1 / 27

Maximum Modulus Principle:

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. R. z 0. C R.

Download Presentation

Maximum Modulus Principle:

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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 z0 in the domain such that |f(z)||f(z0)| for all points z in D. Proof: Assume that |f(z)| does have a maximum value in D. R z0 CR

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

  6. THEOREM 19.17 proof ProofSince f has a simple pole at z = c, its Laurent series is f(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 Cr,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 + rei)|  M. Hence It follows that limr0|I2| = 0 and limr0I2= 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 Cr r x0

  14. CR Cr I2 I1

  15. CR Cr I2 I1

  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])

  19. >> 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) >> mesh(x,y,sin(z)./z)

  20. Cauchy’s Inequality: If f is analytic inside and on CR and M is the maximum value of f on CR, then R z0 CR Proof:

  21. 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. 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. Proof:

  22. C1 The integral does not go to zero on the circle, the integral can’t be solved this way.

  23. Jordan’s Lemma y x

  24. C1 C2

  25. C2

More Related