· complex integration and methods contact: [email protected] 5/30/2012 prerequisites: the...

Complex Integration and Methods

Contact: [email protected] 5/30/2012

Prerequisites: The complex functions handout Concepts of primary interest: ( ) 0f z dz =∫ for f(z) analytic in the interior Laplace property Poles of an analytic function Laurent series Residue theorem: ( ) 2 Res( ( )mf z dz i f zπ= ∑∫

Cauchy’s integral formula 1 ( )( )2 C

f z dzf ai z aπ

=−∫ for f(z) analytic

Sample calculations: SC0: Generating a Laurent series SC3: Finding a residue SC5: Contour morphing to compute a sum SC6: Laplace property SC7: Standard contour integration SC9: Cauchy integral formula SC10: Integrals of trigonometric functions SC11: Integral representation of a special function SC12: Step function as a contour integral Applications Tools of the Trade Evaluating residues Integrals of real functions along the entire real axis: Closing contours with semicircular arcs in the upper and lower halfplane -

Jordan's Lemma Integrals of real function over the positive real axis symmetry and pie wedges. Principal Value integrals Winding number Modified residue theorem

*** Section not proofed.

A basic knowledge of complex methods is crucial for graduate physics. This handout only illustrates

a few of the standard methods, and the developments are not rigorous. For additional study,

download the online text: Mathematical Tools for Physics by James Nearing from

www.physics.miami.edu/nearing/mathmethods/. If you interest swells to the twenty dollar level,

purchase the McGraw-Hill Schaum’s outline Complex Variables by Murray R. Spiegel. Next,

purchase or check out a math-physics text such as Mathematics for Physicists by Susan Lea. If you

find need to be a power applied user, move on to Functions of a Complex Variable by Carrier,

mailto:[email protected]�

5/30/2012 Physics Handout Series.Tank: Complex Integration CI-2

Krook and Pearson (McGraw-Hill 1966) after studying two of the previous suggestions. A less dated

resource is Visual Complex Analysis by Tristan Needham.

The Derivative of a Function of a Complex Variable:

The 1 2 o

2 1

,2 1

( ) ( )( )z z z

f z f zLimitz z→

− −

must be defined independent of the manner in which z1 and z2 approach

zo. If z is represented as x + i y where x, y ∈ the condition requires that

0 0 0 0 0 0 0 0

0 0

( , ) ( , ) ( , ) ( , )x y

f x x y f x y f x y y f x yLimit Limitx i y∆ → ∆ →

+ ∆ − + ∆ −= ∆ ∆

. The conclusion is that

(except for constant functions) differentiable functions of a complex variable must by complex

valued.

Required: f fix y

∂ ∂= −

∂ ∂ [CI.1]

Analytic Functions of a complex variable: A function of a complex variable that is

differentiable in a region is said to be analytic in that region. The function can be

expanded about each point in the region as a Taylor’s series that has a non-zero radius

of convergence.

The Closed-Path Integral Property: The

integral of complex function f(z) around any

closed path bounding a closed region in which

f(z) is analytic at every point is zero. This

property of analytic functions is suggested by

examining the path integral of f(z) around a

small path. The argument and result are

reminiscent of Stokes theorem.

( ) ( , ½ ) ( ½ , )( )

( , ½ )( ) ( ½ , )( )

f z dz f x y y x f x x y i y

f x y y x f x x y i y

≈ − ∆ ∆ + + ∆ ∆

+ + ∆ −∆ + − ∆ − ∆∫

−∆x

i∆y ( , )x y

∆x

x

iy

-i∆y

( )f z dz∫


( , )

( ) ( ) 0x y

f ff z dz i i y xy x

∂ ∂≈ + ∆ ∆ = ∂ ∂

∫ [CI.2]

Cauchy’s integral Theorem: Property [CI.1] of differentiable functions was used to show that the

integral vanishes. This result suggests that:

anypath bounding

region of analyticity

( ) 0f z dz =∫ [CI.3]

That is: the enclosed region in which the function is analytic is simply connected.

The integral of complex function f(z) around any closed path bounding a closed region in which f(z)

is analytic at every point is zero. The enclosed region must be simply connected; f(z) must be analytic

everywhere in the entire enclosed region with no excluded points or patches.

Exercise: List the new words that have been coined in this handout. Attempt to define them.

The Laplace Property: It f(z) is represented as sum u(x,y) + i v(x,y) where u(x,y) and v(x,y) are real-

valued functions, then the condition [CI.1] becomes the Cauchy-Riemann equations:

andu v u vx y y x

∂ ∂ ∂ ∂= = −

∂ ∂ ∂ ∂ [CI.4]

Computing the second derivatives: 2 2 2 2

2 2andu v u vx x y y y x

∂ ∂ ∂ ∂= = −

∂ ∂ ∂ ∂ ∂ ∂ ⇒

2 2

2 2 0u ux y

∂ ∂+ =

∂ ∂

It follows that u(x,y) is a solution of the Laplace equation in two dimensions (as is v(x,y)).

Considered as functions of the real variables x and y, the real and imaginary

parts of an analytic function of a complex variable satisfy the Laplace

equation in two dimensions.

Exercise: Show the v(x,y) satisfies the 2D Laplace equation in the case that v(x,y) is the imaginary

part of an analytic function of a complex variable.


The closed-path integral and Laplace properties of analytic functions are enormously important. The

integral property provides a technique to evaluate integrals that resist conventional techniques, and

the Laplace property provides a tool for solving boundary value problems in two dimensions.

Entire functions: A function that is analytic (has a representation as a convergent Taylor’s series) at

every finite point in the complex plane is called an entire function. A function that is differentiable at

every point will be analytic. A few examples of entire functions are 1, z2, ez and sin(z).

Functions with poles and branch cuts

Examples of functions that fail to be entire are (1 – z)-1, tan(π z), ln(z) and z ½.

The function (1 - z)-1 cannot be expanded about or differentiated at z = 1. The

function has a pole at the isolated point z = 1. The function tan(π z) has isolated

poles for z equal to any integer. The story for ln(z) is a little more complicated.

ln(z) = ln(|z|) + i tan-1(Im(z)/Re(z)) + i n 2 π

The natural logarithm is not single-valued. Each z is mapped to an infinite set of values. As z tracks

around a circular path centered on the origin, the imaginary part of ln(z) increases be smoothly and

continuously by 2π. But, then you are back at the original z with a value of ln(z) that differs from the

original by 2πi. A line must be drawn from the origin out to infinity. The function can be made

continuous and differentiable everywhere in the complex plane except for points on that line. Such

lines are called branch cuts and the endpoint of a branch cut is a branch point. Each such almost

complete plane of values is a sheet of the function. The natural logarithm has an infinite number of

sheets. The screw thread image above represents the imaginary part of the logarithm running to

higher and higher values as the point z travels around a circle about the origin in the complex plane

again and again. For z = |z| eiφ, ln[z] = ln[|z|] + iφ + i n2π. The value of φ is undefined at |z| = 0, and

|z| = 0 is a branch point. The branch cut can run from the branch point to infinity along any non-

intersecting path. The standard choices are the rays φ = 0 and φ = π. The constant φ-rays (relative to

the branch point) are the useful cut choices for the natural logarithm.


The function z ½ has a branch cut. Consider the circle of points |z| = 1 and move around it in the CCW

sense for z = ei φ, z ½ = ei φ/2. Start at z = -1; the square root increases smoothly from i and on to -1 are

z moves around the circle t z = 1. Continuing, the square root approaches –i as the point z = -1 is

approached. As the point z = -1 is approached, z ½ is about -i rather than i. The function is not

continuous and therefore is not differentiable. The cut can run along any line from the origin to

infinity, and a popular choice is a ray out along the negative imaginary axis. The new behavior is that

after two trips around the circle, the value of z ½ does smoothly approach its initial value.

Surface for Square Root: A strained rendition of the

geometry of a function of two sheets with a branch cut.

Note that, as the functions are complex valued, a faithful

representation is not possible. The figure is to suggest that

the values after one circuit fail to match, and that, after a

second circuit, the values do match.

ln[|z|]

ln[|z|]

ln[|z|] + i π

ln[|z|] - i π

Branch cut along the negative real axis. The labeled values are points just above and below the real axis. The values progress continuously across the positive real axis but jump by 2πi crossing the negative axis.

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

ln[|z|]

ln[|z|] + i2π

ln[|z|] + i π

ln[|z|] + i π

Natural Logarithm


Branch cut along the positive real axis. The labeled values are points just above and below the real axis. The values progress continuously across the negative real axis but jump by 2πi crossing the positive axis.


A product of square roots affords one the opportunity to play with the branch cuts. Consider the

function f(z) = [z – z1]½ [z – z2]½ = W1 W2. For definiteness z1 = 1 and z2 = -1. The branch cuts can be

sent left from -1 and right from +1, or they can both run to the right (left) from the branch points.

Exercise: Construct the equivalent branch cut drawings for ( ) 1 1f z z z= − + . Reflect on the

negative sign. Hint: If both cuts are sent to the right (left), the result should have a positive value

above and below the line segment joining the branch points.

|W1| |W2|

(-|W1|)(-|W2|)

Cuts right from -1 and right from +1. The values progress continuously across the real axis outside the branch points but change sign crossing the positive axis for |z| < 1. There is a cut only between -1 and +1!

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxx

(i|W1|)(-|W2|)

(i|W1|)(|W2|)

(i|W1|)(i|W2|)

(i|W1|)(i|W2|)

xxxxxxxxxxxxxxxxxxxxxx

|W1| |W2|

-|W1| |W2|

i|W1| i |W2|

i|W1| (-i |W2|)

[z – 1]½ [z + 1]½ = W1 W2

i|W1| |W2|

i|W1| |W2|

|z|½

|z|½

i |z|½

– i |z|½

Branch cut along the negative real axis. The labeled values are points just above and below the real axis. The values progress continuously across the positive real axis but change sign crossing the negative axis.


|z|½

- |z|½

i |z|½

i |z|½

Square Root


Branch cut along the positive real axis. The labeled values are points just above and below the real axis. The values progress continuously across the negative real axis but change sign crossing the positive axis.

Cuts left from -1 and right from +1. The values progress continuously across the real axis between the branch points but change sign crossing the positive axis for |z| > 1. No cut between -1 and +1!

xxxxxxxxxxxxxxxxxxxx

[z – 1]½ [z + 1]½ = W1 W2


*** A more general discussion of branch cuts and sheets can be found in the references. One should

know that functions that are analytic over a domain map to a range that preserves the local topology.

A connected patch is mapped to a connected patch. Nearby points are mapped to nearby points. Line

segments that do not cross are mapped to line segments that do not cross, and closed paths are

mapped to closed paths.

The function ez maps each strip of width 2π that runs parallel to the real axis into the complete

complex plane. It follows that the inverse function ln(z) must map each point in the complex plane

into a infinite number of points. Each of these points lies in a different sheet of the function ln(z).

The function z2 maps the complex plane into two copies of the complex plane. Its inverse z½ is

therefore double-valued and has a structure with two sheets.

A complex function is said to be analytic on a region R if it is complex differentiable at every point in

R. The terms holomorphic function, differentiable function, and complex differentiable function are

sometimes used interchangeably with "analytic function" (Krantz 1999, p. 16). Many

mathematicians prefer the term "holomorphic function" (or "holomorphic map") to "analytic function"

(Krantz 1999, p. 16), while "analytic" appears to be in widespread use among physicists, engineers,

and in some older texts (e.g., Morse and Feshbach 1953, pp. 356-374; Knopp 1996, pp. 83-111;

Whittaker and Watson 1990, p. 83).

If a complex function is analytic on a region, it is infinitely differentiable in R. A complex function

may fail to be analytic at one or more points through the presence of singularities, or along lines or

line segments through the presence of branch cuts.

A complex function that is analytic at all finite points of the complex plane is said to be entire. A

single-valued function that is analytic in all but possibly a discrete subset of its domain, and at

those singularities goes to infinity like a polynomial (i.e., these exceptional points must be poles and

not essential singularities), is called a meromorphic function.

Weisstein, Eric W. "Analytic Function." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/AnalyticFunction.html

Laurent Series: Complex functions that have an isolated singularity (pole) at a point zo can be

expanded about that point in an extension of Taylor’s series that includes inverse powers. It is called


a Laurent series, and it converges to the value of the function for all points in a neighborhood of z0

except at z0 itself. The function can be expanded around z0 as:

1 101

00 0 0

( ) ... ( )( ) ( ) ( )

p p nnp p

n

a a af z a z zz z z z z z

∞− − + −

−=

= + + + + −− − − ∑ [TS.5]

Given that a-p ≠ 0 (and all a-q = 0 for q > p), f(z) is a singular function at zo with a pole of order p. If

no value of p exists such that the Laurent series converges to the function in a neighborhood of z0

(except at zo itself), then the function has an essential singularity at zo.

A function that is analytic everywhere in a neighborhood of zo except at the point zo has an isolated

singularity at z0. A function that appears to have a singularity at a point zo, but that can be assigned a

value at zo such that it is analytic in a neighborhood of zo has a removable singularity at zo.

The more general Laurent expansion has the form: 0( ) ( )nn

nf z a z z

∞

=−∞

= −∑ and converges in the

annular region r1 < |z – zo| < r2. For example the function 1/(z [z-1]) has the following Laurent

expansions about z = 1:

1

1 1 ( 1) ( 1)( 1) ( 1)(1 [ 1]) n

n nzz z z z

∞

= −

= = − −− − + − ∑ for 0 < | z – zo| < 1

and 2

2 1

1 1 ( 1) ( 1)( 1) ( 1) (1 [ 1] ) n

n nzz z z z

−

−= −∞

= = − −− − + − ∑ for 1 < | z – zo| < ∞

The form that converges in the punctured disk: 0 < | z – zo| < r2 is the most useful. The form of the

Laurent series is Sample calculation SC0 illustrates the construction of a Laurent series using the

binomial theorem.

The function 1/(z [z-1]) has poles at z = 0 and at z = 1. The small | z – 1|

expansion converges in the punctured disk centered on z = 1. It

diverges by the ratio test for | z – 1| > 1. The reason is clear. The pole

at z = 0 lies on the circle | z – 1| = 1. The large | z – 1| expansion

converges in annular region 1 < | z – 1| < ∞. No additional breaks into

annular sub-regions are necessary in this case as no other poles exist

that would disrupt the convergence on the Laurent series.

z = 1 z = 0

0<|z-1|<1

1<|z-1|<∞


Exercise: Find the position of the singularities and their order for: 1sin( )( )( 1)

zf zz

π=

−,

2 2

sin( )( )( 1)

zf zz

π=

−, 3 2( ) sin( )

( 1)( 1)f z z

z zπ π= +

− +, 4

sin( )( ) zf zz

ππ

= and 501/( )( )( ) z zf z e− −= .

order 1 at 1; order 2 at 1; order 1 at 1 and order 2 at -1; removable at 0; essential at zo

Laurent series generation:

Given a function 1 101

00 0 0

( ) ... ( )( ) ( ) ( )

p p nnp p

n

a a af z a z zz z z z z z

∞− − + −

−=

= + + + + −− − − ∑ with a pole of order p

at zo, the function (z – zo) p f(z) is analytic in a neighborhood of zo.

0 0 00

( ) ( ) ( ) ( )p n p mn m p

n p mz z f z a z z a z z

∞ ∞+

−=− =

− = − = −∑ ∑

SC0: The Laurent series for f(z) expanded about zo is (z – zo) –p times the Taylor’s series expansion

of (z - zo) p f(z) about zo. As an example, (z2 + a2)-1 is to be expanded about z = ia.

f(z) = 2 2

1 1( )( )z a z ia z ia

=+ − +

Clearly, f(z) has isolated simple (order 1) poles at ia and –ia.

2 2

1 1 1 1 1( )( ) 2 ( ) ( )z a z ia z ia ia z ia z ia

= = − + − + − +

The Laurent series for f(z) about ia is (z – ia)-1 times the Taylor’s series expansion of (z + ia)-1 about

ia. 1

22 3 1

1 1 1 1 ( 1)( ) ( ) ... ( ) ...( ) ( ) ( ) ( ) ( )

nn

nia ia ia ia

z ia z ia z iaz ia z ia z ia z ia z ia

+

+

−= − − + − + + − +

+ + + + +

12

2 3 1

1 1 1 1 ( 1)( ) ( ) ... ( ) ...( ) (2 ) (2 ) (2 ) (2 )

nn

nz ia z ia z iaz ia ia ia ia ia

+

+

−= − − + − + + − +

+

Multiplying by (z – ia)-1 yields the Laurent series.

12 3 2

1 1 1 1 ( 1)( ) ( ) ... ( ) ...(2 ) ( ) (2 ) (2 ) (2 )

nn

nf z z ia z iaia z ia ia ia ia +

−= − + − + + − +

−

The binomial alternative:

The binomial theorem can be used when a expression like (1 + x)n can be formed with |x| < 1.


1 21 1 1 11 1 ... ( 1) ...

( ) (2 [ ]) 2 2 2 2 2 2

nnz ia z ia z ia z ia

z ia ia z ia ia ia ia ia ia ia

−− − − −= = + = − + − + − +

+ + −

Mathematica 6 Syntax: Series[1/(z + I a),{z, I a,5}]

-(/(2 a))+(z- a)/(4 a2)+( (z- a)2)/(8 a3)-(z- a)3/(16 a4)

-(( (z- a)4)/(32 a5))+(z- a)5/(64 a6)+O[z- a]6

(z - I a)^(-1) Series[(z - I a) (1/(z2 + a2)),{z, I a,5}]

-(/(2 a (z- a)))+1/(4 a2)+( (z- a))/(8 a3)-(z- a)2/(16 a4)

-(( (z- a)3)/(32 a5))+(z- a)4/(64 a6)+O[z- a]5

Residue[1/(z^2+a^2),{ z, I a}]

-(/(2 a))

The zeroes of an analytic function must be isolated (or the function is identically zero).

See F&W App A, page 505

Contour integration and the residue theorem:

1.) The integral of a complex function around a closed path (contour) vanishes if the function is

analytic everywhere in the region bounded by the contour.

Cauchy Integral Theorem:anypath bounding

region of analyticity

( ) 0f z dz =∫

2.) The integrand can be represented by a Laurent series in a neighborhood of each pole enclosed by

the contour.

3.) For a CCW circular contour centered on zo, the integral:

0

0 for 12 for 1( )k

ka dzi a kz z π

≠= =−

∫ ; k integer and a constant

4.) For a circular contour centered on zo, the integral:


1 10 0 11

00 0 0

... ( ) ( ) 2( ) ( ) ( )

p p n nn np p

n n p

a a a a z z dz a z z dz i az z z z z z

π∞ ∞

− − + −−−

= =−

+ + + + − = − = − − −

∑ ∑∫ ∫

5.) The contour need not be circular. The path could be considered to by circular plus a contour

including a region in which the function is analytic. The original contour C is the sum of the path

bounding the light gray region and that bounding the

inner circular disk. The integrand is analytic in the

light gray region so the integral reduced to the result

for that inner path. By comment 4.),

1( ) 2C

f z dz i aπ −=∫ .

That is the path need not to be circular. The value a -1

is called the residue of f(z) at zo, the Laurent

coefficient of the inverse first power of(z – zo).

6.) For a function with a set of isolated poles, any CCW contour can be morphed in to a set of

circular paths running CCW about each pole plus a path enclosing a region in which the function is

analytic. See the Tools of the Trade section. The contour must not touch or cross poles or branch

cuts as it is continuously translated section by section from its initial to its final form. No branch cut

begins at an isolated pole.

7.) For a contour C that is traversed in the CCW sense,

( ) 2 Res[ ( )]j

jCpoles z

f z dz i f zπ= ∑∫ sum over enclosed poles zj

where the zj are the locations of the poles enclosed by the contour.

** One must locate the poles and identify those that lie inside the contour. **

8.) The residue if a function f(z) at an isolated pole zk or order p is:

Res[f(zk)] ] = ( )1

1

1 ( ) ( )( 1)!k

pp

kpz zdLim z z f z

p dz

−

−→

− −

[CI.6]

zo

C


Several of the points listed above are to be proven in homework problems. These points are to be

accepted as facts at this time, and their applications are to be illustrated.

Important Convention: All contours are assumed to wrap around enclosed points

once in the CCW sense. See the discussion of winding number for more details.

Sample Calculations

SC1: The function ez is an entire function. The function should have a convergent Taylor’s series

expansion about every point in the complex plane. Choose point zo.

0 0 0( ) 0

1

( )!

kz z z zz

k

z ze e e ek

∞−

=

−= = ∑

The series is known to converge and to have an infinite radius of convergence.

SC2: Consider the function tan(π z). The function is plotted for a real argument.

1 2 3 4

-10

-7.5

-5

-2.5

2.5

5

7.5

10

The function has poles at each half integer. It is analytic in a neighborhood of zero. The power series

expansion about z = 0 is: 2 3

2 32 3

0 0 0

tan( ) 1 tan( ) 1 tan( )tan( ) tan(0) ...2! 3!z z z

d z d z d zz z z zdz dz dz

π π ππ= = =

= + + + +

3 52 3 5 3 5

0

2 16 1 2tan( ) 0 sec ( ) ... ( ) ( ) ...3! 5! 3 15z

z z z z z z z zπ ππ π π π π π=

= + + + + = + + +

It is noted that tan(πz) passes through zero linearly at z = 0.


SC3: Consider the function [tan(π z)]-1. It has a first order poles at each integer value z. Invoking the

power formula for computing residues, it follows that:

( )1

1 2

Use L'Hospital's Rulewhere: ; ( ) tan( )

1 1 ( ) 1 1Res ( ) ( )tan( z) ( 1)! tan( ) sec ( )k

k

pp

kp

z n f z zz z z n

z n

d z nLim z z f z Limp dz z n

ππ π π π π

−

−

= =

→ →=

−= − = → = −

L’Hospital

A less formal, but effective procedure is to brute force the Laurent expansion about z = 0. Using

3 51 2tan( ) ( ) ( ) ...3 15

z z z zπ π π π= + + + ,

2 4

3 5

11 1 1 1 21 ( ) ( ) ...1 2tan( ) 3 15( ) ( ) ...3 15

z zz zz z z

π ππ ππ π π

− = = + + + + + +

Recalling that [1+ u]-1 ≈ 1 – u + u2 + … and setting u = 2 41 2( ) ( ) ...3 15

z zπ π+ + , we expand

consistently to fourth order. Expanding consistently requires that u be correct to the desired order and

that the expansion be carried out until the power of u times the lower power in u equals the desired

order. For fourth order in z, u must be 2 41 2( ) ( )3 15

z zπ π+ and the expansion must be extended to

order u2. Combining and stirring, 2

2 4 2 41 1 1 2 1 21 ( ) ( ) ( ) ( )tan( ) 3 15 3 15

z z z zz z

π π π ππ π

≈ − + + +

2 4 41 1 1 2 11 ( ) ( ) ( )tan( ) 3 15 9

z z zz z

π π ππ π

≈ − − + + Order(z6)

Note that the z6 and z8 terms are to be discarded. The expansion was only made consistent to fourth order.

The higher order pieces that appear are only some of the terms of that order. One must never retain sixth

order terms unless every sixth order term is found and kept. Expansions must be consistent to an order. 3

31 1 ...tan( ) 3 45

z zz z

π ππ π

≈ − − +

The coefficient of the z -1 power is π-1 verifying the residue value found above.

Exercise: Find the residue of [z2 tan(π z)]-1 at z = 0 using the power formula. What is the order of


the pole? Residue: - π/3

Exercise: Consider [z4 tan(π z)]-1 at z = 0. What is the order of the pole? Find the residue by any

method. Residue: - π3/45

SC4: More residues of [z4 tan(πz)]-1: Find the residue for z equal to a non-zero integer. This is

relatively easy as the function has first order poles at each integer that is not zero. The residue of

[tan(πz)]-1 has been computed to be π-1. At each residue, the active part [tan(πz)]-1 is multiplied by n-4

so the residue is n-4 π-1. Exercise the mega-formula to verify the result.

4 4 4 4 2 40

1 ( ) 1 ( ) 1 1 1Resz tan( z) tan( ) tan( ) sec ( )z n z n

z n

z n z nLim Limz z n z n n nπ π π π π π→ →

= ≠

− −= = → =

SC5. Summing inverse powers of integers: Consider the integral: 4 tan( )C

dzz zπ∫ for a contour that

circles the integer values from 1 to N along the real

axis.

The difference between the contour for the left and right figure is an integral around a contour inside

of which the integral is analytic everywhere. What does such an integral contribute to the overall

result? The integral around the contour on the right is 2πi times the residues enclosed.

4 4 421 1

1 12 Residues 2 2tan( )

N N

Cn n

dz i i iz z n n

π ππ π= =

= = =∑ ∑ ∑∫

If the right end of the contour is moved further and further to the right,

Re

Im

Re

Im

1 N

Contour 1 Contour 2


4 4 431 1

1 12 Residues 2 2tan( )C

n n

dz i i iz z n n

π ππ π

∞ ∞

= =

= = =∑ ∑ ∑∫

One gets double the result if a contour that circles the negative real axis is added. The residues

appear in a matching set.

4 4 441 1

1 12 Residues 2 2 4tan( )C

n n

dz i i iz z n n

π ππ π

∞ ∞

= =

= = =

∑ ∑ ∑∫

Next, two semi-circular arcs at infinity are added. The arcs are assumed to have a radius R, and R is

allowed to grow large. The denominator of the integrand grows large faster than the length of the arc

(πR) so the contribution from the arcs vanishes. This point is to be examined more carefully in a

homework problem.

Adding a segment of an infinite semicircular arc to close a contour is a standard operation.

One must be able to compute the contribution from that addition or establish that it vanishes.

It will vanish if the magnitude argument goes to zero faster than |z|-1. The integrands often

have exponentials with negative real parts of their arguments that grow as |z| grows. Factors of

that nature also drive the arc contributions to zero. See SC12 for an example.

Re

Im

Contour 4

Re

Im

Re

Im

Contour 3


Now, the contour can be shrink to one that circles only the pole at z = 0. It circles CW rather than

CCW so a sign change must be made.

4 4 4 44 5 61

14 2 Res at 0tan( ) tan( ) tan( )C C C

n

dz dz dz i i zz z z z z z n

ππ π π

∞

=

= = = = − =∑∫ ∫ ∫

The residue was found in an exercise above; it is - π3/45.

( )4

4 41 1

345

1 14 2 Res at 0 2 or90n n

i i z in n

ππ π π∞ ∞

= =

= − = = − − =∑ ∑

The process has been quite painful, but the techniques demonstrated are very important as is the

result. It is needed to evaluate an integral related to the blackbody radiation problem.

SC6: Laplace Property. Consider the representation f(z) = f(x+iy) = u(x,y) + i v(x,y) where x, y,

u(x,y) and v(x,y) are real valued. If f(z) is analytic, the functions u and v satisfy the 2D Cartesian

Laplace equation. Consider the function z3, identify u(x,y) and v(x,y) and demonstrate that the satisfy

the Laplace equation.

z3 → (x + i y)3 = x3 + i 3 x2 y – 3 x y2 – i y3 = (x3 – 3 x y2) + i (3 x2 y – y3)

Direct substitution easily demonstrates that the real part u(x,y) = x3 – 3 x y2 and the imaginary part

v(x,y) = 3 x2 y – y3 satisfy the Laplace equation: 2 2 2 2

2 2 2 20 and 0u u v vx y x y

∂ ∂ ∂ ∂+ = + =

∂ ∂ ∂ ∂

Contour 5

Contour 6


SC7: A standard contour integration example. Consider the integral: 2 2

dxx a

∞

−∞ +∫ . It appears

innocuous, but question arise in its evaluation. Revamp it as the complex integration along the real

axis from left to right. 2 21C

dzz a+∫ .

Integrals along the entire real axis: The residue theorem requires a closed contour that perhaps

wraps around a pole. A semicircular are of infinite radius is added in the upper or lower half plane to

close the contour. An argument must be made that the contribution from this are vanishes (or

perhaps can be calculated). Jordan’s lemma (tools of the trade) provides some guidance for arguing

that the contribution from the arc vanishes.

The integrand can be expressed as:

2 2

1 1 1 1 1( )( ) 2z a z ia z ia i a z ia z ia

= = − + − + − +

The second representation shows that the integrand has first order poles at i a and –i a. The pole at i a

is to be of immediate interest. Computing the residue for p = 1.

1 ( ) 1Res( )( ) ( )( ) 2z ia

z ia

z iaLimz ia z ia z ia z ia i a→

=

−= = − + − +

This result could be found by brute forcing a Laurent expansion about i a.

[ ] 2

1 1 1 1 1( )( ) ( )(2 ) ( ) (2 ) (1 )z ia

iaz ia z ia z ia ia z ia z ia ia −= =

− + − + − − +

( ) ( ) 2 3

2

2 21

( )

1 1 1 1 1 1( ) ...

( )( ) (2 ) (2 ) ( ) (2 ) (2 )1 ...z ia z ia

ia ia z iaz ia z ia ia ia z ia ia iaz ia

− −= = − + − +− + −

− + +−

The coefficient of the (z – ia)-1 is (2 i a)-1 as expected.

Closing the contour: Adding a semicircular arc of radius R (→ ∞) in the upper half plane does the

trick. The length of the arc is proportional to R while the integrand vanishes like R -2 so the net

contribution from the added path segment vanishes in the limit R → ∞. Integrating along the real axis

form -∞ to ∞ and closing by returning along the arc corresponds to circling the pole at +ia in the

CCW sense so we are ready to use the residue theorem.


See Jordan’s Lemma in the tools of the trade section for more detailed information about

contributions from semicircular arcs in the upper and lower half-plane in the limit R → ∞.

ia

-iaContour C2

[ ]2 22

12 Res 22C

dz i ia iz a ia a

ππ π = = = + ∫

SC8: Old fashioned integration example. Consider the integral: 2 2

dxx a

∞

−∞ +∫ . Revamp it as the

complex integration along the real axis from left to right. 2 21C

dzz a+∫ and rewrite that as:

1

1 1 12 C

dzi a z ia z ia

− − + ∫ . What issues arise if we use our old fashioned methods?

[ ]2 21 1

1 1 1 1 ln( ) ln( )2 2C C

dz dz z ia z iaz a i a z ia z ia i a

∞

−∞

= − = − − + + − + ∫ ∫ . The problem is deciding

what the final form means. The natural logarithm has branch cuts that require care consideration.

One cannot just jump from the real point −∞ to the real point ∞. The trip must be made continuously

along a path that avoids the cuts.

ia

-ia

For ln(z – ia)

For ln(z + ia)

R ei0

R eiπ

R e-iπ

The path from -∞ on the real axis to ∞ on the

real axis runs below the branch cut for ln(z –

ia) and above the cut for ln(z + ia). Points just

below the branch cut are about |z|e -iπ while

those just above are about |z|e +iπ. Set the

limits as – R and + R.


[ ] [ ]

[ ] [ ]

2 21

1 1ln( ) ln( )2 2

1 1(ln 0) (ln ) (ln 0) (ln )2 2

R R

R RC

dz z ia z iaz a i a i a

R i R i R i R ii a i a

π π

− −= − − +

+

= + − − − + − +

∫

[ ]2 21

1 22C

dz iz a i a a

ππ= =+∫

Note that a more robust process would been to evaluate [ ]ln( ) R

Rz ia

−− as:

[ ] ( ) ( )2 2 1 2 2 1ln( ) ln[ ] tan ln[ ] tanR

Ra az ia R a i R a iR R

− −−

− − − = + + − + + − and

[ ] ( ) ( )2 2 1 2 2 1ln( ) ln[ ] tan ln[ ] tanR

Ra az ia R a i R a iR R

− −−

+ = + + − + + −

The arguments of the inverse tangents have signed numerator and denominator values to set the

quadrant. The branch cut choice for ln(z) is equivalent to the convention that inverse tangent returns

values between -π and π. With that convention, the limit R → ∞ is taken:

[ ] ( ) [ ]1 1ln( ) tan tan 0 ( )R

RR RR

a aLim z ia i Lim i Lim iR R π π− −−→∞ →∞→∞

− − − = − ≈ − − =− and

[ ] ( ) [ ]1 1ln( ) tan tan 0 ( )R

RR RR

a aLim z ia i Lim i Lim iR R π π− −−→∞ →∞→∞

+ = − ≈ − + = −−

These results confirm the answer found previously. Clearly, one must tread lightly around branch

cuts.

SC9: Cauchy’s integral formula: 1 ( )( )2 C

f z dzf ai z aπ

=−∫ for f(z) analytic inside C.

The contour C is a continuous path that winds around the point a once in the CCW sense. If a

function f(z) is analytic on the contour C and at every point enclosed by the contour, then the integral

of f(z)/(z – a) around the contour is 2πi times f(a). First, the contour C is the sum of CA bounding the

light gray region and CB bounding the inner darker gray circular region that is centered on a. The

function f(z)/(z – a) is analytic everywhere inside CA so the contribution from that contour vanishes.

( ) ( )BC C

f z dz f z dzz a z a

=− −∫ ∫


An analytic function is continuous so for every ε, a δ

can be found such that for every point on the inner

contour CB, |f(z) - f(a)| < ε. It follows that:

( ) ( ) 2BC

f z f a dzz a

π ε−<

−∫

In the limit δ → 0,

( ) ( ) 2 ( )B BC C

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

π→ =− −∫ ∫

⇒ 1 ( )( )

2 C

f z dzf ai z aπ

=−∫

Exercise: Show that 2BC

dz iz a

π=−∫ where the contour CB is parameterized as: z = a + δ e i φ for the

parameter range: 0 ≤ φ < 2π.

SC10: Trigonometric integrals and circular contours: The unit circle centered on the origin can be

parameterized as: z = e i φ for the parameter range: 0 ≤ φ < 2π. This leads to representations of the trig

functions of the forms: cos(φ) = (z + z-1)/2 and sin(φ) = (z - z-1)/(2i). In the case on any circular

centered on the origin, z = R eiφ and dz = i R eiφ dφ or, more significantly, dφ = -i dz/z.

2 2 1 2 1 1 3

01cos ( ) ( ) 2 )4 4unit unit

circlel circlel

id z z i z dz z z z dzπ

φ φ − − − −− = + − = + + ∫ ∫ ∫

By inspection, the residue or coefficient of z-1 in the Laurent expansion about 0 is 2.

3

2 1( )f z zz z

= + +

2 2 1 3

0cos 2 ) (2 ) Res[ 0]4 4unit

circlel

i id z z z dz i zπ

φ φ π π− −− − = + + = = = ∫ ∫

This procedure might seem excessive to show that the average value of cos2φ = ½. Apply the

procedure to: 2 22

0 01

1 coscos andnad d

π π

φφ φ φ+∫ ∫ before deciding that it lacks merit. See SC13/14.

a

C = CA + CB

δ CB

CA


General Form: ( )2 1

0

1 1

2 2(sin ,cos ) ,C

z z z zif d i f z dz

πφ φ φ −− −− +→ −∫ ∫ . The contour chosen is a circle

of radius one traveled in the CCW sense and centered on z = 0.

SC11: Contour integral representations of special functions: Many special functions have contour

integral representations that prove convenient for the development of the properties of those special

functions. The reference by Carrier, Krook and Pearson is recommended for further reading.

Consider the integral 2 1

zt

C

e dzz +∫ for the contour C chosen as a CCW circular path of radius 2 centered

on the origin. The integrand has simple (first order) poles at z = i and – i. Examining the pole at i, z t z t

2

e eRes ( )1 ( )( ) 2

it

z iz i

eLim z iz z i z i i→

=

= − = + − +

The residue at –i is – e -i t/2i. Combining the results with the residue theorem, 2

1 sin( )2 1C

zte dz ti zπ

=+∫ .

One of our most familiar special functions has been represented as a contour integral.

SC12. Step Function. The contour C1 is the path from – ∞ to ∞ along the real axis. The pole is at z =

- iη where η is a small (infinitesimal) positive value. The contour is closed with an infinite

semicircular arc in the upper half plane to form C2 or with one in the lower half plane to form C3.

-iηContour C2

1

1 1

2 2 C

izt izt

i i

e dz e dzz i z iπ πη η

∞

−∞

− −→

+ +∫ ∫ ; η small positive

2

1 0for 02 C

izte dz ti z iπ η

−→ = <

+∫

3

1Res[ ]

21

C

tizt

z ii

e dz ez i

ηηπ η

−−

− = −→ = = − ≈ −+∫

The exponential can be written as: e-izt = e+Im(z) t e -i Re(z) t. For t < 0, the factor e+Im(z) t is small for

Im(z) > 0, so the are in the upper half plane (Im(z) > 0) can be added to yield C2 with zero additional

contribution. The pole is not enclosed so the original integral must be zero for t < 0. For t > 0, the

factor e+Im(z) t would grow large in the upper half plane, so the arc in the lower half plane must be


added to close the contour in the case t > 0. The contour C3 is not illustrated, but it winds around the

pole at -iη in the CW sense. CW gives a sign change and e -ηt → 1 as η is infinitesimal. Conclusion:

0 00

) )( )

01

( (

2 2( ) t t

C

iz t t iz t ti ie dz e dz t t ez i z i

η

π πη η∞ − −

−∞

− − − −→ = Θ −

+ +∫ ∫

The unit step Θ(t – to) has a representation as a contour integral in the limit η → 0+. As shown in the

plot below, it requires the limit that η → 0+.

Mathematica:

Integrate[Exp[+ I z ]/(z + I/64 ),{z, - Infinity, Infinity}] = 0 (t < 0 case) t = -1; η = 1/64

Integrate[Exp[- I z ]/(z + I/64 ),{z, - Infinity, Infinity}] = 2164 (t > 0 case) t = +1; η = 1/64

-2 -1 1 2

0.2

0.4

0.6

0.8

1

η = 1/1024

η = 1/4

theta[t_,eta_] :=(I/(2 Pi)) Integrate[Exp[- I z t ]/(z + I eta),{z, - Infinity, Infinity}]

Plot[{Re[theta[t, 1/4]],Re[theta[t, 1/1024]]} ,{t,-2,5}, PlotRange →{-0.1,1.1}, PlotStyle → {{RGBColor[0.8,0,0],

Thickness[0.01]},{RGBColor[0,0,0.8],Thickness[0.005]}}]

SC13. A meaningful trig contour integral

Trigonometric integrals and circular contours: The unit circle centered on the origin can be

parameterized as: z = e i φ for the parameter range: 0 ≤ φ < 2π. This leads to representations of the trig


functions of the forms: cos(φ) = (z + z-1)/2 and sin(φ) = (z - z-1)/(2i). In the case on any circular

centered on the origin, z = R eiφ and dz = i R eiφ dφ or, more significantly, dφ = -i dz/z.

General Form: ( )2 1

0

1 1

2 2(sin ,cos ) ,C


πφ φ φ −− −− +→ −∫ ∫ . The contour chosen is a circle

of radius one traveled in the CCW sense and centered on z = 0.

Apply to procedure to: 2 2

0cos n d

πφ φ∫ → ( )2

11

2n

Cz zi z dz−−+− ∫ .

Special Method: The binomial expansion is to be used and then the a-1 coefficient in the Laurent

expansion of the integrand is to be identified and used as the residue at z = 0.

( ) ( )2 22 0 2 1 2 1 1 (2 1)(2 )! (2 )!1 1

! ! ! !2 2... ... ... ...n nn n n n

C C

n nn n n ni z z z z dz i z z z dz− − − − − +− + + + + = − + + + +∫ ∫

Therefore, the residue a-1 = 2(2 )!1

! !2 nn

n ni− . 2 2

0cos n d

πφ φ∫ = 2

(2 )!! !

22 n

nn n

π

SC14. Another trig example: General Form: ( )2 1

0

1 1

2 2(sin ,cos ) ,C


πφ φ φ −− −− +→ −∫ ∫ .

The contour chosen is a circle of radius one traveled in the CCW sense and centered on z = 0.

Apply to procedure to: 2

01

1 cosa dπ

φ φ+∫ where 0 < a < 1.

1

2 1

0 2]1 1 1

1 cos 1 ½ [ 22C Ca a z z az z ad i z dz i dz

π

φ φ −

−

+ + + + +→− = −∫ ∫ ∫

The integrand has simple poles at z1 = 1 2 1a a− −− + − and z2 = 1 2 1a a− −− − − . As a-1 > 1, |z2| is

greater than one so that pole is not enclosed by the contour. The product of the pole positions

z1 z2 = 1 so | z1| < 1; it is enclosed by the contour.

( )1

22)( )

2 2 1(2( ) i i

a z zaz z a z zf z − −−+ + −= =

( ) ( )1

1 11 22 1

( ))( )2 21 1

(Res[ ( )] ( )z z

i ia az z z zz zf z z z

=− −

− −− ⇒ = − =


( )2

0 2 21241 1

1 cos 2 1aaa d aπ ππ

φ φ− −+ = =

−∫

SC15. Branch Cut Example I: 1

21 1 (1 )dx

x a x− − +∫ where 0 < a < 1.

The technique for this integral depends on the branch cuts for z½ so they should be reviewed before

attacking the problem.

A product of square roots affords one the opportunity to play with the branch cuts. Consider the

function f(z) = 1 11 1z z− +

= W1 W2. For definiteness z1 = 1 and z2 = -1. The branch cuts can be sent

left from -1 and right from +1, or they can both run to the right (left) from the branch points.

|W2|

-|W2|

Cut to the right from -1. The values progress

continuously across the real axis for Re[z] < -

1. There is a sign change upon crossing the

axis for Re[z] > -1.

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

-|W2|

|W2|

i|W2|

i|W2|

[z + 1]-½ = W2

xxxxxxxxxxxxxxxxxxxxxx

-i|W1|

+i|W1|

|W1|

|W1|

[1-z]-½ = W1

|W1|

|W1|

Cut to the right from +1. The values progress continuously across the real axis for Re[z] < 1. There is a sign change upon crossing the axis for Re[z] > 1. Note that this is for [1-z]-½ not [z -1]-½

.


|W1W2|

-|W1W2|Pole @ z= a-1

The goal is to compute I = 1

21 1 (1 )dx

x a x− − +∫ which equals

1 2| | (1 )A

dzW W a zΓ +∫ where ΓA is the

path parallel to the real axis and infinitesimally above it running from -1 to +1.

21 (1 )A

dz Iz a zΓ

≡− +

∫ = 1

21 1 (1 )dx

x a x− − +∫

The ΓB segment is the path parallel to the real axis and infinitesimally below it running from +1 to

-1. The contour traverses the ΓB segment from larger real values to smaller (in the reverse direction)

and the complex integrand equals 1 2

1( | |) (1 )W W a z− +

.

21 (1 )B

dz Iz a zΓ

≡ +− +

∫

There is a net contribution (= 2 I) from ΓA plus ΓB because 1(1 )(1 ) (1 )z z a z− + +

has a discontinuity

(branch cut) as z crosses the real axis between -1 and 1.

The circular turning arc ΓC is described by z = -1 + εiφ. The contribution from this arc is bounded by:

-i|W1| |W2|

i|W1|(- |W2|)

|W1| i |W2|

|W1| (i |W2|)

[1-z]-½ [z + 1]-½ = W1 W2

|W1| |W2|

|W1|(- |W2|) xxxxxxxxxxxxxxxxxxxx

Cuts right from -1 and from +1. The values progress continuously across the real axis outside the branch points but change sign crossing the axis for |z|< 1. There is a cut between -1 and +1 only.


2(1 )C

ii e da

φε φ εεΓ

∝−∫

The contribution from the turning arc vanishes as the limit ε → 0 is taken. Also in this limit, the

contour segments ΓA and ΓB approach z = -1 on the left end.

The contribution from the segments running above and below the real axis for Re[z] > 1 cancel as

there is no discontinuity (branch cut) and the upper and lower segments are traveled in opposite

directions.

The contribution from the arc at ∞ vanishes because the magnitude of the integrand vanishes like R-2

while the arc length grows like R as R → ∞.

1complete 2contour

2 2 Res[ ]1 (1 )

dz I i z az a z

π −≡ = = −− +

∫

1

1complete 2 2 1 2 2contour

1 1 22 2 [ ) 21 (1 ) 1 ( ) 1 1

z a

dz I i z a iz a z z a a z a a

ππ π−

−

−=−

≡ = + = = − + − + − −

∫

21I

aπ

=−

One can also evaluate 1

21 1 (1 )dx

x a x− − +∫ by choosing x = cosθ leading to

0 (1 cos )da

π θθ+∫ .

Compare this result with the previous sample calculation. SC14: 2

0 221

1 cos 1a d aπ π

φ φ+ =−∫

SC16. Branch Cut Example II: The integral 30 (1 )dx

x∞

+∫ cannot be extended over the entire real

axis because the integrand has a pole on the negative real axis. It can be computed by considering

13

ln[ ](1 )C

z dzz+∫ . The logarithm is chosen with a branch cut along the positive real axis. Note that the

branch point is at z = 0 so there is no contribution from the portions of the horizontal segments to the


left of z = 0. The contribution from the small turning circle vanishes as its radius goes to zero. The

contributions from the arc at infinity vanish following Jordan’s lemma.

13 3 30 0

ln[ ] (ln[ ] [ln[ ] 2 ] 2(1 ) (1 ) (1 )C

z dz x x i dx dxiz x x

π π∞ ∞− +

= = −+ + +∫ ∫ ∫

Complete the problem by exercising the residue theorem.

Pole @ z= -1

Ln[z] + i 2π

Ln[z]

Contour C1

Pole @ z= -1

Contour C2

The integral 30 (1 )dx

x∞

+∫ can also be evaluated using a pie wedge contour C2. The value of z3 repeats

for a rotation 2π/3. Choose a 1200 pie wedge.

( ) ( )2

2 /32 /3 2 /3

3 3 2 /3 3 3 30 0 0 01 1

(1 ) (1 ) (1 [ ] ) (1 ) (1 )

ii i

iC

dz du e du du dxe ez u ue u x

ππ π

π

∞ ∞ ∞ ∞= + − = − = −

+ + + + +∫ ∫ ∫ ∫ ∫

As an added benefit, the contour encloses just one pole, and the residue is easily computed using the

derivative rule for simple poles.

[ ] 22 /3

231

3( ) ( ) 1Res ( ) Res( ) ( ) 3k

k

k k

k k i

i

z zz z e

g z g zf zh z h z z

eπ

π

==

= = → ′

=

SC17. A standard contour integration example. Consider the integral: 2 2

cos( )x dxx a

∞

−∞ +∫ . Express it as

a complex integration along the real axis from left to right. 2 21 1

cos( ) ( )½( )( )

iz iz

C C

z dz e e dzz a z ia z ia

−+=

+ − +∫ ∫ ..


ia

-iaContour C2

Integrals along the entire real axis: The residue theorem requires a closed contour that perhaps wraps around a pole. A semicircular arc of infinite radius is added in the upper or lower half plane to close the contour. An argument must be made that the contribution from this arc vanishes (or perhaps that it can be calculated). Jordan’s lemma (tools of the trade) provides some guidance for arguing that the contribution from the arc vanishes. An exponential eiz (= eiRe[z] e-Im[z] must be closed in the upper half plane so that e-Im[z] is small.

The integrand can be expressed as:

2 2

( ) ( ) 1 ( ) ( )2( ) 4( )( ) 4

iz iz iz iz iz iz iz ize e e e e e e ez a z ia z ia i a z ia z ia

− − − − + + + += = − + − + − +

The second representation shows that the integrand has first order poles at i a and –i a. Jordan’s

lemma directs us to close terms with eiz in the numerator in the upper half plane and those with e-iz in

the lower half plane. The factor [z – ia]-1 has a pole in the upper half plane while [z + ia]-1 has one

in the lower half plane. We need to find the residues of: 1 1and4 4

iz ize ei a z ia i a z ia

− − − +

as those are

the only terms that contribute.

1 1Res ; Res4 4 4 4

iz a iz a

z ia z ia

e e e ei a z ia ia i a z ia ia

− − −

= =−

= − = − − +

The integral run from -∞ το ∞ along the real axis making the contour closing around the upper half plan CCW and the on closing I the lower half plane CW.

{ }2 21

cos( ) 2 Res[ ] Res[ ] 24 4

a a

C

az dz e e ei ia ia iz a ia ia a

ππ π− − − −

= − − = − = + ∫

Exercise: The residues for SC17 were calculated indirectly. Use the standard equation for the residue

of a first order pole to compute the residues of 2 2 2 2and2( ) 2( )

iz ize ez a z a

−

+ + at ia and at –ia

respectively. Res[f(zk)] ] = ( ) ( )1

1

1 ( ) ( ) ( ) ( )( 1)!k k

pp

k kpz z z zdLim z z f z Lim z z f z

p dz

−

−→ →

− → − −

.


Exercise: Explain why the exponentials eiz and e-iz dictate that the contour be closed in the upper and lower half planes respectively.

Exercise: What value do you expect for the integral 2 2

sin( )x dxx a

∞

−∞ +∫ ? Explain. Use contour methods to

verify your expectation.

Tools of the Trade

Evaluation of residues:

1.) Read the residue off the Laurent expansion. The residue is the a-1 coefficient in the Laurent

expansion.

2.) Use the Cauchy integral formula approach.

[ ] ( )1

1

1Res ( ) ( ) ( )( 1)!k k

pp

kpz z z zdf z Lim z z f z

p dz

−

−= →

= − −

For a first order pole, the equation simplifies to :

[ ] [ ]Res ( ) ( ) ( )k k

kz z z zf z Lim z z f z

= →= −

3.) For the special case that f(z) has a first order pole and the function f(z) is the ratio of two analytic

functions (= g(z)/ h(z)), poles occur at the zeroes of h(z). In the case that the pole is of order one, h(z)

has a first order zero at zk and g(zk) ≠ 0,

( ) ( ) ( ) ... ( ) ( ) ( ) ...( )( )( ) ( ) ( ) ( ) ... 0 ( ) ( ) ...

k k k k k k

k k k k k

g z g z z z g z g z z zg zf zh z h z h z z z h z z z

′ ′+ − + + − += = →

′ ′+ − + + − +

Using [ ] ( )Res ( ) ( ) ( )k k

kz z z zf z Lim z z f z

= →= − ,


[ ] ( ) ( ) ( ) ... ( )( )Res ( ) Res ( )( ) 0 ( ) ( ) ... ( )k

kk

k k k kk

k k kz z z z

z z

g z g z z z g zg zf z Lim z zh z h z z z h z= →

=

′ + − += = − = ′ ′+ − +

More complicated relations can be derived for the residue of higher order poles. Note that he

notation h′ has been used to represent dh/dz.

Residue examples: The function f(z) = z2/(1 + z4) has simple poles at z = / 4 3 / 4;i ie eπ π± ± . In order to

evaluate the residue at eiπ/4, one must find 2 / 4 2

/ 4/ 4 3 / 4 3 / 4 / 4 / 4 3 / 4 / 4 3 / 4 / 4 / 4/ 4

( )( )

( )( )( )( ) ( )( )( )

ii

i i i i i i i i i iiz e

z eLimit z e

z e z e z e z e e e e e e e

ππ

π π π π π π π π π ππ − − − −→− =

− − − − − − −

The vector picture of complex numbers provides some relief. / 4 2

/ 4 3 / 4 / 4 3 / 4 / 4 / 4

/ 4

/ 4

( )( )( )( ) 42 (2 ) ( 2 )

i

i i i i i i

i

i

ee e e e e e

i ee i

π

π π π π π π

π

π− −

−

− − −= =

A product of differences of the pole positions

appears in the denominator. Each factor is

represented by a vector on a plot of the pole

locations. The factor (z1 – z2) is the vector from

z2 to z1 which is the real number 2½. The factor

(z1 – z3) is the vector from z3 to z1 or 2

eiπ/4. 1 3/ 42 iz z e π− =

Exercise: What vector (complex number)

represents (z1 – z4)? … (z1 – z3)?

The residue is more easily computed if one uses the derivative rule for simple poles.

[ ] ( ) ( ) ( ) ... ( )( )Res ( ) Res ( )( ) 0 ( ) ( ) ... ( )k

kk

k k k kk

k k kz z z z

z z

g z g z z z g zg zf z Lim z zh z h z z z h z= →

=

′ + − += = − = ′ ′+ − +

Given f(z) = z2/(1 + z4), g(z) = z2 and h(z) = 1 + z4. The residue for the pole at eiπ/4 becomes:

[ ] / 4/ 4/ 4/ 4

2

3

/ 4( ) 1Res ( ) Res( ) 4 4 4i

iii ee

i

z ez e

g z zf zh z z z

eπ

πππ

π−

==

= = = =

z1 z2

z3 z4

1 12

12

z1 – z2 = 2

1 3/ 42 iz z e π− =


The derivative rule can be extended to second order poles at the cost of a huge list of conditions and

an increase in complexity. The second order pole rule is:

If a function f(z) has a second order pole at zo and f(z) is of the form g(z)/p(z) where g(z) and p(z) are

analytic in a neighborhood of zo. Further, that the analytic function (z – zo)-2 p(z) = h(z) is known,

then: [ ]0

0

2

( ) ( ) '( )Res ( )( ) ( ( )) z z

z zg z g z h zf zh z h z =

=

′ = −

(restricted 2nd order pole result)

Consider z2/(z2 + a2)2 which has second order poles at z = ± ia. The residue at ia is to be found.

Matching up, g(z) = z2 and h(z) = (z + ia)2.

2 2

2 2 2 2 2 4

( ) ( ) '( ) 2 ( ) 2( )Res( ) ( ) ( ( )) ( ) ( )z ia z iaz ia

z g z g z h z z z z iaz a h z h z z ia z ia= ==

′ += − = − + + +

2 2

2 2 2 2 4

2 ( ) 2(2 ) 1 1 1Res( ) (2 ) (2 ) 2 4 4z ia

z ia ia ia iz a ia ia ia a=

− = − = − = +

A modest saving of effort has been achieved.

Integrals of real functions along the entire real axis:

Closing contours with semicircular arcs in the upper and lower halfplane - Jordan's Lemma

An integral along the real axis is often converted to an integral over a closed contour by adding a

semicircular arc of radius R (→ ∞) in the upper or lower half-plane. The length of the arc is

proportional to R so, if the magnitude of the integrand vanishes faster than R-1, the net contribution

from the added path segment vanishes. Integrating along the real axis form - ∞ to ∞ and returning

along the arc corresponds to circling the poles in the upper half-plane in the CCW sense.


Pole in upperhalf plane

Contour C2Pole in lower

half plane

C1

-R +R

x

y

Jordan’s Lemma provides conditions under

which the contribution from the infinite

semicircular arc vanishes.

1( )

C

iazg z e dz∫ → 0 for Re[a] = 0 if |g(z)| → 0

faster than R-1 as R → ∞.

1( )

C

iazg z e dz∫ → 0 for Re[a] > 0 if |g(z)| → 0

uniformly as R → ∞. For R such that |g(z)| < ε,

the contribution is of order ε/Re[a] → 0.

In the case that the form 1

( )C

iazg z e dz−∫ with Re[a] > 0 is considered, the contour must be closed in

the lower half plane to ensure that the contribution from the semicircular arc at infinite vanishes

(unless g(z) also vanishes faster than some power of R). The sign in the preceding i in the argument

of the exponential and the sign of the real part of a dictate the half plane in which the contour is

closed. Review sample calculations SC7 and SC12.

Winding Number: A complex function that is analytic except at a set of isolated points is said to

have poles. Consider a closed integration path that is tangled around poles located at za, zb, zc and zd.

We play a game that allows one to alter the path by stretching and moving the path as anyway

desired as long as the path is not pulled onto or across a pole. With these restrictions, the number of

times that the path winds around each pole in the CCW sense is (a topological) invariant. It stays the

same for all path variations that do not violate the rules. The usual winding numbers are 0, +1 and -1,

but any integer value is possible.

The winding numbers na, nb, nc and nd for each pole appear below the figures.

za

zb

zc

zd

za

zb

zc

zd


na = -1, nb = +1, nc = +2, nd = 0 na = -2, nb = +2, nc = nd = -1

One method to identify the winding number is to distort the path into essentially complete circles

around each pole plus paired path segments running close to one another in opposite directions. The

left side drawing could be distorted to be:

The winding numbers are seen to be, na = -1, nb = +1, nc = +2 and nd = 0.

A less tiresome method that identifies the winding number is to draw a semi-infinite straight line

from each pole out to infinity. Increment the winding number each time the path crosses that line

from right to left as you look out from the pole along the line and decrement it each time it crosses

left to right.

The path crosses the za line once from left to

right ⇒ na = -1. The path does not cross the

upper zd line ⇒ nd = 0. It crosses the lower zd

line once from right to left and once from left to

right ⇒ nd = 0. As you move out along the zb

line the changes run as +1, -1, -1, +1 and

+1. ⇒ nb = 1. Moving out the zc line, both crossings are right to left ⇒ nc = 2.

The stretching and moving method is very important as it is this procedure that enhances the value of

complex integration along contours (paths). An integral along a path can be replaced by one along a

stretched and moved path as long as the path is not pulled onto or across a pole.

za

zb

zc

zd

za

zb

zc

zd

za

zb

zc

zd


Modified Residue Theorem:

( ) 2 Res( ( )m mf z dz i f zπ η= ∑∫

The symbol ηm represents the winding number of the contour about the pole at zm.

Problems

1.) Compute 0( )nna z z dz−∫ for a circular path of radius R about the point zo. For this path, z = zo +

R e iφ where n is an integer (possibly negative) and an is a constant. Show that the integral vanishes

except for the case that n = -1. In that case show that 1 0 11( ) 2a z z dz i aπ− −

−− =∫ . Hint: what is dz?

2.) Examine the integrand of 45 tan( )C

dzz zπ∫ to verify that adding the two semicircular arcs at infinity

to the contour makes a negligible contribution to the integral.

3.) Explain why the integral 45 tan( )C

dzz zπ∫ can be shifted to 46 tan( )C

dzz zπ∫ , the one over contour 6

in the sample calculation section. Begin by preparing a drawing that identifies contour 5 as contour 6

plus another contour. These contours appear in Sample Calculation 5 (SC6).

4.) Give the form of the Laurent expansion about zo for a function f(z) that has a pole at zo of order p.

Show that the procedure

Res[f(zk)] ] = ( )1

1

1 ( ) ( )( 1)!k

pp

kpz zdLim z z f z

p dz

−

−→

− −

yields a -1, the coefficient for the (z – zo)-1 factor in the expansion

5.) Find the location, order and residue for each pole of the following functions:

a.) 2 4z

z −, b.)

2

2( 2)z

z −, c.)

izez π−

, d.) tan( )

izez zπ


6.) Evaluate 2

1

1n n

∞

=∑ using a contour integration.

7.) WINDING NUMBER: A complex function that is analytic except at a set of isolated points is

said to have poles. Consider a closed integration path that is tangled around poles located at za, zb, zc

and zd. We play a game that allows one to alter the path by stretching and moving the path as anyway

desired as long as the path is not pulled onto or across a pole. With these restrictions, the number of

times that the path winds around each pole in the CCW sense is a (topological) invariant. It stays the

same for all path variations that do not violate the rules. The usual winding numbers are 0, +1 and -1,

but any integer value is possible.

Find the winding numbers na, nb, nc and nd for each pole in the figures below.

na = -1, nb = +1, nc = +2, nd = 0 na = -2, nb = +2, nc = nd = -1

The left side drawing could be distorted to be:

Attempt to distort the right side drawing to a similar form to reveal the winding numbers. Note the

circles are complete except for infinitesimal gaps and that the non-circular parts are close paths pairs

traversed in opposite directions.

za

zb

zc

zd

za

zb

zc

zd

za

zb

zc

zd

za

zb

zc

zd


8.) Laplace Property. Consider the representation f(z) = f(x+iy) = u(x,y) + i v(x,y) where x, y, u(x,y)

and v(x,y) are real valued. If f(z) is analytic, the functions u and v satisfy the 2D Cartesian Laplace

equation. For each function below, identify u(x,y) and v(x,y) and demonstrate that the satisfy the

Laplace equation.

a.) z2 b.) z4 c.) z -1 d.) ekz

9.) Repeat the sample calculation SC7, but choose a contour that encircles the pole at z = - i a. If

your attempt is not completely successful, read about the winding number in the Tools of the Trade

section.

10.) Show that sin(z)/z has a removable singularity at z = 0.

11.) a.) Find the residue of [z2 tan(π z)]-1 at z = 0 using the power formula. What is the order of the

pole?

b.) Consider [z4 tan(π z)]-1 at z = 0. What is the order of the pole? Find the residue by any

method. Residues: - π/3 ; - π3/45

12.) Develop an argument supporting Cauchy’s integral formula that is based on the residue theorem.

See SC9.

13.) Use standard integration methods to argue that: 1

( )( ) 1( ) ( 1) ( )n nC C

dfdz dzf z dz

z a n z a −=− − −∫ ∫

. The contour C

is a continuous path that winds around the point a once in the CCW sense. The function f(z) is

analytic on the contour C and at every point enclosed by the contour.

14.) Use the result of the previous problem to conclude that: 1

( ) 1( ) !

n

n nCz a

f z dz d fz a n dz+

==

−∫ . The contour

C is a continuous path that winds around the point a once in the CCW sense. The function f(z) is

analytic on the contour C and at every point enclosed by the contour.


15.) Use the residue theorem conclude that: 1

( ) 1( ) !

n

n nCz a

f z dz d fz a n dz+

==

−∫ . The contour C is a

continuous path that winds around the point a once in the CCW sense. The function f(z) is analytic on

the contour C and at every point enclosed by the contour. Use the arguments of SC9 to switch to the

contour CB and then represent f(z) by it general Taylor’s series expansion about a.

16.) Compute a.) 2 2

0cos n d

πφ φ∫ and b.)

2

01

1 cosa dπ

φ φ+∫ . Expect to use the binomial theorem and

the quadratic formula. Assume n is a positive integer and a is a real number between 0 and 1. Take

care to locate the poles of the integrand. Class them as enclosed or not enclosed by the contour.

likely answers: 2

(2 )!22 ! !n

nn n

π

; 2

21 a

π−

17.) a.) Compute 4 4dz

z a∞

−∞ +∫ where a is a real positive number. Nearing (reference 5) suggests using

a graphical representation of the zj – zi, the pole separations. 32 a

π

b.) Compute 4 4

ikze dzz a

∞

−∞ +∫ where a is a real positive number. / 23 42cos[( ) ]ka kae

aππ − −

18.) Consider an integral of the form: 2

0 1 sin cosd

a bπ φ

φ φ+ +∫ . Describe the method to solve this

integral. Given that a and b are real, what limits would you impose on those values to ensure that the

integral is finite.

19.) Consider an integral of the form: 2

0

(cos )1 sin cos

n da b

π φ φφ φ+ +∫ . Describe the method to solve this

integral. Given that a and b are real, what limits would you impose on those values to ensure that the

integral is finite.


20.) Compute: 2

0 cos sind

a b c iπ φ

φ φ+ +∫ for a, b, c real. Explain why a + b ≠ 0 is an adequate

condition to ensure that the integral is finite. Compute: 2

0 8 28cos 4 sind

iπ φ

φ φ− −∫

21.) The left-side contour can be continuously distorted into the right-side contour. Note the circles

are complete except for infinitesimal gaps and that the non-circular parts are close paths pairs

traversed in opposite directions.

Imagine the gaps and paths in the limit: gap width → 0. The straight line segments become pairs of

straight line segments traversed in opposite directions. Consider the net value of ( )f z dz∫

integrating over one such path pair given that f(z) is analytic except for poles at za, zb, zc and zd. Can

its value be neglected?

22. For the figure pairs below, the black dots represent the locations of poles of a function f(z) that is

analytic everywhere else in the complex plane.

a.) Represent contour 2 as contour 1 plus a contour Cadd with the property that f(z) is analytic

Re

Im

Re

Im

1 N

Contour 1 Contour 2

za

zb

zc

zd

za

zb

zc

zd


everywhere on and inside that contour. What is the value of ( )addC

f z dz∫ ?

b.) Represent contour 6 as contour 5 plus a contour Cadd with the property that f(z) is analytic

everywhere on and inside that contour. What is the value of ( )addC

f z dz∫ ?

23. a.) Compute the Laurent series for (z2 + a2)-2 about z = ia.

b.) Compute the residue of (z2 + a2)-2 at z = ia. What is the order of the pole? Compare the residue

with the coefficient of the (z – ia)-1 term in the Laurent expansion of (z2 + a2)-2 about z = ia.

(z - I a)^(-2) Series[(z - I a)^2 (1/(z2 + a2)^2),{z, I a,5}]

-(1/(4 a2 (z- a)2))-(/(4 a3 (z- a)))+3/(16 a4)+( (z- a))/(8 a5)

-(5 (z- a)2)/(64 a6)-((3 (z- a)3)/(64 a7))+O[z- a]4

Residue[1/(z^2+a^2)2,{ z, I a}]

-(/(4 a3))

24.) Develop Laurent series for:

a.) sin( )( )

zz

ππ−

about z = π. b.) 2

sin( )( )

zz

ππ−

about z = π. c.) 2

cos( )zzπ about z = 0.

Contour 5

Contour 6


d.) 2

sin( ) zz ez z

− about z = 0.

25.) a.) Develop the Laurent series for the function 1/(z [z-1]) about z = 1 that converges for small

|z - 1| and identify the bounds on it annulus of convergence.

b.) Develop the Laurent series for the function 1/(z [z-1]) about z = 1 that converges for large |z – 1|

and identify the bounds on it annulus of convergence.

26.) The inverse of a Laplace transform can be computed directly using complex integration methods

and the Bromwich integral 1( ) (2 ) ( )C

tzf t i f z e dzπ −= ∫ where ( )f z is L[f(t)]s=z. The contour consists

of a straight path up parallel to the imaginary axis and to the right of all the poles of ( )f z and

which is closed by a large circular arc closing on the left thus enclosing all the poles of ( )f z .

Compute the inverse Laplace transforms of: a.) (s – k)-1 b.) (s – k)-2 c.) (s – a)-1(s – b)-1 d.) s (s2 +

k2)-1 e.) k (s2 + k2)-1

Answers: ekt, t ekt, (a – b)-1 [e-bt – e-at], cos(kt), sin(kt)

The arc closing to the left does not contribute so long as t > 0. For t > 0, the contour must be closed

on the right leading to a result of 0. The t > 0 case is all that is of direct interest.

27.) Given 3

2

2 2

221( , )kmi kx tax t e dk

k aπ

∞

−∞

− Ψ = + ∫

. Compute Ψ(x,0). Note that the result

depends on the sign of x.

28.) As a step to computing some Fourier transforms, one needs to evaluate 2( )t ike dt

∞ − +

−∞∫ . One

knows that the real integral 2xe dx π

∞ −

−∞=∫ . The desired integral corresponds to the complex

integration 2ze dz

∞ −

−∞∫ along the path z = x + ik for -∞ < x < ∞. Consider the contour path z = x for -

L < x < L ; z = L + iy for 0 < y < k; z = x + ik for L > x > -L; z = -L + iy for k > y > 0. Take the limit

L→ ∞.


29.) Evaluate: (1 )

0

n ik xx e dx∞ − +∫ . Consider the target value to be 1

!(1 )n

nik ++

. Define α = tan-1(k).

Use the contour 0 to R along the real axis; the arc of radius R from θ = 0 to a; and in along the

line from (R, α) to (0, 0). Take the limit R → ∞, and argue that the contribution from the circular

arc vanishes.

References:

1. K. F. Riley, M. P. Hobson and S. J. Bence, Mathematical Methods for Physics and Engineering,

2nd Ed., Cambridge, Cambridge UK (2002).

2. The Wolfram web site: mathworld.wolfram.com/

3. Susan Lea, Mathematics for Physicists, Thomson (2004).

4. Murray R. Spiegel, Complex Variables, McGraw-Hill Schaum’s Outline (1964).

5. James Nearing, Mathematical Tools for Physics, www.physics.miami.edu/nearing/mathmethods/, [email protected] 6. A. L. Fetter and J. D. Walecka, Theoretical Mechanics of Particles and Continua, McGraw-

Hill (1980). See appendix A.

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