Kenneth Kuttler

15.3. CAUCHY INTEGRAL FORMULA FOR A CYCLE 381

|z| > r, it must be the case that z ∈ H and so for such z, the bottom description of g(z)found in 15.7 is valid. Therefore, it follows lim|z|→∞ ∥g(z)∥ = 0 and so g is bounded andanalytic on all of C. By Liouville’s theorem, g is a constant. Hence, the constant can onlyequal zero.

For z ∈Ω\∪mk=1γ∗k , since it was just shown that h(z) = g(z) = 0 on Ω

0 = h(z) =1

2πi

∑k=1

∫γk

φ (z,w)dw =1

2πi

∑k=1

∫γk

f (w)− f (z)w− z

dw =

12πi

∑k=1

∫γk

f (w)w− z

dw− f (z)m

∑k=1

n(γk,z) . ■

In case it is not obvious why g is analytic on H, use the formula. For z ∈ H, z /∈ γ∗k forany k. The issue reduces to showing that z→ ∑

mk=1

∫γk

f (w)w−z dw is analytic. You can show

this by taking a limit of a difference quotient and argue that the limit can be taken insidethe integral. Taking a difference quotient and simplifying a little, one obtains∫

γk

f (w)w−(z+h)dw−

∫γk

f (w)w−z dw

h=∫

γk

f (w)(w− z)(w− (z+h))

considering only small h, the denominator is bounded below by some δ > 0 and also f (w)is bounded on the compact set γ∗k , | f (w)| ≤M. Then for such small h,∣∣∣∣∣ f (w)

(w− z)(w− (z+h))− f (w)

(w− z)2

∣∣∣∣∣=

∣∣∣∣ 1w− z

(w− (z+h))− 1

(w− z)

)f (w)

∣∣∣∣≤ ∣∣∣∣ 1w− z

∣∣∣∣ 1δ

it follows that one obtains uniform convergence as h→ 0 of the integrand to f (w)(w−z)2 for any

sequence h→ 0 and by Theorem 14.4.7, the integral converges to∫

γk

f (w)(w−z)2 dw.

The following is an interesting and fairly easy corollary.

Corollary 15.3.3 Let Ω be an open set (note that Ω might not be simply connected)and let γk : [ak,bk]→ Ω, k = 1, · · · ,m, be closed, continuous and of bounded variation.Suppose also that ∑

mk=1 n(γk,z) = 0 for all z /∈ Ω and ∑

mk=1 n(γk,z) is an integer for z ∈

∩mk=1

(Ω\ γ∗k

). Then if f : Ω→ X is analytic, ∑

mk=1

∫γk

f (w)dw = 0.

Proof: This follows from Theorem 15.3.2 as follows. Let g(w) = f (w)(w− z) wherez ∈Ω\∪m

k=1γk ([ak,bk]) . Then by this theorem,

0 = 0m

∑k=1

n(γk,z) = g(z)m

∑k=1

n(γk,z) =m

∑k=1

12πi

∫γk

g(w)w− z

dw =1

2πi

∑k=1

∫γk

f (w)dw. ■

What if Ω is simply connected? Can one assert something interesting in this casebeyond what is said above? Yes, and this is a very important result. Recall what it meantfor an open set Ω ⊆ C to be simply connected. It meant that Ω is connected and ΩC isconnected in the extended complex plane Ĉ.

15.3. CAUCHY INTEGRAL FORMULA FOR A CYCLE 381\z| > r, it must be the case that z € H and so for such z, the bottom description of g (z)found in 15.7 is valid. Therefore, it follows Tim) -\ 520 ||g (z)|| = 0 and so g is bounded andanalytic on all of C. By Liouville’s theorem, g is a constant. Hence, the constant can onlyequal zero.For z € Q\ Ut, ¥%{, since it was just shown that h(z) = g(z) =OonQ| fW=f@= =a | ol zw)d "=a % w-—Z dws1 m fw) moni & i pat FG) Lenz).In case it is not obvious why g is analytic on H, use the formula. For z € H, z ¢ ¥ forany k. The issue reduces to showing that z > )7"_| Jy LO) ayy is analytic. You can showthis by taking a limit of a difference quotient and argue that the limit can be taken insidethe integral. Taking a difference quotient and simplifying a little, one obtainsIno w— a zany aw — I, w— dw f (w)h 4 =a) (w— (eh)considering only small , the denominator is bounded below by some 6 > 0 and also f (w)is bounded on the compact set 7;,|,f (w)| <M. Then for such small h,f(w) __f(w)(w—z)(w—(cth)) (w—2)?=| (eee eo Ish simit follows that one obtains uniform convergence as h — 0 of the integrand to a “Ww-Zzsequence h — 0 and by Theorem 14.4.7, the integral converges to Sy, ( £ os dwfor anyThe following is an interesting and fairly easy corollary.Corollary 15.3.3 Let Q be an open set (note that Q might not be simply connected)and let Y;,: [ax, bg] 4 Q, k = 1,--+,m, be closed, continuous and of bounded variation.Suppose also that V7) n(Y%,z) = 0 for all z € Q and VL, n(Y%,Z) is an integer for z €MW, (Q\ yf). Then if f : Q— X is analytic, Le, Sy, f(w)dw=0.Proof: This follows from Theorem 15.3.2 as follows. Let g(w) = f (w) (w—z) whereZE€ Q\ UL IY, ([ax, be]) . Then by this theorem,m m_ _ m ~0=0Y) n(%,z) —_ g(z 2) yn (Yq2) ~ Doni = == Y |. w) dw. ak=1 k=1 k=What if Q is simply connected? Can one assert something interesting in this casebeyond what is said above? Yes, and this is a very important result. Recall what it meantfor an open set Q C C to be simply connected. It meant that Q is connected and QL isconnected in the extended complex plane C.