Now the following is the general Cauchy integral formula.

Theorem 15.6.1Let Γ be a simple closed rectifiable curve and let γ be an oriented parametrization for
Γ which has inside U_{i}. Also let f be continuous on U_{i}∪ Γ and differentiable in U_{i}. Then ifz ∈ U_{i},

∫
n (γ,z)f (z) = -1- f (w-)dw
2πi γ w − z

Proof: Consider the function

{ f(w)−f(z)-
g(w ) ≡ w−z if w ⁄= z (15.33)
f ′(z) if w = z

(15.33)

This is clearly continuous on U_{i}∪ Γ. It is also clear that g^{′}

(w)

exists if w≠z. It remains to consider whether
g^{′}

(z)

exists. Then from the Theorem 15.5.3, we can write f

(z + h)

as a power series in h whenever h is
suitably small.

f(z+h)−f(z) ′ ( ( ) )
----h-----−-f-(z) = 1- 1- f ′(z)h + 1-f′′(z)h2 + 1-f′′′(z)h3 + ⋅⋅⋅ − f′(z)
h h (h( 2! 3! ) )
1- ′ 1-′′ 1- ′′′ 2 ′
= h f (z)+ 2!f (z)h + 3!f (z)h + ⋅⋅⋅ − f (z)
1 ′′ 1 ′′′
= 2!f (z)+ 3!f (z)h+ higher order terms

Thus the limit of the difference quotient exists and is

1-
2!

f^{′′}

(z)

. It follows that

1 ∫ 1 ∫ f (w) 1 ∫ f (z)
0 = --- g(w)dw = --- -----dw − --- -----dw
2πi∫γ 2πi γw − z 2πi γw − z
= -1- f-(w-)dw − f (z)n(γ,z)■
2πi γ w − z

The following is a spectacular application. It is Liouville’s theorem.

Theorem 15.6.2Suppose f is analytic on ℂ and that

∥f (z)∥

is bounded for z ∈ ℂ. Then f isconstant.

Proof: It was shown above that if γ_{r} is a counter clockwise oriented parametrization for the circle of
radius r, then

∫
′ -1- -f-(w)--
f (z) = 2πi γr(w − z)2dw if |z| < r

and so

|f′(z)| ≤ 1-C2πr 1-
2π r2

where

|f (z)|

< C for all z and this is true for any r so let r →∞ and you can conclude that f^{′}

(z)

= 0 for
all z ∈ ℂ. However, this shows that f^{(k)
}

(z)

= 0 for all z and for each k ≥ 1. Thus the power series for
f

This leads right away to the shortest proof of the fundamental theorem of algebra.

Theorem 15.6.3Let p

(z)

be a non constant polynomial with complex coefficients. Then p

(z)

= 0
for some z ∈ ℂ. That is, p

(z)

has a root in ℂ.

Proof:Suppose not. Then 1∕p

(z)

is analytic on ℂ. Also, the leading order term dominates the others
and so 1∕p

(z)

must be bounded. Indeed, lim_{|z|
→∞}

(1∕|p(z)|)

= 0 and the continuous function
z → 1∕

|p(z)|

achieves a maximum on any bounded ball centered at 0. By Liouville’s theorem, this quotient
must be constant. However, by assumption, this does not take place. Hence there is a root of p