Diapositiva 1

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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
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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).
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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.

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Fig 19.13
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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 ? ?
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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)

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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 gt 0 for which g(c rei?) ? M.
• Hence It follows that limr?0I2 0 and
  limr?0I2 0.We complete the proof.

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Example 5

• Evaluate the Cauchy principal value of
• Solution Since the integral is of form (3), we
  consider the contour integral

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Fig 19.14

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

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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

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Example 5 (3)

• Now
• Therefore,

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Example 5 (4)

• Using e-1i e-1(cos 1 i sin 1), then

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Indented Paths
Cr
r
x0
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CR
Cr
I2
I1
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CR
Cr
I2
I1
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Contour Integration Example

• The graphical interpretation

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gtgt x-10pi0.110pi gtgt plot(x,sin(x)./x) gtgt
grid on gtgt axis(-10pi 10pi -0.4 1)
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gtgt x-10pi0.110pi gtgt plot3(x,zeros(size(x))
,sin(x)./x) gtgt grid on gtgt axis(-10pi 10pi -1 1
-0.4 1)
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gtgt x-10pi0.110pi gtgt y -30.13.' gtgt
zones(size(y))xi.(yones(size(x))) gtgt
mesh(x,y,cos(z)./z)
gtgt mesh(x,y,sin(z)./z)
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Cauchys Inequality If f is analytic inside and
on CR and M is the maximum
value of f on CR, then
R
z0
CR
Proof
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As R goes to infinity, then f(z) must go to
zero, everywhere. Then f(z) must be constant.
Liouvilles Theorem If f is entire and bounded
in the complex plane, then f(z) is
constant throughout the plane.
Gausss 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
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C1
The integral does not go to zero on the circle,
the integral cant be solved this way.
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Jordans Lemma
y
x
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C1
C2
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C2