1. Maximum Modulus Principle: If f is analytic and not constant in a given domain D, then |f(z)| has no maximum value in D. That is, there is no z 0 in the domain such that |f(z)|  |f(z 0 )| for all points z in D. Proof: Assume that |f(z)| does have a maximum value in D. z0z0 R CRCR

2. Alternatively Theorem: If f is analytic, continuous and not constant in a closed bounded region D, then the maximum value of |f(z)| is achieved only on the boundary of D. Some other aspects of the maximum modulus theorem: Assume that f(z) is not 0 in a region R. Then if f(z) is analytic in R, then so is 1/f(z). Result: minimum of |f(z)| also occurs on the boundary. Since the max and min of |f(z)| are on the boundary, so is the max and min of u(x,y). Same applies to v(x,y).

3. Indented Contour The complex functions f(z) = P(z)/Q(z) of the improper integrals (2) and (3) did not have poles on the real axis. When f(z) has a pole at z = c, where c is a real number, we must use the indented contour as in Fig 19.13.

4. Fig 19.13

5. Suppose f has a simple pole z = c on the real axis. If Cr is the contour defined by THEOREM 19.17 Behavior of Integral as r → 

6. THEOREM 19.17 proof Proof Since f has a simple pole at z = c, its Laurent series isf(z) = a -1 /(z – c) + g(z) where a -1 = Res(f(z), c) and g is analytic at c. Using the Laurent series and the parameterization of C r, we have (12)

7. THEOREM 19.17 proof First we see Next, g is analytic at c and so it is continuous at c and is bounded in a neighborhood of the point; that is, there exists an M > 0 for which |g(c + re i  )|  M. Hence It follows that lim r  0 |I 2 | = 0 and lim r  0 I 2 = 0. We complete the proof.

8. Example 5 Evaluate the Cauchy principal value of Solution Since the integral is of form (3), we consider the contour integral

9. Fig 19.14 f(z) has simple poles at z = 0 and z = 1 + i in the upper half-plane. See Fig 19.14.

10. Example 5 (2) Now we have (13) Taking the limits in (13) as R   and r  0, from Theorem 19.16 and 19.17, we have

11. Example 5 (3) Now Therefore,

12. Example 5 (4) Using e -1+i = e -1 (cos 1 + i sin 1), then

13. Indented Paths x0x0 CC 

14. CC CRCR I1I1 I2I2

15. CC CRCR I1I1 I2I2

16. Contour Integration Example The graphical interpretation

17. >> x=[-10*pi:0.1:10*pi]; >> plot(x,sin(x)./x) >> grid on >> axis([-10*pi 10*pi -0.4 1])

18. >> x=[-10*pi:0.1:10*pi]; >> plot3(x,zeros(size(x)),sin(x)./x) >> grid on >> axis([-10*pi 10*pi -1 1 -0.4 1])

20. >> mesh(x,y,sin(z)./z) >> x=[-10*pi:0.1:10*pi]; >> y = [-3:0.1:3].'; >> z=ones(size(y))*x+i.*(y*ones(size(x))); >> mesh(x,y,cos(z)./z)

21. Cauchy’s Inequality: If f is analytic inside and on C R and M is the maximum value of f on C R, then z0z0 R CRCR Proof:

22. As R goes to infinity, then f’(z) must go to zero, everywhere. Then f(z) must be constant. Liouville’s Theorem: If f is entire and bounded in the complex plane, then f(z) is constant throughout the plane. Proof: Gauss’s Mean Value Theorem: If f is analytic within and on a given circle, its value at the center is the arithmetic mean of its values on the circle.

23. C1C1 The integral does not go to zero on the circle, the integral can’t be solved this way.

25. x y Jordan’s Lemma

26. C1C1 C2C2

27. C2C2