Complex Integration and Other Methods.Pdf
Complex Integration and Methods
Prerequisites: The complex functions handout Concepts of primary interest: ∫ f() z dz = 0 for f(z) analytic in the interior Laplace property Poles of an analytic function Laurent series Residue theorem: fzdz()= 2π i Res(() fz ∫ ∑ m 1f () z dz Cauchy’s integral formula fa()= for f(z) analytic 2π i∫C za− 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,
Contact: [email protected] 5/30/2012
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:
fz()21− fz () The Limit must be defined independent of the manner in which z1 and z2 approach zz, → z 12 o()zz21−
zo. If z is represented as x + i y where x, y ∈ the condition requires that
fx(0+∆ xy , 0 ) − fx (, 00 y ) fx (,00 y+∆ y ) − fx (,00 y ) Limit = Limit . The conclusion is that ∆→xy00 ∆∆x ∆→ iy (except for constant functions) differentiable functions of a complex variable must by complex valued. ∂∂ff Required: = −i [CI.1] ∂∂xy 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 iy −∆x 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 (,xy ) ∆ property of analytic functions is suggested by -i∆y i y examining the path integral of f(z) around a ∫ f() z dz small path. The argument and result are reminiscent of Stokes theorem. ∆x
x
fzdzfxy( )≈ ( , − ½ ∆ y ) ∆+ x fx ( + ½ ∆ xyiy , )( ∆ ) ∫ +fxy( , + ½ ∆ y )( −∆ x ) + fx ( − ½ ∆ xy , )( − iy ∆ )
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∂∂ff ∫ f() z dz≈ i +(i ∆∆ y x ) = 0 [CI.2] ∂∂yx(,)xy Cauchy’s integral Theorem: Property [CI.1] of differentiable functions was used to show that the integral vanishes. This result suggests that: ∫ f() z dz = 0 [CI.3] anypath bounding region of analyticity 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: ∂∂uv ∂ u ∂ v = and = − [CI.4] ∂∂xy ∂ y ∂ x Computing the second derivatives: ∂∂22uv ∂ 2 u ∂ 2 v ∂∂22uu = and = − ⇒ +=0 ∂x22 ∂∂ xy ∂ y ∂∂ yx ∂∂xy22 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.
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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
-1 ½ Examples of functions that fail to be entire are (1 – z) , tan(π z), ln(z) and z . 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. -1 Im(z) ln(z) = ln(|z|) + i tan ( /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.
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Natural Logarithm
ln[|z|] + i π ln[|z|] ln[|z|] + i π ln[|z|] xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx ln[|z|] + i π ln[|z|] + i2π ln[|z|] - i π ln[|z|]
Branch cut along the positive real axis. The Branch cut along the negative real axis. The labeled values are points just above and labeled values are points just above and below the real axis. The values progress below the real axis. The values progress continuously across the negative real axis continuously across the positive real axis but but jump by 2πi crossing the positive axis. jump by 2πi crossing the negative axis.
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.
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Square Root
½ ½ i |z|½ |z| i |z|½ |z| xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
i |z|½ - |z|½ – i |z|½ |z|½
Branch cut along the positive real axis. The Branch cut along the negative real axis. The labeled values are points just above and labeled values are points just above and below the real axis. The values progress below the real axis. The values progress continuously across the negative real axis continuously across the positive real axis but but change sign crossing the positive axis. change sign crossing the negative axis.
A product of square roots affords one the opportunity to play with the branch cuts. Consider the ½ ½ function f(z) = [z – z1] [z – z2] = 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.
½ ½
½ ½ [z – 1] [z + 1] = W1 W2 [z – 1] [z + 1] = W1 W2
i|W | i |W | |W | |W | |W1| |W2| 1 2 i|W1| |W2| 1 2 (i|W1|)(i|W2|) (i|W1|)(|W2|) xxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxx i|W1| |W2| i|W1| (-i |W2|) -|W1| |W2| (i|W1|)(i|W2|) (i|W1|)(-|W2|) (-|W1|)(-|W2|)
Cuts left from -1 and right from +1. The Cuts right from -1 and right from +1. The values progress continuously across the real values progress continuously across the real axis between the branch points but change axis outside the branch points but change sign crossing the positive axis for |z| > 1. sign crossing the positive axis for |z| < 1. No cut between -1 and +1! There is a cut only between -1 and +1!
Exercise: Construct the equivalent branch cut drawings for fz()=−+ 1 z 1 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.
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*** 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
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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:
∞ aa− −+ a =pp +1 ++−1 + − n fz() pp−1 ... ∑ an () z z0 [TS.5] ()()zz−−00 zz () zz − 0n=0
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.
∞ n The more general Laurent expansion has the form: fz()=∑ an ( z − z0 ) and converges in the n=−∞ 1 annular region r1 < |z – zo| < r2. For example the function /(z [z-1]) has the following Laurent expansions about z = 1:
∞ 11 n n = =−−∑ ( 1) (z 1) for 0 < | z – zo| < 1 zz(− 1) ( z − 1)(1 +− [ z 1]) n= −1
11−2 = =−−n n ∞ and 21− ∑ ( 1) (z 1) for 1 < | z – zo| < zz(− 1) ( z − 1) (1 +− [ z 1] ) n= −∞
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. 1 The function /(z [z-1]) has poles at z = 0 and at z = 1. The small | z – 1| 1<|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 z = 1 at z = 0 lies on the circle | z – 1| = 1. The large | z – 1| expansion z = 0 0<|z-1|<1 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.
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sin(π z ) Exercise: Find the position of the singularities and their order for: fz()= , 1 (z − 1)
sin(π z ) π sin(π z ) −−()1/(zz ) fz()= , fz() = + sin(π z ) , fz()= and fz()= e 0 . 2 (z − 1) 2 3 (zz2 −+ 1)( 1) 4 π z 5
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:
∞ aa− −+ a =pp +1 ++−1 + − n Given a function fz() pp−1 ... ∑ an () z z0 with a pole of order p ()()zz−−00 zz () zz − 0n=0
p at zo, the function (z – zo) f(z) is analytic in a neighborhood of zo.
∞∞ p np+ m ()()()zz−0 fz =∑∑ azzn −= 00 amp− () zz − np=−=m0
–p SC0: The Laurent series for f(z) expanded about zo is (z – zo) times the Taylor’s series expansion
p 2 2 -1 of (z - zo) f(z) about zo. As an example, (z + a ) is to be expanded about z = ia. 11 f(z) = = z22+ a( z −+ ia )( z ia ) Clearly, f(z) has isolated simple (order 1) poles at ia and –ia. 1 1 11 1 = = − z22+ a()()2()() 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 1 1 1 (− 1) n+1 = − − + −2 ++ −n + 23()z ia (z ia ) ... n+ 1(z ia ) ... ()()()z++ ia z iaia z + ia ia ()z + ia ia ()z+ ia ia 1 1 1 1 (− 1) n+1 = −(z − ia ) + ( z − ia )2 ++ ... (z − ia )n + ... (z+ ia ) (2 ia ) (2 ia )23 (2ia ) (2ia )n+ 1 Multiplying by (z – ia)-1 yields the Laurent series. 1 1 1 1 (− 1) n f() z = − +(z − ia )1 ++ ... (z − ia )n + ... (2ia ) ( z− ia ) (2 ia )23 (2 ia ) (2ia )n+2 The binomial alternative: The binomial theorem can be used when a expression like (1 + x)n can be formed with |x| < 1.
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−12n 1 11z− ia 1 z −− ia z ia n z − ia = =1 + =1 − + −... +− ( 1) +... ()(2[])22222z+ ia ia +− z ia ia ia ia ia ia 2ia 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) 3 5 4 6 5 -(( (z- a) )/(32 a ))+(z- a) /(64 a )+O[z- a]
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: ∫ f() z dz = 0 anypath bounding region of analyticity
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: a dz 0 fork ≠ 1 = ; k integer and a constant ∫ k ()zz− 0 2π ia for k = 1
4.) For a circular contour centered on zo, the integral:
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∞∞ aa− −+ a pp+1 +++−=... −1 a() z znn dz a( z −= z ) dz 2π i a ∫∫pp−1 ∑∑nn001− ()()zz−−00 zz () zz − 0n=0 np=− 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 C 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.),
f() z dz= 2π i a− . zo ∫C 1
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,
fzdz( )= 2π i Res[ fz ( )] sum over enclosed poles zj ∫C ∑ j poles z j
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:
p−1 1 d p Res[f(zk)] ] = Lim (z− z ) f () z [CI.6] → p−1 ( k ) zzk (p− 1)! dz
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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.
∞ k − ()zz− ez = eez00() zz = e z 0∑ 0 k =1 k! 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.
10
7.5
5
2.5
1 2 3 4
-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:
dztan(ππ ) 1 d23 tan( z ) 1 d tan( π z ) π =+++23 + tan(z ) tan(0) zz23 z... dz z=0 2! dz zz=003! dz =
2ππ35 16 1 2 tan(πz )=+ 0 ππ sec2 ( zz ) + z35 + z +=...ππ z + ( z )3 + ( π z ) 5 + ... z=0 3! 5! 3 15 It is noted that tan(πz) passes through zero linearly at z = 0.
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-1 SC3: Consider the function [tan(π z)] . It has a first order poles at each integer value z. Invoking the power formula for computing residues, it follows that:
1 1 dp−1 ()zn− 1 1 = −p = →= Res Lim p−12(()()z zk f z) Lim π zz→→k − zn π π ππ tan( z) zn= (p 1)! dz tan(z ) sec ( n ) where:zk = nfz ; ( ) = tan(π z ) Use L'Hospital's Rule L’Hospital A less formal, but effective procedure is to brute force the Laurent expansion about z = 0. Using 12 tan(ππzz )=+++ ( π z )35 ( π z ) ... , 3 15 −1 1 1 11 24 2 = =1 ++ (ππzz ) ( ) + ... ππ1235 tan(zz ) ππzz++( ) ( π z ) + ... 3 15 3 15 12 Recalling that [1+ u]-1 ≈ 1 – u + u2 + … and setting u = (ππzz )24++ ( ) ... , we expand 3 15 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 12 order. For fourth order in z, u must be ()ππzz24+ () and the expansion must be extended to 3 15 order u2. Combining and stirring, 2 111224 12 24 ≈−1 ()ππzz + () + () ππ zz + () tan(ππzz ) 3 15 3 15
11121 2 44 6 ≈1 −− ()πz () ππ zz + () + Order(z ) tan(ππzz ) 3 15 9 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. 11ππ3 ≈−−zz3 +... tan(ππzz ) 3 45
-1 -1 The coefficient of the z power is π verifying the residue value found above.
2 -1 Exercise: Find the residue of [z tan(π z)] at z = 0 using the power formula. What is the order of
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the pole? Residue: - π/3
4 -1 Exercise: Consider [z tan(π z)] at z = 0. What is the order of the pole? Find the residue by any 3 method. Residue: - π /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.
1 ()zn−− 1 ()1zn 1 1 Res =Lim =Lim →= 4π zn→→4π4zn π42 π ππ 4 z tan( z) zn= ≠0 ztan( zn ) tan( zn ) sec( nn )
dz SC5. Summing inverse powers of integers: Consider the integral: for a contour that ∫C zz4 tan(π ) circles the integer values from 1 to N along the real axis.
Im Im
1 N
Re Re
Contour 1 Contour 2
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. dz NN11 = 2ππi Residues= 2 ii = 2 ∫C 2 4 ∑ ∑∑44 zztan(ππ ) nn=11n= n If the right end of the contour is moved further and further to the right,
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Im Im
Re Re Contour 3
dz ∞∞11 = 2ππi Residues= 2 ii = 2 ∫C3 4 ∑ ∑∑44 zztan(ππ ) nn=11n= n One gets double the result if a contour that circles the negative real axis is added. The residues appear in a matching set.
Im
Re Contour 4
dz ∞∞11 = 2ππi Residues= 2 2 ii= 4 ∫C 4 4 ∑ ∑∑44 zztan(ππ ) nn=11n= n 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.
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Contour 5
Contour 6
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. dz dz dz ∞ 1 = = ==−=4i 2π iz Res at 0 ∫∫∫CCC456444∑ 4 zztan(πππ ) zz tan( ) zz tan( ) n=1 n
3 The residue was found in an exercise above; it is - π /45.
∞∞4 113 π =−ππ ==−−π = 4i∑∑44 2 iz Res at 0 2 i( 45) or nn=11nn= 90 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: ∂∂22uu ∂∂22 vv +=0 and +=0 ∂∂xy22 ∂∂ xy22
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∞ dx SC7: A standard contour integration example. Consider the integral: . It appears ∫−∞ xa22+ innocuous, but question arise in its evaluation. Revamp it as the complex integration along the real dz axis from left to right. . ∫C1 za22+
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: 1 1 11 1 = = − z22+ a( z −+ ia )( z ia ) 2 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 ()z− ia 1 Res = Lim = −+ z→ ia −+ (z ia )( z ia ) z= ia (z ia )( z ia ) 2 i a
This result could be found by brute forcing a Laurent expansion about i a. 1 1 11 1 = = − + − +− − z− ia ()()()(2z ia z ia z ia ia[ z ia] )()(2) z ia ia (1+ 2ia )
1 11 2 11 1 1 = −+z−− ia z ia += − + −+ 1 ( 22ia ) ( ia ) ... 23(z ia ) ... (z−+ ia )( z ia ) (2 ia ) ()z− ia (2ia ) ( z− ia ) (2 ia ) (2 ia ) -1 -1 The coefficient of the (z – ia) is (2 i a) 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.
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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 → ∞.
dz 1 π = 2ππi Res[ ia] = 2 i = ∫C 2 z22+ a 2ia a ia
Contour C2 -ia
∞ dx SC8: Old fashioned integration example. Consider the integral: . Revamp it as the ∫−∞ xa22+ dz complex integration along the real axis from left to right. and rewrite that as: ∫C1 za22+ 1 11 − dz . What issues arise if we use our old fashioned methods? 2i a∫C1 z−+ ia z ia
dz 1 11 1 ∞ = −dz =[ln( z −− ia ) ln( z + ia )] . The problem is deciding ∫∫CC11z22+ a22 i a z −+ ia z ia i a −∞ 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. The path from -∞ on the real axis to ∞ on the real axis runs below the branch cut for ln(z – For ln(z – ia) ia ia) and above the cut for ln(z + ia). Points just -iπ -iπ R e R ei0 below the branch cut are about |z|e while those just above are about |z|e +iπ. Set the iπ R e limits as – R and + R. -ia For ln(z + ia)
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dz 11RR =[ln(z −− ia )] [ ln(z + ia )] ∫C1 z22+ a22 ia −−RRia
11 =[(lnRi +− 0) (ln Ri −ππ )] −[ (ln Ri +− 0) (ln Ri + )] 22ia ia dz 1 π =[2π i] = ∫C1 z22+ a2 ia a
− R Note that a more robust process would been to evaluate [ln(z ia )] −R as:
R [ln(zia− )] = ln[ Ra22 ++ ] i tan −− 1−−aa −ln[Ra22 ++ ] i tan 1 and −R ( RR) ( − )
R [ln(zia+ )] = ln[ Ra22 ++ ] i tan−− 1aa − ln[Ra22 ++ ] i tan 1 −R ( RR) ( − ) 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:
R Limln( z− ia ) = i Lim tan−−11−−aa− tan ≈i Lim 0 −− (ππ ) =i and [ ] −R ( ) [ ] RR→∞ R→∞ RR − →∞
R Limln( z+ ia ) = i Lim tan−−11aa− tan ≈i Lim 0 −+ (ππ ) =−i [ ] −R ( ) [ ] RR→∞ R→∞ RR − →∞ These results confirm the answer found previously. Clearly, one must tread lightly around branch cuts.
1f () z dz SC9: Cauchy’s integral formula: fa()= for f(z) analytic inside C. 2π i∫C za− 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
f(z) of /(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
f(z) function /(z – a) is analytic everywhere inside CA so the contribution from that contour vanishes. f() z dz f () z dz ∫∫= CCza−−B za
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An analytic function is continuous so for every ε, a δ C = CA + CB
CA can be found such that for every point on the inner
contour CB, |f(z) - f(a)| < ε. It follows that: − δ f() z f () a dz CB ∫ < 2πε CB za− a In the limit δ → 0, f() z dz f () a dz ∫∫→=2π if () a CCBBza−− za 1f () z dz ⇒ fa()= π ∫C − 2 i za
dz i φ Exercise: Show that ∫ = 2π i where the contour CB is parameterized as: z = a + δ e for the CB za− parameter range: 0 ≤ φ < 2π.
SC10: Trigonometric integrals and circular contours: The unit circle centered on the origin can be φ parameterized as: z = e i 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
iφ iφ dz centered on the origin, z = R e and dz = i R e dφ or, more significantly, dφ = -i /z.
2π cos2φφd =1 ()()z + z−12 − i z − 1 dz =−i z ++2 z−−1 z 3 dz ) ∫∫0 unit ∫unit 44circlel circlel By inspection, the residue or coefficient of z-1 in the Laurent expansion about 0 is 2. 21 fz()=++ z zz3
2π cos2φφd =−−iiz ++2 z−−13 z dz ) = (2)Res[π i z == 0] π ∫∫0 unit 44circlel
This procedure might seem excessive to show that the average value of cos2φ = ½. Apply the
22ππ procedure to: cos2n φφdd and 1 φbefore deciding that it lacks merit. See SC13/14. ∫∫001+a cosφ
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2π −−11 General Form: f (sinφ ,cos φφ ) d→− i f zz−+, zz z−1 dz . The contour chosen is a circle ∫∫0 C ( 22i ) 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. ezt dz Consider the integral for the contour C chosen as a CCW circular path of radius 2 centered ∫C z2 +1 on the origin. The integrand has simple (first order) poles at z = i and – i. Examining the pole at i,
eezt zt eit Res =−=Lim() z i 2 + zi→ −+ z 1 zi= (ziz )( i ) 2 i zt -i t 1 e dz The residue at –i is – e /2i. Combining the results with the residue theorem, = sin(t ) . 21π iz∫C 2 + 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.
−−izt izt 11∞ e dz e dz → ; η small positive 22ππii∫∫−∞ zi++ηηC1 zi
1 e−izt dz →=<0fort 0 2πηi∫C 2 zi+
− 1 eizt dz → =−=−Res[ziη ] =−e−ηt ≈−1 2π i ∫C3 zi+ η Contour C2 -iη
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
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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: −− −− ∞ iz(( t t00))iz t t iie dz e dz −−η ()tt0 → =Θ−()tte0 22ππ∫∫−∞ zi++ηηC1 zi
+ 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: 1 Integrate[Exp[+ I z ]/(z + I/64 ),{z, - Infinity, Infinity}] = 0 (t < 0 case) t = -1; η = /64 2 1 Integrate[Exp[- I z ]/(z + I/64 ),{z, - Infinity, Infinity}] = 1 64 (t > 0 case) t = +1; η = /64
1 1 η = /1024
0.8
1 0.6 η = /4
0.4
0.2
-2 -1 1 2
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 φ parameterized as: z = e i for the parameter range: 0 ≤ φ < 2π. This leads to representations of the trig
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functions of the forms: cos(φ) = (z + z-1)/2 and sin(φ) = (z - z-1)/(2i). In the case on any circular
iφ iφ dz centered on the origin, z = R e and dz = i R e dφ or, more significantly, dφ = -i /z.
2π −−11 General Form: f (sinφ ,cos φφ ) d→− i f zz−+, zz z−1 dz . The contour chosen is a circle ∫∫0 C ( 22i ) of radius one traveled in the CCW sense and centered on z = 0.
2π −1 2n Apply to procedure to: cos2n φφd → − i zz+ z−1 dz . ∫0 ∫C ( 2 )
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.
112nn(2nn )! 021− − 211( n− (2 )! − −+21 n) −i22nn z ++...z ++ ... z z dz =− i z ++...z ++ ... z dz 22∫∫CC( nn!! ) ( nn!! )
2π 1 (2n )! 2n 2π (2n )! Therefore, the residue a-1 = − i 2n . cos φφd = 2n 2 nn!! ∫0 2 nn!!
2π −−11 SC14. Another trig example: General Form: f (sinφ ,cos φφ ) d→− i f zz−+, zz z−1 dz . ∫∫0 C ( 22i ) The contour chosen is a circle of radius one traveled in the CCW sense and centered on z = 0.
2π Apply to procedure to: 1 dφ where 0 < a < 1. ∫0 1+a cosφ
2π 11−1 1 dφ →− i −1 z dz=−2 i 2 dz ∫∫0 1+a cosφ CC1++ ½az [ z] ∫az++2 z a
−−12 −−12 -1 The integrand has simple poles at z1 = −+aa −1 and z2 = −−aa −1 . As a > 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. −−22ii1 fz()=2 = ( a ) az++2 z a (zz− 1)(zz− 2 )
⇒=−−−22ii11= Res[fz (11 )] ( z z )( aa) ( ) − (zz− 1)(zz− 2 ) ()zz12 zz= 1
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2π 11φ = 4π = 2π d ( a ) − ∫0 1+a cosφ 2 a 2−1 1−a2
1 dx SC15. Branch Cut Example I: where 0 < a < 1. ∫− 1 1−+x2 (1 ax ) 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) = 11 = W W . For definiteness z = 1 and z = -1. The branch cuts can be sent 11−+zz 1 2 1 2 left from -1 and right from +1, or they can both run to the right (left) from the branch points.
-½ -½ [1-z] = W1 [z + 1] = W2
|W | -i|W | |W2| 1 |W1| 1 i|W2| |W2| xxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
|W1| |W1| +i|W1| i|W2| -|W2| -|W2|
Cut to the right from -1. The values progress Cut to the right from +1. The values progress continuously across the real axis continuously across the real axis for Re[z] < - for Re[z] < 1. There is a sign change upon crossing the axis for Re[z] > 1. Note that 1. There is a sign change upon crossing the -½ -½ this is for [1-z] not [z -1] axis for Re[z] > -1. .
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-½ -½ [1-z] [z + 1] = W1 W2
|W1| i |W2| |W1| |W2| -i|W1| |W2| xxxxxxxxxxxxxxxxxxxx |W1| (i |W2|) |W1|(- |W2|) i|W1|(- |W2|) |W1W2|
Pole -|W1W2| Cuts right from -1 and from +1. The values @ z= a-1 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.
1 dx dz The goal is to compute I = which equals where ΓA is the ∫− ∫Γ 1 2 A 1−+x (1 ax ) |WW12 | (1+ az ) path parallel to the real axis and infinitesimally above it running from -1 to +1.
dz 1 dx ≡ I = ∫Γ ∫− A 1−+z2 (1 az ) 1 1−+x2 (1 ax )
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) 1 and the complex integrand equals . (−+ |WW12 | ) (1 az ) dz ≡+I ∫Γ B 1−+z2 (1 az ) 1 There is a net contribution (= 2 I) from ΓA plus ΓB because has a discontinuity (1−+z )(1 z ) (1 + az ) (branch cut) as z crosses the real axis between -1 and 1.
iφ The circular turning arc ΓC is described by z = -1 + ε . The contribution from this arc is bounded by:
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ieεφiφ d 2 ∝ ε ∫Γ C ε (1− a )
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 → ∞.
dz ≡=2I 2π i Res[ za =−−1 ] ∫complete 2 contour 1−+z (1 az )
dz 1 12π ≡=2I 2[ππ iza +−1 ) =2i = ∫complete 2 21− 2 2 contour 1−+z (1 az ) 1−z a ( a + z ) −1 a −1 1 −a za=− π I = 1− a2
1 dx π dθ One can also evaluate by choosing x = cosθ leading to . ∫− ∫ 1 1−+x2 (1 ax ) 0 (1+ a cosθ )
2π π Compare this result with the previous sample calculation. SC14: 1 dφ = 2 ∫0 1+a cosφ 1−a2
∞ dx SC16. Branch Cut Example II: The integral cannot be extended over the entire real ∫0 (1+ x3 ) axis because the integrand has a pole on the negative real axis. It can be computed by considering ln[z ] dz . The logarithm is chosen with a branch cut along the positive real axis. Note that the ∫ 3 C1 (1+ z ) branch point is at z = 0 so there is no contribution from the portions of the horizontal segments to the
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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. ln[z ] dz∞∞ (ln[ x ]−+ [ln[ x ] 2π i ] dx dx = = −2πi ∫∫33∫ 3 C1 (1++zx ) 00(1 ) (1 + x ) Complete the problem by exercising the residue theorem.
Ln[z]
Pole Ln[z] + i 2π Pole @ z= -1 @ z= -1
Contour C1 Contour C2
∞ dx 3 The integral can also be evaluated using a pie wedge contour C2. The value of z repeats ∫0 (1+ x3 )
2π 0 for a rotation /3. Choose a 120 pie wedge.
π dz ∞∞du ei2 /3 du ∞du ∞dx =+−=−=−11eeii2ππ /3 2 /3 ∫3 ∫∫3i2π /3333( ) ∫( ) ∫ C2 (1+z )00 (1 ++u ) (1 [ ue ] ) 0(1+u ) 0(1 +x ) As an added benefit, the contour encloses just one pole, and the residue is easily computed using the derivative rule for simple poles.
gz() gz () 1 i2π = kk= → = 1 3 Res[ fz ( )] = Res 2 e zzk ′ 3 hz()kk h () z 3 z i2π /3 zz= k e
∞ cos(x ) dx SC17. A standard contour integration example. Consider the integral: . Express it as ∫−∞ xa22+ cos(z ) dz ( eiz+ e− iz ) dz a complex integration along the real axis from left to right. = ½ .. ∫∫CC11z22+ a( z −+ ia )( z ia )
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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 ia 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 Contour C2 exponential eiz (= eiRe[z] e-Im[z] must be closed in -ia the upper half plane so that e-Im[z] is small.
The integrand can be expressed as:
()()1()()eeiz+− iz eeiz + − iz eeiz ++−− iz ee iz iz = = − 22 2(z+ a ) 4( z −+ ia )( z ia ) 4 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
11eeiz − −iz in the lower half plane. We need to find the residues of: and as those are 44i a z−+ ia i a z ia the only terms that contribute.
11eeiz −a e −−iz ea Res =−=; Res − 4i a z−+ ia 44 ia i a z ia 4ia z= ia z=− 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.
cos(z ) dz e−−aa− eπ e−a =2ππi Res[ ia ] − Res[ −= ia ] 2 i − = ∫ 22 { } C1 z+ a 44ia ia a
Exercise: The residues for SC17 were calculated indirectly. Use the standard equation for the residue eeiz −iz of a first order pole to compute the residues of and at ia and at –ia 2(za22++ ) 2( za 22 ) p−1 1 d p respectively. Res[f(zk)] ] = Lim (z− z ) f () z →− Lim( ( z z ) f () z ) . →→p−1 ( kk) zzkk(p− 1)! dz zz
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Exercise: Explain why the exponentials eiz and e-iz dictate that the contour be closed in the upper and lower half planes respectively.
∞ sin(x ) dx Exercise: What value do you expect for the integral ? Explain. Use contour methods to ∫−∞ xa22+ 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 d p−1 = − p Res[ fz ()] Limp−1 ((z zk ) fz ()) zz= k → zzk (p− 1)! dz For a first order pole, the equation simplifies to : = − Res[ fz ()] = Limz[ ( zk ) fz ()] zzk zz→ k 3.) For the special case that f(z) has a first order pole and the function f(z) is the ratio of two analytic g(z) functions (= / 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, gz() gz( )+ g′′ ( z )( z −+ z ) ... gz ( ) + g ( z )( z −+ z ) ... fz()= = k kk→ k kk hz( ) hz (k )+ h′′ ( z kk )( z −+ z ) ... 0 +h ( zkk )( z −+ z ) ... = − Using Res[ fz ()] = Lim(( z zk ) fz ()) , zzk zz→ k
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gz() gz()+ g′ ()( z z −+ z )... gz () ==−=k kk k Res[ f ( z )] = Res Lim() z zk zzk zz→ k ′′ hz() 0+hz (kk )( z −+ z ) ... hz (k ) zz= k
More complicated relations can be derived for the residue of higher order poles. Note that he dh notation h′ has been used to represent /dz.
Residue examples: The function f(z) = z2/(1 + z4) has simple poles at z = ee±±iiππ/4; 3 /4 . In order to evaluate the residue at eiπ/4, one must find
ze2 ()iπ /4 2 Limit() z−= eiπ /4 π iπ/4 i3 π /4 −− i3 π /4 i π/4 iii π/4 3 ππ /4 /4− ii 3 ππ /4 /4 − i π /4 ze→ i /4 (ze−− )( ze )( ze − )( ze − ) ( e − e )( e − e)( e − e ) The vector picture of complex numbers provides some relief.
()eiπ /4 2 ie−iπ /4 = = (eeeeiiiπ/4−− 3 ππ /4 )( /4−− ii 3 ππ /4)( ee /4 − i π /4 ) 2 (2eiiπ /4 )( 2 ) 4 A product of differences of the pole positions appears in the denominator. Each factor is z1 – z2 = 2 z z 2 1 represented by a vector on a plot of the pole
1 1 locations. The factor (z1 – z2) is the vector from 2 ½ z2 to z1 which is the real number 2 . The factor 1 2 (z1 – z3) is the vector from z3 to z1 or 2 iπ/4 iπ /4 e . zz13−=2 e
z3 z4 iπ /4 zz13−=2 e 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.
gz() gz()+ g′ ()( z z −+ z )... gz () ==−=k kk k Res[ f ( z )] = Res Lim() z zk zzk zz→ k ′′ hz() 0+hz (kk )( z −+ z ) ... hz (k ) zz= k Given f(z) = z2/(1 + z4), g(z) = z2 and h(z) = 1 + z4. The residue for the pole at eiπ/4 becomes:
gz() z2 1 e−iπ /4 Resfz ( )π = Res = = = [ ] ze= i /4 3 hz()ze= iπ /4 4 zeiπ /4 4 zeiπ /4 4
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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:
g(z) If a function f(z) has a second order pole at zo and f(z) is of the form /p(z) where g(z) and p(z) are -2 analytic in a neighborhood of zo. Further, that the analytic function (z – zo) p(z) = h(z) is known,
g′() z gz () h '() z then: Resfz ( ) = − (restricted 2nd order pole result) [ ] zz= 2 0 hz() (()) hz zz= 0
Consider z2/(z2 + a2)2 which has second order poles at z = ± ia. The residue at ia is to be found. 2 2 Matching up, g(z) = z and h(z) = (z + ia) .
z22g′() z g () z h '() z 2z ( z )2( z+ ia ) =−=− Res 2 22 2 2 4 (z+ a )h () z (()) h z (z++ ia ) ( z ia ) z= ia z= ia z= ia
z22 2ia ( ia ) 2(2 ia ) 1 1 1 −i Res = − = −= 2+ 22 2 4 (z a ) z= ia (2ia ) (2 ia ) ia 2 4 4 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.
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Jordan’s Lemma provides conditions under y which the contribution from the infinite C1 semicircular arc vanishes.
g() z eiaz dz → 0 for Re[a] = 0 if |g(z)| → 0 ∫C1 Pole in upper -1 → ∞ half plane faster than R as R . g() z eiaz dz → 0 for Re[a] > 0 if |g(z)| → 0 x ∫C1 -R Contour C2 +R uniformly as R → ∞. For R such that |g(z)| < ε, ε Pole in lower the contribution is of order /Re[a] → 0. half plane In the case that the form g() z e−iaz dz with Re[a] > 0 is considered, the contour must be closed in ∫C1 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.
zb zb
za za zc zc
zd zd
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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:
zb zb
za za zc zc
zd zd
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 zb right ⇒ na = -1. The path does not cross the za zc upper zd line ⇒ nd = 0. It crosses the lower zd line once from right to left and once from left to
zd 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.
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Modified Residue Theorem: fzdz()= 2πη i Res(() fz ∫ ∑ mm
The symbol ηm represents the winding number of the contour about the pole at zm.
Problems 1.) Compute a() z− zn dz for a circular path of radius R about the point z . For this path, z = z + ∫ n 0 o o iφ R e 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 a() z−= z−1 dz 2π i a . Hint: what is dz? ∫ −−10 1
dz 2.) Examine the integrand of to verify that adding the two semicircular arcs at infinity ∫C5 zz4 tan(π ) to the contour makes a negligible contribution to the integral.
dz dz 3.) Explain why the integral can be shifted to , the one over contour 6 ∫C5 zz4 tan(π ) ∫C6 zz4 tan(π ) 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
p−1 1 d p Res[f(zk)] ] = Lim (z− z ) f () z → p−1 ( k ) zzk (p− 1)! dz -1 yields a -1, the coefficient for the (z – zo) factor in the expansion
5.) Find the location, order and residue for each pole of the following functions: z z2 eiz eiz a.) , b.) , c.) , d.) z2 − 4 (z − 2)2 z −π zztan(π )
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∞ 1 6.) Evaluate using a contour integration. ∑ 2 n=1 n 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.
zb zb
za za zc zc
zd zd
na = -1, nb = +1, nc = +2, nd = 0 na = -2, nb = +2, nc = nd = -1 The left side drawing could be distorted to be:
zb zb
za za zc zc
zd zd
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.
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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. 2 4 -1 kz a.) z b.) z c.) z d.) e
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.
sin(z) 10.) Show that /z has a removable singularity at z = 0.
2 -1 11.) a.) Find the residue of [z tan(π z)] at z = 0 using the power formula. What is the order of the pole? 4 -1 b.) Consider [z tan(π z)] at z = 0. What is the order of the pole? Find the residue by any 3 method. Residues: - π/3 ; - π /45
12.) Develop an argument supporting Cauchy’s integral formula that is based on the residue theorem. See SC9.
f() z dz 1 ()df dz 13.) Use standard integration methods to argue that: = dz . The contour C ∫∫CC()(1)()za−nn n −− za−1 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.
f() z dz 1 dn f 14.) Use the result of the previous problem to conclude that: = . The contour ∫C nn+1 ()z− a n ! dz za= 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.
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f() z dz 1 dn f 15.) Use the residue theorem conclude that: = . The contour C is a ∫C nn+1 ()z− a n ! dz za= 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.
2π 2π 16.) Compute a.) cos2n φφd and b.) 1 dφ . Expect to use the binomial theorem and ∫0 ∫0 1+a cosφ 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.
(2n )! 2π π likely answers: 2 2n ; 2nn !! 1− a2
∞ 17.) a.) Compute dz where a is a real positive number. Nearing (reference 5) suggests using ∫−∞ za44+ π a graphical representation of the zj – zi, the pole separations. 2 a3
∞ ikz π b.) Compute e dz where a is a real positive number. e−ka /2cos[(ka )−π ] ∫−∞ za44+ a3 2 4
2π dφ 18.) Consider an integral of the form: . Describe the method to solve this ∫0 1++ab sinφφ cos integral. Given that a and b are real, what limits would you impose on those values to ensure that the integral is finite.
n 2π (cosφφ ) d 19.) Consider an integral of the form: . Describe the method to solve this ∫0 1++ab sinφφ cos integral. Given that a and b are real, what limits would you impose on those values to ensure that the integral is finite.
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2π dφ 20.) Compute: for a, b, c real. Explain why a + b ≠ 0 is an adequate ∫0 a++ bcosφφ ci sin
2π dφ condition to ensure that the integral is finite. Compute: ∫0 8−− 28cosφφ 4i sin 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.
zb zb
za za zc zc
zd zd
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.
Im Im
1 N
Re Re
Contour 1 Contour 2
a.) Represent contour 2 as contour 1 plus a contour Cadd with the property that f(z) is analytic
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everywhere on and inside that contour. What is the value of f() z dz ? ∫Cadd
Contour 5
Contour 6
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 f() z dz ? ∫Cadd
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: sin(π z ) sin(π z ) cos(π z ) a.) about z = π. b.) about z = π. c.) about z = 0. ()z −π ()z −π 2 z2
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sin(ze ) z d.) − about z = 0. zz2 1 25.) a.) Develop the Laurent series for the function /(z [z-1]) about z = 1 that converges for small |z - 1| and identify the bounds on it annulus of convergence. 1 b.) Develop the Laurent series for the function /(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 −1 tz and the Bromwich integral f( t )= (2π i ) f ( z ) e dz where fz() is L[f(t)]s=z. The contour consists ∫C
of a straight path up parallel to the imaginary axis and to the right of all the poles of fz() and
which is closed by a large circular arc closing on the left thus enclosing all the poles of fz(). 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.
3 k2 2 ∞ ikx− t a 1 2m 27.) Given Ψ=(,)x t e dk . Compute Ψ(x,0). Note that the result π ∫−∞ ka22+ depends on the sign of x.
∞ −+()t ik 2 28.) As a step to computing some Fourier transforms, one needs to evaluate e dt . One ∫−∞
∞ 2 knows that the real integral e− x dx = π . The desired integral corresponds to the complex ∫ −∞
∞ 2 integration e− z 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→ ∞.
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∞ n−+(1 ik ) x n! -1 29.) Evaluate: x e dx . Consider the target value to be + . Define α = tan (k). ∫0 (1+ik )n 1
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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