THE CAUCHY INTEGRAL FORMULA AND THE
FUNDAMENTAL THEOREM OF ALGEBRA
D. ARAPURA
1. Cauchy’s formula
We indicate the proof of the following, as we did in class.
THEOREM 1. Let f (z) be an analytic function defined on a simply connected region D enclosed by a piecewise smooth curve C going once around counterclockwise.
If w is in D, then
Z
f (z)
1
dz
f (w) =
2πi C z − w
We need start with the following basic fact from real analysis
THEOREM 2. If f (x) is a continuous function on [a, b], with |f (x)| ≤ M on the
same interval then
Z b
|
f (x)dx| ≤ M (b − a)
a
COROLLARY 1. If f (z) is a continuous function on a circle C of radius r, with
|f (x)| ≤ M on C then
Z
|
f (z)dz| ≤ M (2πr)
C
Proof of theorem 1. Let CR be a small circle of radius R centered at w contained
f (z)
is analytic when z 6= w, Cauchy’s theorem discussed earlier in
in D. Since z−w
class shows that
Z
Z
f (z)
f (z)
dz =
dz
z
−
w
z
−w
Cr
C
So it is enough to show that the right side goes to 2πif (w) as r → 0. Since
Z
f (w)
2πif (w) =
dz
z
−w
Cr
by an easy computation, it is enough to show that
Z
g(z)
lim
dz = 0
r→0 C z − w
r
whereg(z) = f (z) − f (w). Note that since f is continuous, g(z) → 0 as z → w.
This means that the maximum value M (r) of |g(z)| on Cr goes to zero as r → 0
By corollary 1 have that
Z
g(z)
M
|
dz| ≤
(2πr) → 0
z
−
w
r
Cr
1
2
D. ARAPURA
2. Cauchy’s formula for derivatives
We want to establish the following refined theorem
THEOREM 3. Let f (z) be an analytic function defined on a simply connected region D enclosed by a piecewise smooth curve C going once around counterclockwise.
Then the derivatives to all orders f 0 (w), f 00 (w), . . . f (n) (w) . . . exists for w ∈ D and
Z
1
f (z)
f 0 (w) =
dz
2πi C (z − w)2
...
Z
n!
f (z)
f (n) (w) =
dz
2πi C (z − w)n
This is completely different from the real case; a real valued differentiable function need not have a second derivative. And certainly, we won’t have a formula
like the one above. In spite of the differences, it might be helpful to start with the
following basic fact in real analysis which allows us to differentiate under the under
the integral sign.
THEOREM 4. Let g(x, y) be a real valued function differentiable in x, for x ∈
(c, d), and continuous in y, for y ∈ [a, b], then
Z b
G(x) =
g(x, y)dy
a
is differentiable for x ∈ (c, d) with
G0 (x) =
Z
b
gx (x, y)dy
a
Proof. Differentiablity means that
g(x + h, y) − g(x, h)
− gx (x, y) → 0
h
as h → 0 for x ∈ (c, d). This means for the absolute value of this quantity can
be made smaller than any number we choose, say for h close enough to 0. It is
enough to show that
#
"
Z b
G(x + h) − G(x)
lim
−
gx (x, y)dy = 0
h→0
h
a
The left side, after the limit, can be expanded to
Z b
g(x + h, y) − g(x, h)
− gx (x, y) dy
h
a
The absolute value is bounded above by
Z b
g(x + h, y) − g(x, h)
− gx (x, y) dy < (b − a)
h
a
Since, we choose as small as we like for h small enough, we can make this go to
zero as h → 0.
CAUCHY’S FORMULA
3
We turn to an outline of the proof of theorem 3. Consider
f (z)
z−w
Treating this as a function of w, we can take the complex derivative in w to get
f (z)
(z − w)2
Differentiating again yields
2
f (z)
(z − w)3
and so on. Now use theorem 1
f (w) =
1
2πi
Z
C
f (z)
dz
z−w
An argument similar to the proof of theorem 4 shows that we can differentiate
under the integral sign to get
Z
1
f (z)
0
f (w) =
dz
2πi C (z − w)2
Differentiating again
2
f (w) =
2πi
00
Z
C
f (z)
dz
(z − w)3
etc.
3. Fundamental theorem of algebra
By applying corollary 1 to
f (n) (w) =
n!
2πi
Z
C
f (z)
dz
(z − w)n
we obtain
THEOREM 5. Suppose that f (z) is analytic inside and on the circle C of radius
R centered at w. Suppose that |f (z)| ≤ M on C, then
|f (n) (w)| ≤
n!M
Rn
for z inside C.
An entire function is one that is analytic on the whole plane C. An entire
function f (z) is bounded if |f (z)| ≤ M for some M and all z ∈ C.
THEOREM 6 (Liouville’s theorem). A bounded entire function is constant
Proof. Using theorem 5 shows that
M
R
in a circle of radius R centered at 0. As R shows that |f 0 (z)| gets arbitrarily small.
So it must be zero. This means that f 0 (z) = 0 so f (z) is constant.
|f 0 (z)| ≤
THEOREM 7 (Fundamental theorem of algebra). Any nonconstant polynomial
p(z), with complex coefficients, has a zero (or root) in C
4
D. ARAPURA
Before starting the proof, recall that the triangle inequality says that given a, b ∈
C
|a + b| ≤ |a| + |b|
We can turn this into a lower bound, which we will call the reverse triangle inequality (but it’s not standard)
(1)
|a + b| ≥ |a| − |b|
by noticing that
|a| = |(a + b) − b| ≤ |a + b| + |b|
We will also need a fact from real analysis: a continuous real valued function on
the plane takes a maximum value on any set of the form {(x, y) ∈ R2 | x2 +y 2 ≤ R}.
In particular, it is bounded. We will refer to this as the Heine-Borel theorem,
although it is really a very special case of it.
Proof. There is no harm in assuming that the leading coefficient is 1, so that
p(z) = z n + an−1 z n−1 + . . . a0 , n > 0
We prove this by contradiction and suppose that p(z) has no zeros. Then f (z) =
1/p(z) will be an entire function. If we can show that it is bounded, then f (z) is
constant. This means that p(z) is constant, which contradicts what we assumed
about it.
Set
T (z) = an−1 z −1 + . . . a0 z −n
then
p(z)
= 1 + T (z)
zn
Note that when |z| > C, by the triangle inequality
|T (z)| ≤ |an−1 |C −1 + . . . |a0 |C −n
By choosing C large, we can assume that |T (z)| ≤ 1/2. Therefore for |z| > C, from
the reverse triangle inequality, we have
p(z) 1
1
z n = |1 + T (z)| ≥ 1 − 2 = 2
This implies that
2
2
|f (z)| ≤ n < n
|z|
C
On the other |f (z)| is continuous and therefore bounded on the set |z| ≤ C by the
Heine-Borel theorem.