The Cauchy Integral Formula

Transcription

Cauchy Integral Formula
Infinite Differentiability
Fundamental Theorem of Algebra
Maximum Modulus Principle
Bernd Schröder
Bernd Schröder
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Louisiana Tech University, College of Engineering and Science
Introduction
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Introduction
1. One of the most important consequences of the Cauchy-Goursat
Integral Theorem is that the value of an analytic function at a
point can be obtained from the values of the analytic function on
a contour surrounding the point
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Introduction
a contour surrounding the point (as long as the function is
defined on a neighborhood of the contour and its inside).
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Introduction
2. The result itself is known as Cauchy’s Integral Theorem.
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Introduction
2. The result itself is known as Cauchy’s Integral Theorem.
3. Among its consequences is, for example, the Fundamental
Theorem of Algebra, which says that every nonconstant complex
polynomial has at least one complex zero.
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Introduction
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Introduction
4. Once we have introduced series, another consequence is the fact
that every analytic function is locally equal to a power series.
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Introduction
This very powerful result is a cornerstone of complex analysis.
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Introduction
5. In this presentation we will at least be able to prove that analytic
functions have derivatives of any order.
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Introduction
5. In this presentation we will at least be able to prove that analytic
functions have derivatives of any order.
6. ... and the above are only highlights of the consequences of
Cauchy’s Integral Theorem.
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Theorem.
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Theorem. Cauchy’s Integral Formula.
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Theorem. Cauchy’s Integral Formula. Let C be a simple closed
positively oriented piecewise smooth curve, and let the function f be
analytic in a neighborhood of C and its interior.
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Theorem. Cauchy’s Integral Formula. Let C be a simple closed
positively oriented piecewise smooth curve, and let the function f be
analytic in a neighborhood of C and its interior. Then for every z0 in
the interior of C we have that
f (z0 ) =
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2πi
f (z)
dz.
C z − z0
Z
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Example.
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Example. Let’s first check out if the theorem works for f (z) = 1 and
the circle C(r, z0 ) of radius r around z0 .
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1
2πi
1
dz
C(r,z0 ) z − z0
Z
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1
2πi
1
dz =
C(r,z0 ) z − z0
Z
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1
2πi
Z 2π
0
1
ireit dt
(z0 + reit ) − z0
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1
2πi
1
dz =
C(r,z0 ) z − z0
Z
=
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2π
1
1
ireit dt
2πi 0 (z0 + reit ) − z0
Z 2π
1
1
ireit dt
2πi 0 reit
Z
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1
2πi
1
dz =
C(r,z0 ) z − z0
Z
=
=
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2π
1
1
ireit dt
2πi 0 (z0 + reit ) − z0
Z 2π
1
1
ireit dt
2πi 0 reit
Z
1 2π
dt
2π 0
Z
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1
2πi
1
dz =
C(r,z0 ) z − z0
Z
=
=
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2π
1
1
ireit dt
2πi 0 (z0 + reit ) − z0
Z 2π
1
1
ireit dt
2πi 0 reit
Z
1 2π
dt = 1
2π 0
Z
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1
2πi
1
dz =
C(r,z0 ) z − z0
Z
=
=
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2π
1
1
ireit dt
2πi 0 (z0 + reit ) − z0
Z 2π
1
1
ireit dt
2πi 0 reit
Z
1 2π
dt = 1 = f (z0 )
2π 0
Z
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Proof of Cauchy’s Integral Formula.
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D
-
C(r, z0 )
r
]
?
O
C
-
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1
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D
r
-
C(r, z0 )
r
]
?
O
C
-
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1
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D
r
r
-
C(r, z0 )
r
]
?
O
C
-
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1
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D
r
r
r-
C(r, z0 )
r
]
?
O
C
-
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1
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D
r
r
rr C(r, z )
r 0?
]
O
C
-
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1
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D
r
r
I r rC(r, z0 )
r
?
O
]
C
-
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1
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D
r
r
R
I r rC(r, z0 )
r
?
O
]
C
-
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1
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D
r
r
R
I r rC(r, z0 )
r
?
O
]
C
-
1
f (z)
dz =
C z − z0
Z
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f (z)
dz.
C(r,z0 ) z − z0
Z
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Z
f
(z)
C z − z0 dz − 2πif (z0 )
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Z
f
(z)
C z − z0 dz − 2πif (z0 )
Z
Z
f (z)
f (z0 ) = dz −
dz
C(r,z0 ) z − z0
C(r,z0 ) z − z0
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Z
f
(z)
C z − z0 dz − 2πif (z0 )
Z
Z
f (z)
f (z0 ) = dz −
dz
C(r,z0 ) z − z0
C(r,z0 ) z − z0
Z
f (z) − f (z0 ) dz
= z − z0
C(r,z0 )
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Z
f
(z)
C z − z0 dz − 2πif (z0 )
Z
Z
f (z)
f (z0 ) = dz −
dz
C(r,z0 ) z − z0
C(r,z0 ) z − z0
Z
f (z) − f (z0 ) dz
= z − z0
C(r,z0 )
Z
f (z) − f (z0 ) ≤
z − z0 d|z|
C(r,z )
0
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Z
f
(z)
C z − z0 dz − 2πif (z0 )
Z
Z
f (z)
f (z0 ) = dz −
dz
C(r,z0 ) z − z0
C(r,z0 ) z − z0
Z
f (z) − f (z0 ) dz
= z − z0
C(r,z0 )
Z
f (z) − f (z0 ) Z
1
d|z| ≤ max f (z) − f (z0 )
≤
d|z|
z − z0
C(r,z0 )
C(r,z0 )
C(r,z0 ) r
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Z
f
(z)
C z − z0 dz − 2πif (z0 )
Z
Z
f (z)
f (z0 ) = dz −
dz
C(r,z0 ) z − z0
C(r,z0 ) z − z0
Z
f (z) − f (z0 ) dz
= z − z0
C(r,z0 )
Z
f (z) − f (z0 ) Z
1
d|z| ≤ max f (z) − f (z0 )
≤
d|z|
z − z0
C(r,z0 )
C(r,z0 )
C(r,z0 ) r
1
= max f (z) − f (z0 )2πr
r
C(r,z0 )
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Z
f
(z)
C z − z0 dz − 2πif (z0 )
Z
Z
f (z)
f (z0 ) = dz −
dz
C(r,z0 ) z − z0
C(r,z0 ) z − z0
Z
f (z) − f (z0 ) dz
= z − z0
C(r,z0 )
Z
f (z) − f (z0 ) Z
1
d|z| ≤ max f (z) − f (z0 )
≤
d|z|
z − z0
C(r,z0 )
C(r,z0 )
C(r,z0 ) r
1
= max f (z) − f (z0 )2πr = 2π max f (z) − f (z0 )
r
C(r,z0 )
C(r,z0 )
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Z
f
(z)
C z − z0 dz − 2πif (z0 )
Z
Z
f (z)
f (z0 ) = dz −
dz
C(r,z0 ) z − z0
C(r,z0 ) z − z0
Z
f (z) − f (z0 ) dz
= z − z0
C(r,z0 )
Z
f (z) − f (z0 ) Z
1
d|z| ≤ max f (z) − f (z0 )
≤
d|z|
z − z0
C(r,z0 )
C(r,z0 )
C(r,z0 ) r
1
r
C(r,z0 )
C(r,z0 )
and if we choose r small enough, we can make the last term arbitrarily
small.
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Z
f
(z)
C z − z0 dz − 2πif (z0 )
Z
Z
f (z)
f (z0 ) = dz −
dz
C(r,z0 ) z − z0
C(r,z0 ) z − z0
Z
f (z) − f (z0 ) dz
= z − z0
C(r,z0 )
Z
f (z) − f (z0 ) Z
1
d|z| ≤ max f (z) − f (z0 )
≤
d|z|
z − z0
C(r,z0 )
C(r,z0 )
C(r,z0 ) r
1
r
C(r,z0 )
C(r,z0 )
small. But that means that the original difference must be zero.
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Z
f
(z)
C z − z0 dz − 2πif (z0 )
Z
Z
f (z)
f (z0 ) = dz −
dz
C(r,z0 ) z − z0
C(r,z0 ) z − z0
Z
f (z) − f (z0 ) dz
= z − z0
C(r,z0 )
Z
f (z) − f (z0 ) Z
1
d|z| ≤ max f (z) − f (z0 )
≤
d|z|
z − z0
C(r,z0 )
C(r,z0 )
C(r,z0 ) r
1
r
C(r,z0 )
C(r,z0 )
small. But that means that the original difference must be zero.
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1
Analyzing f (z0 ) =
2πi
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f (z)
dz.
γ z − z0
Z
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1
Analyzing f (z0 ) =
2πi
f (z)
dz.
γ z − z0
Z
1. Note that the right side is a function of z0 .
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1
Analyzing f (z0 ) =
2πi
f (z)
dz.
γ z − z0
Z
1
2. For z fixed,
is differentiable in z0 .
z − z0
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1
Analyzing f (z0 ) =
2πi
f (z)
dz.
γ z − z0
Z
1
2. For z fixed,
z − z0
3. So if we could move the derivative into the integral, we could get
a formula for f 0 .
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1
Analyzing f (z0 ) =
2πi
f (z)
dz.
γ z − z0
Z
1
2. For z fixed,
z − z0
a formula for f 0 .
1
4. And, anticipating that the new integrand will involve
,
(z − z0 )2
there is no reason to think that the process should stop there.
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1
Analyzing f (z0 ) =
2πi
f (z)
dz.
γ z − z0
Z
1
2. For z fixed,
z − z0
a formula for f 0 .
1
4. And, anticipating that the new integrand will involve
,
(z − z0 )2
there is no reason to think that the process should stop there.
5. So for analytic functions, being once differentiable should imply
that we have derivatives of any order.
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Warning.
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Warning. The next result (as motivated on the preceding panel) does
not work for functions of a real variable.
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not work
of a real variable. Consider
for functions
x2 ; for x > 0,
f (x) =
−x2 ; for x ≤ 0.
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not work
for functions
x2 ; for x > 0,
Its derivative is 2|x|
f (x) =
−x2 ; for x ≤ 0.
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not work
for functions
x2 ; for x > 0,
Its derivative is 2|x|, which is not
f (x) =
−x2 ; for x ≤ 0.
differentiable at 0.
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Theorem.
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Theorem. Cauchy’s Integral Formula
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Theorem. Cauchy’s Integral Formula (extended).
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Theorem. Cauchy’s Integral Formula (extended). Let C be a simple
closed positively oriented piecewise smooth curve, and let the
function f be analytic in a neighborhood of C and its interior.
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Theorem. Cauchy’s Integral Formula (extended). Let C be a simple
closed positively oriented piecewise smooth curve, and let the
function f be analytic in a neighborhood of C and its interior. Then
for every z0 in the interior of C and every natural number n we have
that f is n-times differentiable at z0 and its derivative is
f
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(n)
n!
(z0 ) =
2πi
f (z)
dz.
n+1
C (z − z0 )
Z
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Proof.
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Proof. Induction on n.
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Base step, n = 0.
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Base step, n = 0. This is Cauchy’s Integral Formula.
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Induction step, n → (n + 1).
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f (n) (w) − f (n) (z0 )
w→z0
w − z0
lim
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f (n) (w) − f (n) (z0 )
w→z0
w − z0
Z
Z
f (z)
f (z)
1
n!
n!
= lim
dz −
dz
w→z0 w − z0
2πi C (z − w)n+1
2πi C (z − z0 )n+1
lim
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f (n) (w) − f (n) (z0 )
w→z0
w − z0
Z
Z
f (z)
f (z)
1
n!
n!
= lim
dz −
dz
w→z0 w − z0
2πi C (z − w)n+1
2πi C (z − z0 )n+1
Z
n!
1
f (z)
f (z)
= lim
−
dz
n+1
w→z0 2πi w − z0 C (z − w)
(z − z0 )n+1
lim
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f (n) (w) − f (n) (z0 )
w→z0
w − z0
Z
Z
f (z)
f (z)
1
n!
n!
= lim
dz −
dz
w→z0 w − z0
2πi C (z − w)n+1
2πi C (z − z0 )n+1
Z
n!
1
f (z)
f (z)
= lim
−
dz
n+1
w→z0 2πi w − z0 C (z − w)
(z − z0 )n+1
lim
n!
= lim
w→z0 2πi
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Z
f (z)
C(z0 ,r)
1
(z−w)n+1
− (z−z1 )n+1
0
w − z0
dz
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f (n) (w) − f (n) (z0 )
w→z0
w − z0
Z
Z
f (z)
f (z)
1
n!
n!
= lim
dz −
dz
w→z0 w − z0
2πi C (z − w)n+1
2πi C (z − z0 )n+1
Z
n!
1
f (z)
f (z)
= lim
−
dz
n+1
w→z0 2πi w − z0 C (z − w)
(z − z0 )n+1
lim
1
Z
− (z−z1 )n+1
n!
(z−w)n+1
0
= lim
f (z)
dz
w→z0 2πi C(z0 ,r)
w − z0
Z
n!
1
=
f (z)(−(n + 1))
(−1) dz
2πi C(z0 ,r)
(z − z0 )n+2
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n!
2πi
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Z
f (z)(−(n + 1))
C(z0 ,r)
1
(−1) dz
(z − z0 )n+2
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1
(−1) dz
n+2
(z
−
z
C(z0 ,r)
0)
Z
f (z)
(n + 1)!
=
dz
n+2
2πi
C(z0 ,r) (z − z0 )
n!
2πi
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Z
f (z)(−(n + 1))
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1
(−1) dz
n+2
(z
−
z
C(z0 ,r)
0)
Z
f (z)
(n + 1)!
=
dz
n+2
2πi
C(z0 ,r) (z − z0 )
Z
(n + 1)!
f (z)
=
dz
2πi
(z
−
z0 )n+2
C
n!
2πi
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Z
f (z)(−(n + 1))
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1
(−1) dz
n+2
(z
−
z
C(z0 ,r)
0)
Z
f (z)
(n + 1)!
=
dz
n+2
2πi
C(z0 ,r) (z − z0 )
Z
(n + 1)!
f (z)
=
dz
2πi
(z
−
z0 )n+2
C
n!
2πi
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Z
f (z)(−(n + 1))
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Corollary.
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Corollary. Let f be an analytic function on an open domain.
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Corollary. Let f be an analytic function on an open domain. Then all
derivatives of f are analytic on this domain, too.
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derivatives of f are analytic on this domain, too. Moreover, the
component functions (the real and imaginary parts) have continuous
partial derivatives of all orders throughout the domain.
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Proof.
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Proof. By the preceding result, we see that f has derivatives of all
orders.
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orders. That means all derivatives are differentiable
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orders. That means all derivatives are differentiable and thus all
derivatives are analytic on the domain.
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derivatives are analytic on the domain. Therefore real and imaginary
part have
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part have (by repeated application of the Cauchy-Riemann equations)
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continuous partial derivatives of all orders throughout the domain.
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continuous partial derivatives of all orders throughout the domain.
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Theorem.
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Theorem. Morera’s Theorem.
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Theorem. Morera’s Theorem. Let f be a continuous complex
function on an open set so that for Zevery simple closed curve C
f (z) dz = 0.
contained in the open set we have
C
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f (z) dz = 0. Then f is analytic in
the open set.
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C
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the open set.
C
Proof.
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the open set.
C
Proof. By theorem from an earlier presentation, f has an
antiderivative F.
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the open set.
C
antiderivative F. By the preceding theorem, all derivatives of F
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the open set.
C
(including f ) are analytic.
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the open set.
C
(including f ) are analytic.
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Theorem. Cauchy’s Inequality.
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Theorem. Cauchy’s Inequality. Let f be analytic on a circle CR (z0 )
of radius R centered at z0 and in the circle’s interior.
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of radius R centered at z0 and in the circle’s interior. If |f | is bounded
n!M
R
by MR on the circle, then for all n we have f (n) (z0 ) ≤
.
Rn
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n!M
R
.
Rn
Proof.
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n!M
R
.
Rn
Proof.
n! Z
f
(z)
(n)
dz
f (z0 ) = 2πi CR (z0 ) (z − z0 )n+1 Bernd Schröder
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n!M
R
.
Rn
Proof.
n! Z
f
(z)
(n)
dz
f (z0 ) = 2πi CR (z0 ) (z − z0 )n+1 Z
f (z) n!
d|z|
≤
2π CR (z0 ) (z − z0 )n+1 Bernd Schröder
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n!M
R
.
Rn
Proof.
n! Z
f
(z)
(n)
dz
f (z0 ) = 2πi CR (z0 ) (z − z0 )n+1 Z
f (z) n!
d|z|
≤
2π CR (z0 ) (z − z0 )n+1 Z
n!
MR
≤
d|z|
2π CR (z0 ) Rn+1
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n!M
R
.
Rn
Proof.
n! Z
f
(z)
(n)
dz
f (z0 ) = 2πi CR (z0 ) (z − z0 )n+1 Z
f (z) n!
d|z|
≤
2π CR (z0 ) (z − z0 )n+1 Z
n!
MR
≤
d|z|
2π CR (z0 ) Rn+1
n!
MR
≤
2πR n+1
2π
R
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n!M
R
.
Rn
Proof.
n! Z
f
(z)
(n)
dz
f (z0 ) = 2πi CR (z0 ) (z − z0 )n+1 Z
f (z) n!
d|z|
≤
2π CR (z0 ) (z − z0 )n+1 Z
n!
MR
≤
d|z|
2π CR (z0 ) Rn+1
n!MR
n!
MR
≤
2πR n+1 =
Rn
2π
R
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n!M
R
.
Rn
Proof.
n! Z
f
(z)
(n)
dz
f (z0 ) = 2πi CR (z0 ) (z − z0 )n+1 Z
f (z) n!
d|z|
≤
2π CR (z0 ) (z − z0 )n+1 Z
n!
MR
≤
d|z|
2π CR (z0 ) Rn+1
n!MR
n!
MR
≤
2πR n+1 =
Rn
2π
R
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Theorem.
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Theorem. Liouville’s Theorem.
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Theorem. Liouville’s Theorem. Let f be an entire function.
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Theorem. Liouville’s Theorem. Let f be an entire function. (That is,
f is analytic in C.)
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f is analytic in C.) If f is bounded, then f is constant.
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Proof.
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Proof. Let |f | be bounded by B.
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Proof. Let |f | be bounded by B. Let R > 0 be arbitrary.
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Proof. Let |f | be bounded by B. Let R > 0 be arbitrary. By Cauchy’s
Inequality with n = 1, for every z0 in C we have
0
f (z0 )
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0
f (z0 ) ≤ B
R
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0
f (z0 ) ≤ B → 0 (R → ∞).
R
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0
f (z0 ) ≤ B → 0 (R → ∞).
R
Because R was arbitrary, we infer that f 0 (z0 ) = 0.
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0
f (z0 ) ≤ B → 0 (R → ∞).
R
Because R was arbitrary, we infer that f 0 (z0 ) = 0. Because z0 was
arbitrary, we have f 0 = 0.
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0
f (z0 ) ≤ B → 0 (R → ∞).
R
arbitrary, we have f 0 = 0. But f 0 = 0 implies that f is constant.
Bernd Schröder
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0
f (z0 ) ≤ B → 0 (R → ∞).
R
arbitrary, we have f 0 = 0. But f 0 = 0 implies that f is constant.
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Theorem.
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Theorem. Fundamental Theorem of Algebra.
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Theorem. Fundamental Theorem of Algebra. Every nonconstant
complex polynomial has at least one complex zero.
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Proof.
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Proof. Let p be a complex polynomial without zeros.
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Proof. Let p be a complex polynomial without zeros. Then
analytic in the complex plane.
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1
p
is
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Proof. Let p be a complex polynomial without zeros. Then 1p is
analytic in the complex plane. We claim that 1p is bounded on C.
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analytic in the complex plane. We claim that 1p is bounded on C. To
1
see this claim, first note that lim
=0
z→∞ p(z)
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1
= 0, because lim p(z) = ∞.
z→∞ p(z)
z→∞
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1
z→∞ p(z)
z→∞
1
Thus p is bounded outside a circle C(R, 0) for sufficiently large R.
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1
z→∞ p(z)
z→∞
1
But the only way 1p can be unbounded inside C(R, 0) is for the
denominator p(z) to go to zero somewhere
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1
z→∞ p(z)
z→∞
1
denominator p(z) to go to zero somewhere, which was excluded by
hypothesis.
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1
z→∞ p(z)
z→∞
1
hypothesis. Thus 1p is bounded on C.
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1
z→∞ p(z)
z→∞
1
Now by Liouville’s Theorem we infer that 1p is constant.
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1
z→∞ p(z)
z→∞
1
Now by Liouville’s Theorem we infer that 1p is constant. Hence p is
constant.
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1
z→∞ p(z)
z→∞
1
Now by Liouville’s Theorem we infer that 1p is constant. Hence p is
constant.
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Lemma.
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Lemma. If f is analytic in some neighborhood |z − z0 | < ε and
|f (z)| ≤ |f (z0 )| for all z in that neighborhood
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Lemma. If f is analytic in some neighborhood |z − z0 | < ε and
|f (z)| ≤ |f (z0 )| for all z in that neighborhood, then in fact f (z) = f (z0 )
for all z in that neighborhood.
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Proof.
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Proof. By Cauchy’s Formula, for all circles C(z0 , r) with r < ε we
have that
Z
f (z)
f (z0 ) = 1
2πi C(z ,r) z − z0 dz
0
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have that
Z
f (z)
f (z0 ) = 1
0
Z
f (z) 1
d|z|
≤
2π C(z0 ,r) z − z0 Bernd Schröder
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have that
Z
f (z)
f (z0 ) = 1
0
Z
f (z) 1
d|z|
≤
2π C(z0 ,r) z − z0 Z
f (z0 )
1
d|z|
≤
2π C(z0 ,r) r
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have that
Z
f (z)
f (z0 ) = 1
0
Z
f (z) 1
d|z|
≤
2π C(z0 ,r) z − z0 Z
f (z0 )
1
d|z| = f (z0 )
≤
2π C(z0 ,r) r
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have that
Z
f (z)
f (z0 ) = 1
0
Z
f (z) 1
d|z|
≤
2π C(z0 ,r) z − z0 Z
f (z0 )
1
d|z| = f (z0 )
≤
2π C(z0 ,r) r
and the latter integral will be strictly smaller
than
f (z0 ) if there is
even a single point z on C(z0 , r) where f (z) < f (z0 ).
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have that
Z
f (z)
f (z0 ) = 1
0
Z
f (z) 1
d|z|
≤
2π C(z0 ,r) z − z0 Z
f (z0 )
1
d|z| = f (z0 )
≤
2π C(z0 ,r) r
than
f (z0 ) if there is
even a single point z on C(z0 , r) where f (z) < f (z0 ). Thus for all
r < ε and all |z − z0 | = r we must have f (z) = f (z0 ).
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have that
Z
f (z)
f (z0 ) = 1
0
Z
f (z) 1
d|z|
≤
2π C(z0 ,r) z − z0 Z
f (z0 )
1
d|z| = f (z0 )
≤
2π C(z0 ,r) r
than
f (z0 ) if there is
r < ε and all |z − z0 | = r we must have f (z) = f (z0 ). But if |f | is
constant for |z − z0 | < ε, then so is f .
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have that
Z
f (z)
f (z0 ) = 1
0
Z
f (z) 1
d|z|
≤
2π C(z0 ,r) z − z0 Z
f (z0 )
1
d|z| = f (z0 )
≤
2π C(z0 ,r) r
than
f (z0 ) if there is
r < ε and all |z − z0 | = r we must have f (z) = f (z0 ). But if |f | is
constant for |z − z0 | < ε, then so is f .
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Theorem.
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Theorem. Maximum Modulus Principle.
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Theorem. Maximum Modulus Principle. Let f be analytic and not
constant on an open domain.
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constant on an open domain. Then f does not assume a maximum
value on the domain.
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value on the domain. That is, there is no z0 in the domain so that
|f (z)| ≤ |f (z0 )| for all z in the domain.
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Proof.
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Proof. Suppose for a contradiction that such a z0 does exist.
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Proof. Suppose for a contradiction that such a z0 does exist. By the
preceding lemma, f would be constant on a disk around z0 .
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preceding lemma, f would be constant on a disk around z0 . Moreover,
we could choose the radius of the disk arbitrarily large, as long as it
stays inside the domain.
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stays inside the domain. Now let z1 be in the domain.
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stays inside the domain. Now let z1 be in the domain. Then there is an
arc C from z0 to z1 and the arc has a positive (minimum) distance
from the boundary of the domain.
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from the boundary of the domain. Now there is a disk of maximum
radius around z0 on which f is equal to f (z0 ).
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radius around z0 on which f is equal to f (z0 ). Thus the point on C that
is in this disk and closest to z1 (on the arc C) has the same properties
as z0 .
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as z0 . Continue with a disk of maximum radius around this new point.
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Repeat until we have that f (z1 ) = f (z0 ).
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Repeat until we have that f (z1 ) = f (z0 ). This would prove that f is
constant, a contradiction.
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Repeat until we have that f (z1 ) = f (z0 ). This would prove that f is
constant, a contradiction.
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Visualization
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Visualization
z0
u
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Visualization
z0
u
u
z1
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Visualization
z0
u
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z1
Bernd Schröder
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Visualization
z0
u
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u
z1
Bernd Schröder
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Visualization
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u
C
u
z1
Bernd Schröder
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Visualization
z0
u
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z1
Bernd Schröder
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Visualization
z0
u
C
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z1
Bernd Schröder
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Visualization
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Bernd Schröder
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Visualization
z0
u
C
u
z1
Bernd Schröder
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Visualization
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z0
u
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Bernd Schröder
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Visualization
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Bernd Schröder
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Visualization
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Bernd Schröder
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Visualization
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Bernd Schröder
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Bernd Schröder
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Bernd Schröder
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Visualization
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Bernd Schröder
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Visualization
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Bernd Schröder
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Visualization
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Bernd Schröder
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Visualization
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Bernd Schröder
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Visualization
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Bernd Schröder
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Visualization
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Bernd Schröder
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Visualization
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Bernd Schröder
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Visualization
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Bernd Schröder
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Visualization
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Bernd Schröder
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Visualization
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Bernd Schröder
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Visualization
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Bernd Schröder
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Visualization
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Bernd Schröder
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Corollary.
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Corollary. Let f be continuous on a closed and bounded region and
let it be analytic and nonconstant in the interior of the region.
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let it be analytic and nonconstant in the interior of the region. Then
the largest value of |f | will be assumed at some point on the boundary
of the region and it will not be reached in the interior.
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Proof.
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Proof. Direct consequence of the fact that continuous functions will
assume a maximum on closed and bounded regions and the Maximum
Modulus Principle.
Bernd Schröder
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Proof. Direct consequence of the fact that continuous functions will
assume a maximum on closed and bounded regions and the Maximum
Modulus Principle.
Bernd Schröder
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The Cauchy Integral Formula

Transcription

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