1. October 21 Residue theorem 7.1 Calculus of residues Chapter 7 Functions of a Complex Variable II 1 Suppose an analytic function f (z) has an isolated singularity at z 0. Consider a contour integral enclosing z 0. z0z0 The coefficient a -1 =Res f (z 0 ) in the Laurent expansion is called the residue of f (z) at z = z 0. If the contour encloses multiple isolated singularities, we have the residue theorem: z0z0 z1z1 Contour integral =2  i ×Sum of the residues at the enclosed singular points

2. 2 Residue formula: To find a residue, we need to do the Laurent expansion and pick up the coefficient a -1. However, in many cases we have a useful residue formula(Problem 6.6.1):

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5. 5 Residue at infinity: Stereographic projection: Residue at infinity: Suppose f (z) has only isolated singularities, then its residue at infinity is defined as Another way to prove it is to use Cauchy’s integral theorem. The contour integral for a small loop in an analytic region is One other equivalent way to calculate the residue at infinity is By this definition a function may be analytic at infinity but still has a residue there.

6. 6 Read: Chapter 7: 1 Homework: 7.1.1 Due: October 28

7. 7 Cauchy principle value: Suppose f (z) has an isolated singularity z 0 lying on a closed contour C. The contour integral is then not well defined. To solve this problem, we can remove a small segment of the contour for a distance of  on each side of the singularity and create a new contour C(  ). We define the Cauchy principle value of as October 24 Cauchy principle value z0z0 C()C() S  C Let us first see on what condition this limit exists. We draw a semicircle path S with radius  around z 0. Suppose contour C=C(  )+S does not enclose any singularity, then

8. 8 2) We assumed that the contour is smooth at z =z 0. If not the value should be 3) It is easy to remember: when the singularity is on the contour, it contributes half as if it were inside the contour. 4) Similar definitions exist for open contour integrals. Example: Also for infinity integration limits. Example: More about Cauchy principle values: 1) If C originally encloses isolated singularities, then

9. 9 Example: 1 C i -i R

10. 10 Read: Chapter 7: 1 Homework: 7.1.4 Due: November 4

11. 11 Cauchy’s integral theorem and Cauchy’s integral formula revisited: (in the view of the residue theorem): October 26 Evaluation of definite integrals -1 7.1 Calculus of residues

12. 12 L’Hospital’s rule: I. Integrals of trigonometric functions : C r=1

13. 13 C r=1 z+z+ z-z- C

14. 14 C r=1 z+z+ z-z-

15. 15 C r=1 z+z+ z-z- z0z0

16. 16 Read: Chapter 7: 1 Homework: 7.1.7,7.1.8,7.1.10 Due: November 4

17. 17 October 31 Evaluation of definite integrals -2 7.1 Calculus of residues II. Integrals along the whole real axis: Assumption 1: f (z) is analytic in the upper or lower half of the complex plane, except isolated finite number of poles. ∩ R Condition for closure on a semicircular path: Assumption 2: when |z| , | f (z)| goes to zero faster than 1/|z|. Then,

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19. 19 III. Fourier integrals: Assumption 1: f (z) is analytic in the upper half of the complex plane, except isolated finite number of poles. Jordan’s lemma: ∩ R

20. 20 Question: How about Answer: We can go the lower half of the complex plane using a clockwise contour.

21. 21 Question: How about Answer:

22. 22 Read: Chapter 7: 1 Homework: 7.1.11,7.1.12,7.1.13,7.1.14,7.1.16 Due: November 11

23. 23 November 2 Evaluation of definite integrals -3 7.1 Calculus of residues IV. Rectangular contours: Exponential and hyperbolic forms is a periodic function. We may use rectangular contours and hope that the integral reappears in some way on the upper contour line. R ii

24. 24 R i  i3  This integral can also be done by using the line y=i  and the fact

25. 25 V. Sector contours For functions involving we may use sector contours and hope that the integral reappears in some way on the upper radius of the sector. R 2/n2/n

26. 26 VI. Contours avoiding branch points When the integrands have branch points and branch cuts, contours need to be designed to avoid the branch points and the branch cuts. CC L+L+ L-L- ia -ia

27. 27 CC L+L+ L-L- ia -ia

28. 28 Read: Chapter 7: 1 Homework: 7.1.17 (b),7.1.18,7.1.19,7.1.21,7.1.22,7.1.25,7.1.26 Due: November 11

29. 29 Reading: Dispersion relations 7.2 Dispersion relations Suppose f (z) = u(z)+iv(z) is analytic, and in the upper half plane, then C R x0x0 These are the dispersion relations. u(z) and v(z) are sometimes called a Hilbert transform pair.

30. 30 Symmetry relations:

31. 31 In optics, the refractive index can be chosen to be a complex number where its real part is the normal refractive index, determined by the electric permittivity  (  ), and its imaginary part is responsible for absorption, determined by the electric conductivity  (  ): These are the Kramers-Kronig relations. These relations state that knowledge of the absorption coefficient of a material over all frequencies will allow one to obtain the (normal) refractive index of the material for any frequency, and vise versa.

32. 32 Hi, Everyone: My previous student Mahendra Thapa brought to me this problem, which may be in Arfken 3 rd edition: It is used in the black body radiation: If you can crack this integral (using contour integral on the complex plane) by yourself, please come to show me.