51.7. THE GENERAL CAUCHY INTEGRAL FORMULA 1637

Proof: Suppose B(z0,δ ) has no points of f (B′ (a,r)) . If f (B′ (a,r)) is not dense, thensuch a ball exists. Then for z ∈ B′ (a,r) , | f (z)− z0| ≥ δ > 0. It follows from Theorem51.7.8 that 1

f (z)−z0has a removable singularity at a. Hence, there exists h an analytic func-

tion such that for z near a,

h(z) =1

f (z)− z0. (51.7.22)

There are two cases. First suppose h(a) = 0. Then ∑∞k=1 ak (z−a)k = 1

f (z)−z0for z near

a. If all the ak = 0, this would be a contradiction because then the left side would equal zerofor z near a but the right side could not equal zero. Therefore, there is a first m such thatam ̸= 0. Hence there exists an analytic function, k (z) which is not equal to zero in someball, B(a,ε) such that

k (z)(z−a)m =1

f (z)− z0.

Hence, taking both sides to the −1 power,

f (z)− z0 =1

(z−a)m

∞

∑k=0

bk (z−a)k

and so 51.7.21 holds.The other case is that h(a) ̸= 0. In this case, raise both sides of 51.7.22 to the−1 power

and obtainf (z)− z0 = h(z)−1 ,

a function analytic near a. Therefore, the singularity is removable. This proves the theorem.This theorem is the basis for the following definition which classifies isolated singular-

ities.

Definition 51.7.10 Let a be an isolated singularity of a complex valued function, f . When51.7.21 holds for z near a, then a is called a pole. The order of the pole in 51.7.21 is M. Iffor every r > 0, f (B′ (a,r)) is dense in C then a is called an essential singularity.

In terms of the above definition, isolated singularities are either removable, a pole, oressential. There are no other possibilities.

Theorem 51.7.11 Suppose f : Ω→ C has an isolated singularity at a ∈ Ω. Then a is apole if and only if

limz→a

d ( f (z) ,∞) = 0

in Ĉ.

Proof: Suppose first f has a pole at a. Then by definition, f (z) = g(z)+∑Mk=1

bk(z−a)k

for z near a where g is analytic. Then

| f (z)| ≥ |bM||z−a|M

−|g(z)|−M−1

∑k=1

|bk||z−a|k

=1

|z−a|M

(|bM|−

(|g(z)| |z−a|M +

M−1

∑k=1|bk| |z−a|M−k

)).