an analytic function, f(z), in some connected region. In a simply connected regionall contours enclose points within that region, and all derivatives exist, and are continuousat every point within, and on, a contour. Cauchy’s theorem states that;

∮

P

f(z) = 0

The contour is illustrated in Figure 6. It is a closed path, P . We note that this meansthat the integral only depends on its end points. In physical terms, the function, f , can beconsidered a potential function. We develop the integral as follows.

J =∑

n

f(zn) dz →∮

P

dz f(z)

J =∮

P

(u+ iv)(dx+ idy) =∮

P

(u dx− v dy) + i∮

P

(u dx + v dy)

Each of these integrals ia a real integral along the contour. We will convert these integralsusing Stokes theorem. Recall that Stokes theorem has the form;

17

zz z

zz

z

12 3

45

6

z Planex

y Contour P

Figure 13: The integral of an analytic function over a contour in a simply connected region

∮

~A · d~l =∫

(~∇× ~A) · d~σ =∫

(Ax dx + Ay dy)

In the above dσ is the differential surface area, dx dy, enclosed by the contour d~l. This isillustrated in Figure 14.

A (x,y+ y/2)∆x

A (x, y − y/2)∆x

x

y

A(x,y)

Figure 14: An illustration of stokes theorem over a differential element

The line integral around the differential area is ;

∫

~A · d~l =∑ Ax(x, y − dy/2) − Ax(x, y + dy/2)

dydx dy+

Ay(x+ dx/2, y) − Ay(x− dx/2, y)dy

dx dy

18

∫

~A · d~l =∑

[∂Ay∂x

= ∂Ax∂y

]d σ

Now for a 2-D vector in the (x, y) plane with ~A(x, y) = Ax(x, y) x̂ + Ay(x, y) ŷ integratedaround the differential loop enclosing and area, d~σ

~∇ × ~A = ẑ [∂Ay∂x

− ∂Ax∂y

]

Then make the identification that u = Ax and v = Ay. Substitute into∮

f(z) dz, above.

∮

C

[u dx − v dy] = −∫

d x d y [∂v∂x

+ ∂u∂y

]

Or if the identification that u = Ay and v = Ax is used, one obtains;

∮

C

[v dx + u dy] = −∫

d x d y [∂u∂x

+ ∂v∂y

]

Applying the Cauchy-Riemann conditions, each integral vanishes so that,∮

C

f(z) dz = 0.

This is not so surprising as analytic functions can represent a potential, and potential func-tions have no curl as they lead to conservative forces. The result is independent of theintegral path, depending only on its end points. Now extend this theorem to multiply con-nected regions. Introduce a multiply connected region as shown in Figure 15 and draw apath which connects the regions.

Not AnalyticRegion

Not ValidContour

ValidContour

x x

y y

Figure 15: A multiply connected region and a path which can be used to integrate withinthe regions

Because the enclosed region shown in the Figure 15 is not analytic, Cauchy’s Theoremcannot be applied using the contour shown on the left side of the figure. However, thecontour on the right is a valid contour if applied to all points enclosed by the curve. Asan example, consider Ampere’s law,

∮

~B · d~l. If the hatched region above contains currentfiliments, this region has singularities and so it is not analytic. You know that for the contour

19

on the left in Figure 15 one would obtain,∮

~B · d~l = I (not 0) where I is the enclosedcurrent. The contour on the right is a valid contour so that we expect

∮

~B · d~l = 0 in theenclosed region. Thus the sum of the integrals over each line segment must vanish, ie somecomponents are positive and some are negative. If the functions are not single valued thenwe need to supply a cut line and integrate on a single valued sheet.Cauchy’s theorem implies that valid contours may be changed without changing the valueof the resulting integral. This allows us to choose any contour which is appropriate for aspecific problem.

9 Cauchy Integral Formula

We suppose a function, f(z), which is analytic and single valued. Place a simple pole in thedomain at a point z = a. Then consider the integral;

∮

C

dzf(z)z − a

Let (z − a) = ρeiφ

∮

C

f(z)ρ eiφ

iρ eiφ dφ = i∮

f(z) dφ

We choose a circular contour about the pole z = a. Remember that any contour aroundthe non-analytic region is as good as any other. The resulting integral was evaluated earlier.Choose to let ρ→ 0 so that the value of f(z) is evaluated at z = a. The result is;

∮

C

dzf(z)z − a = 2πi f(a)

In the case there there are a number of poles we must sum over the all the singulatities ascan be seen by drawing contours around each pole separately. Thus suppose the functionf(z) is analytic except for a finite number of simple poles. The integral formula then becomes;

∮

C

dz f(z) = 2π i∑

Residues

In the above, a Residue is obtained by extracting one of the poles from f(z) and evaluatingthe remaining function at the position of the extracted pole.

10 Higher order poles

In this section we extend the Cauchy Integral Formula to poles of higher order. Consider aderivative of the formula for a simple pole.

20

df(z)dz

|z=a = limh→ 0 f(z + h) − f(z)h |z=a

df(z)dz

|z=a = limh→ 0 12π i h∮

dz f(z)[ 1z − a− h −

1z − a ]

df(z)dz

|z=a = limh→ 0 12π i∮

dzf(z)

(z − a− h)(z − a) =1

2π i

∮

dzf(z)

(z − a)2

Now suppose the pole is on the contour. This is shown in Figure 16. We choose a contour asshown, which goes around the pole. Remember that we integrate moving in a direction givenby the right hand rule. In this case we assume f(z) is analytic and f(z) → 0 as z → |∞|

z = a

z plane

Contour C

Contour A

Figure 16: The contour chosen to evaluate the integral when the pole lies on the contour

The integral below is defined as the principle value of f(z) as z → a.

J = P∞∫

−∞

dzf(z)z − a

The integral over the contour C vanishes as f(z) is everywhere analytic within and on thecontour. From the definition for the principle value;

J =a−ǫ∫

−∞

f(x)x− a dx +

∞∫

a+ǫ

f(x)x− a dx

This is then rewritten as;

J =∮

C+A/2

f(z)z − a dx −

∫

−A/2

f(x)z − a dz

21

In the notation above, A/2 represents the upper half circular arc and −A/2 represents thelower half circular arc integrated in the negative direction. Thus

J = 0 − [−2π i/2] f(a) = iπf(a)

For higher order poles we can use induction to obtain the formula. Above, we obtained theexpression for the first derivative. For a proof by induction, assume that the formula hasthe form;

f (n)(z0) =n!

2π i

∮

dzf(z)

(z − z0)n+1

In the above, fn represents the nth derivative. Then for fn+1;

fn+1(z0) = limh→ 0 fn(z0 + h) − fn(z0)

h

fn+1(z0) =n!

2π i

∮ dzh

[f(z)

z − z0 − h −f(z)z − z0 ]

fn+1(z0) =(n+ 1)!

2π i

∮

dzf(z)

[z − z0]n+2

11 Example

We are to demonstrate limn→ ∞ [(n/π) 11 + n2x2

] is a representation of a delta function.

That is;

∞∫

−∞

dx f(x) [(n/π) 11 + n2x2

] → f(0) as n→ ∞

We integrate in the complex plane choosing the contour in Figure 17. Note the function

(n/π) 11 + n2x2

has poles at z = ±i/n. The integral in question is along the real axis. Closethe contour by a circular loop in the upper half plane. The integral around this loop vanishesas lim z → ∞. Thus ;

∮

dz f(z) (n/π) 11 + n2x2

=∞∫

−∞

dz f(z) (n/π) 11 + n2x2

By the Cauchy Integral Theorem;

∞∫

−∞

dz f(z) [(n/π) 11 + n2x2

] =

2π i f(i/n) (1/nπ)(n/2π i) = f(i/n) → f(0) as limn→ ∞

22

−i/n

i/n

z Plane

x

y

Figure 17: The contour of integration to demonstrate that the example in the text is arepresentation of the delta function

12 Green’s Function Integral

The following integral is a representation of a delta function source for the time dependent,scalar wave function.

G(~x, t) = 14π3

∫

dk3∫

dω ei~k·~R eiωτ

k2 − (1/c2)(ω + iǫ)2

In this expression ~R = ~r−~r0 and τ = t− t0. The radial distance from the point sourceis R and the time relative to the source is τ . To implement causality, a signal cannot occurfor τ < 0 so G = 0 for all τ < 0. The ω integral is along the real ω axis. We will do this inthe complex ω plane as shown in Figure 18. To satisfy causality we move the poles whichlie on the real ω axia at ±ck to just below the real ω axis as shown in the figure. Then whenτ < 0 we close the integration coutour in the upper half plane. Note that the integrandevaluated on this loop as |ω| → ∞ vanishes, and there are no singularities within the loop.Thus by the Cauchy Theorem the result is zero, and satisfies causality. However if τ > 0then we must close the integration contour by a circle in the lower half plane to make thevalue of the integrand vanish as |ω| → ∞. In this case the loop contains poles which give anon zero result. The ω integral then becomes;

J =∮

dω ei~k·~Re−iωτ

(−ck + ω)(cl + ω)

J = −2π i ei~k·~R[e−ickτ

2ck− e

ickτ

2ck]

23

= −i ck+εω= −i ck−εω

Complex ω Plane

Im ( )

Re ( )ω

ωτ > 0

τ < 0

Figure 18: An integral to evaluate a delta function source with two poles showing how toimplement causality

J = 2π i ei~k·~R sin(ckτ)

ck

In the above the direction of integration along the lower arc gives the negative sign for theintegral. This result must be then integrated over dk3 to complete the evaluation of theGreen’s function. We do not do this here, but it results in a delta function satisfying bothcausality and relativistic retardation.

13 Branch Points and Lines

Suppose two paths in a complex plane do not yield the same value of the function. Thatis, the function is multi-valued. As an example. consider the map; f(z) = z1/2. In polarcoordinates this is written;

f(z) = r1/2 eiφ/2

The function is multi-valued because;

1) For 0 < φ < 2π f(z) = r1/2 eiφ/2

2) For 2π < φ < 4π f(z) = r1/2 ei(φ+2π)/2 = −r1/2 eiφ/2

Obviously, the number of values of the function depends on the value of the exponent. For

24

example, using z1/4 instead of z1/2 gives 4 values instead of the 2 values originally considered.The graphical representation is shown in Figure 19. The map transforms a→ A and b→ B.Crossing the axis (branch line) e→ f moves into the 2nd sheet, represented by E → F andF → E. However, moving from E to F without crosssing the branch line stays on the samesheet and the function is evaluated on the same branch. The two sheets represent the twobranches of the function. The points at z = 0 and z = ∞ are singular and represents abranch point. The value of f(z) is common at all branch points and branch lines, so thesheets are joined along the branch line. The Cauchy Integral Theorem and Formula can beapplied on each sheet independently.

a b

c

e

f

A

B

C

EF

F E

Branch

Branch

Branch

Branch

BranchLine

Line

Line Point

Point

x

y

u

v

Figure 19: A figure showing geometrically the map f(z) = z1/2 with branch cuts and pathson the sheets

As another example, consider the map f(z) =√z2 − 1, Figure 20. This map has branch

points at z = ±1. The point at ∞ is not a branch point because if z = 1/z0 then;

f =√

(1/z0)2 − 1 =

√

1 − z20z0

So that there is a simple pole as z0 → 0. The branch line then runs from −1 to +1, althoughthis line could take the long way around through ∞ and not the shorter line as shown inFigure 20. To remain on one sheet a contour must move around the branch line. The mapin the z and w planes is given by;

u = [(x2 + y2)2 − [2(x2 − y2) − 1]]1/2 cos(φ/2)

u = [(x2 + y2)2 − [2(x2 − y2) − 1]]1/2 sin(φ/2)

cos(φ) =x2 − y2 − 1

[(x2 + y2)2 − (2(x2 − y2) − 1)]1/2

sin(φ) =2xy

[(x2 + y2)2 − (2(x2 − y2) − 1)]1/2

25

Figure 20: A figure showing geometrically the map f(z) =√z2 − 1 with the branch cuts

For −1 < x < 1 amd y = 0√z2 − 1 is imaginary and φ = π/2. For x > 1 or X < 1 and

y = 0√z2 − 1 is real and φ = π (x < 0) or φ = 0 (x > 0). As a physical interpretation, the

lines u = constant could represent equi-potentials and then v represents lines of force.

Another example of a map is given by;

z = −(a/π)[w + 1 + ew]

x = (a/π)[u+ 1 + eu cos(v)]

y = (a/π)[v + eu sin(v)]

Maxwell used this map to obtain the field at the edge of a capacitor. The equi-potential andfield lines are shown in Figure 21. Where are the branch lines and points? This is an exampleof a specific type of transformation (map) called a Schwartz-Cristoffel transformation. Wedo not present this further here, as today it is easier to let a computer determine the solutionto Laplace’s equation using finite element analysis. But file this away. Some day you mayneed to look it up. The Schwartz-Cristoffel transformation maps the interior of a polygon inthe z plane into a polygon in the w plane, changing the angles at the verticies of the polygon.

14 Laurant Expansion

It is not possible to expand a function about a singular point using a Taylor expansion.However, we can develop an expansion using using a contour as shown in Figure 22. Thesum of the integral along the cut lines joining the two arcs vanishes since the direction of

26

Figure 21: A figure showing geometrically a map which provides the equi-potentials and thefields at the edge of a capacitor

integration is opposite. Thus we can apply the Cauchy Theorem and Formula for pointswithin the arcs, as the function is analytic within this region. Thus we write;

f(a+ h) = 12π i

∮

dzf(z)

z − (a+ h)This provides the value of function at a+ h. Integration over the contours C1 and C2 yieldthe following expression;

f(a+ h) = 12π i [∮

C1

dzf(z)

z − (a+ h) −∮

C2

dzf(z)

z − (a+ h) ]

The negative sign is due to the direction on the integration. Now on C1 h < (z − a) and forC2 h > (z − a). This allows a Taylor expansion for each of these integrands.

For C2

1[(z − a) − h] = (−1/h)[1 + · · · +

(z − a)nhn

+ · · · ]

For C1

1[(z − a) − h] = [−1/(z − a)][1 + · · · +

hn

(z − a)n + · · · ]

Then substitute into the integral and apply the Cauchy Formula.

27

Contour

x

y

R1

R2

h

C12C

Figure 22: Contours for the Laurant series expansion about the point z = a. The Red circlesindicate the radii from a to the singularities. The contour required to obtain the Laurantexpansion parameters is shown in blue.

12π i

∮

C2

dxf(z)

(z − a− h) =

12π i

∑

n

(1/hn)∮

c2

dz f(z) (z − a)n

12π i

∮

C1

dxf(z)

(z − a− h) =

− 12π i∑

n

(1/(z − a))∮

c1

dz f(z) hn

(z − a)n

These are combined to give a series in positive and negative powers in (z − a). The generalform is;

f(a+ h) =∞∑

n=−∞

bn hn

If f(z) is analytic within the radius of convergence from z0, we can directly apply the CauchyFormula. This gives;

bn =1

2π i

∮

dzf(z)

(z − a)n+1 =f (n)

n!Substitute the above series expression for f(z) into the expression for f(z0) and evaluate theintegral using the Cauchy Formula for several terms in the seeries.

For n = 0

28

a0 =1

2π i

∮

dz a0z − z0 = f(z0)

For n = −1

a−1 =1

2π i

∮

dz a−1(z − z0) = 0

This is because a1(z − z0) is analytic. Note that all coefficients having negative values of nwill vanish by the Cauchy Integral Theorem.

For n = +1

a1 =1

2π i

∮

dza1(z − z0)(z − z0)2

= f (1)(z0)

This is the 2nd term in the Taylor expansion of f(z − z0) evaluated at z = z0 so that onesees that a Taylor series develops.

In the case where there is a pole, suppose a function of the formg(z)

(z − z0)m with g(z) analytic.The series expansion takes the from above, and runs from positive to negative values of n.

f(z) =∞∑

N=−∞

aN (Z − Z0)n

We also know that for g(z) analytic

fm(z0) =n!

2π i

∮

dzg(z)

(z − z0)m+1

So a Taylor develops in inverse powers of z − z0 and runs through positive powers until theexpansion becomes analytic, which at higher values of n the expansion coefficients vanish.

Now consider an example, f(z) = 1(z − R1)(z −R2) . There are two poles, z = R1 and

z = R2. A Taylor series about z = 0 (here a = 0) has a radii of convergence R1 and R2. Butwe can use a Laurant expansion to obtain expansions in other regions where the function isanalytic.

f(z) =∞∑

N=−∞

an (Z − Z0)n

f(z) = 1R1R2[ 1z − R1 −

1z − R2 ]

1 ) For |z| < R1 and |z| < R2

The series representation is a Taylor expansion since f(z) is analytic within the circle ofconvergence for |z| < R1 < R2.

29

f(z) = 1R1R2[1 − R1 +R2R1R2 +

2[1 +R2/R1 +R1/R2]2!R21R

22

· · · ]

2) For R1 < |Z| < R2

This requires the Laurant expansion using the contour between the singularities. The seriesexpansion runs through negative values of n to positive of values until the series functionbecomes analytic. This again develops as a set of Taylor expansions.

1(z − R1) =

1z(1 − R1/z) = (1/z)[1 + (R1/z) + (R1/z)

2 + · · · ]

1(z − R2) =

1R2(1 − z/R2) = (1/R2)[1 + (z/R2) + (z/R2)

2 + · · · ]

Sunstitute into the expression for f(z) above.

3) For |Z| > R1 > R2

The series develops by Taylor expansion in a similar way to 2).

1(z − R1) =

1R1(1 − z/R1) = (1/R1)[1 + (z/R1) + (z/R1)

2 + · · · ]

1(z − R2) =

1R2(1 − z/R2) = (1/R2)[1 + (z/R2) + (z/R2)

2 + · · · ]

15 Analytic Continuation

The knowledge of a function can be given only in a form which is valid up to the radius of

convergence in the complex plane. As an example,∞∫

0

dt e−zt does not converge for all values

of z. However it is possible by comparing the function within the circle of convergence, toextend the representation of the function around the singularity, Figure 23. This is calledanalytic continuation. as another example, consider;

f(z) = 1 + z + z2 + · · ·

Obviously this converges for |z| < 1. However, the function f(z) = 11 − z also can be ex-pressed by the above series for |z| < 1. However, this function is valid for |z| > 1. Note herethe singularity at |z| = 1, and from the last section we can obtain a Laurant expansion. Inmost cases we can extend the function by a power series. If we extend the function by twodifferent paths and arrive at two different values of the function, the function is multi-valuedand we have extended the function onto a different sheet. There was a branch cut which

30

R

R1

2SingularPoint

Figure 23: A geometric representation of annalytic continuation

was crossed. Also remember that an analytic function can always be represented by a Taylorexpansion within its radius of convergence.

16 Method of Steepest Descent

The method of steepest descent (saddle-point method) is used to find the optimum ex-pansion of an asymptotic-series representation of a function. We discuss asymptotic seriesin a later chapter, however an asymptotic series representation of a function, f(z), is written;

f(z) = g(z)[A0 + A1/z + A2/z2 + · · · ]

where the behavior of g(z) is known as z → ∞ and;

lim|z|→∞[zn[f(z)g(z)

−n∑

p

Apzp

]] = 0

The series may diverge, so that one selects only a few, usually no more than 2, terms. It isthen necessary to obtain an optimal expansion. Begin by assuming an integral representationof the function. This allows an asymptotic series if the integral over a path, C, has the form;

J(z) =b∫

a

dt ezf(t)

The integral must vanish at the ends of the contour. Although this appears restrictive, itcan be used in many cases of physical interest. For example, the gamma function;

31

Γ(z + 1) =∞∫

0

dt e−t tz = zz+1∞∫

0

dt ez(ln(t)−t)

The above is obtained by substituting t→ tz. Using the integral for J(z) above, note that forlarge values of complex |z|, the integran has rapid oscillations, making the integral difficultto evaluate. Thus it is desirable to deform the contour to keep the real component of thefunction large while maintaining a constant imaginary component. In general, the contourruns through regions where Re[zf(t)] is either greater or less than 0. Regions where it is> 0 are clearly most important and we look for regions where this is a maximum. We alsolook for regions where Im[zf(t)] is constant. However, to complete the contour, we minimizeRe[zf(t)] and allow rapid oscillations induced by the imaginary component. We wish to finda t0 where;

dfdt

= 0

and the imaginary component is constant. The point in the complex plane where this occursis illustrated in Figure 24. Now |f(t)| cannot have a true maximum or minimum, but canhave a saddle point since the function is analytic. Therefore

Re[t]

Im[t]

t = to

ContourSurfaceFlat

Re[f(t)]

Figure 24: The surface in which the integral is evaluated showing the saddle point over whichwe take the path for the integral

∂2U∂x2

+ ∂2U∂y2

= 0

This means that the slope in the x direction is the negative of the slope in the y direction.The surface has a saddle point over which we take the integration. Now we wish Im[zf(z)]to be constant. Thus write z = u+ iv. Along the path to be chosen,

dv = ~∇v · d~s = |~∇v| |d~s| cos(θ)

32

The above the path is d~s = dx x̂ y ŷ, and the path perpendicular to the is d~s′ = −dx ŷ + dy x̂so that d~s · d~s′ = 0. Then ;

dv = 0 = ∂v∂x

dx + ∂v∂y

dy

Use the Cauchy-Riemann conditions.

0 = −∂u∂y

dx + ∂u∂x

dy

0 = |~∇u| |d~s′| cos(θ)

However, if we change to the path d~s which is perpendicular to d~s′, then this path liesalong the gradient and has the maximum change of u. So we see that trhe point of constantv represents a point of maximum change of u.

The first term in the series expansion is obtained from the following evaluation.

J(z) = eiIm[z f(z)]∫

C

dt eRe[z f(t)]

In the neighborhood of the saddle point f ′(t0) = 0. By Taylor expansion;

f(t) = f(t0) + [(t− t0)2/2] f ′ ′(t0) + · · ·

Choose the path so the integral is a decreasing exponential.

τ = [√

ei(π+φ)f ′′(t0)](t− t0)

where z = |z|eiφ

J(z) = ezf(t0)

√

ei(π+φ) f ′′(t0)

∫

C

dτ e−|z|/2 τ2

As |z| → ∞ the integral becomes steeper and less of the contour becomes inportant. Forsufficiently large |z| replace the contour by the integral between −∞ and ∞.

J(z) = ezf(t0)

√

ei(π+φ) f ′′(t0)

∞∫

−∞

dτ e−|z|/2 τ2

The result becomes

J(z) = ez f(t0)√

2πz eiπ f ′′(t0)

33

As an example consider the Gamma Function, Γ.

Γ(z + 1) =∞∫

0

dτ e−τ τ z

Integrate by parts to obtain the result;

Γ(z + 1) = zz+1∞∫

0

dt e−tz tz = zz+1∞∫

0

dt ez(ln(t)−t)

Identify f(t) = ln(t)− t so that f ′(t) = 1/t− 1. This is set equal to 0 to find t0 = 1. Thus;

f(t0) = −1f ′(t0) = 0f ′′(t0) = −1τ = [ei(π+φ) f ′′(t0)](t− t0)

τ = (t− 1) eiφ/2 0 < τ < ∞

The result is Stirling’s approximation for the factorial function.

Γ(z + 1) → |z→∞√

2π zz+1/2 e−z

17 Dispersion Relations

When we considered the Green’s function integral for the scalar wave function in the fre-quency domain, we noted that causality required that no wave could develop before thewave event ocurred at the source. This means that the function must be analytic for τ < 0where τ measures the time of the wave at a position in space relative to the event occurranceat the source. When the wave propagation was analyzed in frequency space, there was animaginary exponential eiωτ so that the integration could be completed in the complex plane.Then integration around a closed loop where τ < 0 must vanish from causality and by theCauchy Theorem this means that the function is analytic in the upper half of the complexω plane. Thus many physical functions must have similar properties. This behavior can beexploited to define general properties of the functions.

It is importaht to understand the analytic properties of an integrand. Suppose a functiong(E) decreases at least as fast as 1/E as |E| → ∞ for Im(E) > 0. Then we have;

|g(E)| ≤ A|E| |E| → ∞

34

For a path along the real axis, close the contour in the upper half plane. For reference, seeFigure 18 which was given earlier. Thus;

g(E) = 1iπ P∞∫

−∞

dE ′g(E ′)E ′ − E

In the above integral, P represents the principle value of the integral. Then the real andimaginary components of the function are related. We already know that there should bea relation between real and imaginary components of analytic functions from the Cauchy-Riemann relations.

Re[g(E)] = 1π P∞∫

−∞

dE ′Im[g(E ′)]E ′ −E

Im[g(E)] = −1π P∞∫

−∞

dE ′Re[g(E ′)]E ′ − E

These are dispersion relations, and entered physics with the work of Kronig and Kramersin optics. The name relates to the conntection between wavelength and frequency. Thusthe index of refraction may have an imaginary component which relates to absorption of thewave. The dispersion relation connects the real component to an integral of the imaginarycomponent. This has been exploited in other equations describing physical processes. Inaddition suppose we have the symmetry, g(−E) = g∗(E) then;

Re[g(E)] = Ee[g(−E)] Im[g(E)] = −Im[g(−E)]

So that Re[g(e)] is an even function and Im[g(E)] is an odd function. This is called crossingsymmetry in quantum mechanics. Use this symmetry to write;

Re[g(E)] = 1π P0∫

−∞

dEIm[g(E)]E − E0 +

1π P

∞∫

0

dEIm[g(E)]E − E0

Re[g(E)] = 2π P∞∫

0

dEE0 Im[g(E)]E2 − E20

A similar expression can be obtained for the imaginary component. As a simple example,look at the motion of a charge, q, acted on by a field, Ee−iωt. The equation of motion is;

m[ẍ+ γẋ+ ω0x] = −qEe−iω t

The solution has the form x = x0eiω t and the magnitude of the dipole moment of the motion

is p = qx. The polarization of the medium, P, is the dipole moment per unit volume, so if Nmolecules per unit volume with α electrons of charge q are moving due to the field, we obtain;

p = qx =q2Ee−iω t

m[ω2 − ω20 + iωγ]

35

P =q2Nα

m[ω2 − ω20 + iωγ]E e−iω t

The dielectric constant, ǫ, of the material is obtained from;

D = ǫE = ǫ0E + P

ǫ = ǫ0n2

In the above D is the electric displacement, ǫ0 is the permitivity of free space, and n is theindex of refraction. The expression for n2 has a resonant behavior with poles below the realaxis [ω = −i(γ/2) ±

√

ω20 − (γ/2)2]. It is analytic above the real axis as it must be to obeycausality. Note however, that n2 → 1 as ω → ∞. There is then some algebra to obtain thefinal dispersion relation result.

Re[ǫ(ω)] = 1 + (2π)P∞∫

0

dω′ω′ Im[ǫ(ω′)]ω′ 2 − ω2

Im[ǫ(ω)] = −(2π)P∞∫

0

dω′ω Re[ǫ(ω′)] − 1

ω′ 2 − ω2

These are the Kronig-Kramers relations.

36