1748 CHAPTER 55. COMPLEX MAPPINGS

Definition 55.4.1 Let f be analytic on B(a,r) and let β ∈ ∂B(a,r) . Then β is called aregular point of f if there exists some δ > 0 and a function, g analytic on B(β ,δ ) suchthat g = f on B(β ,δ )∩B(a,r) . Those points of ∂B(a,r) which are not regular are calledsingular.

a

β

Theorem 55.4.2 Suppose f is analytic on B(a,r) and the power series

f (z) =∞

∑k=0

ak (z−a)k

has radius of convergence r. Then there exists a singular point on ∂B(a,r).

Proof: If not, then for every z∈ ∂B(a,r) there exists δ z > 0 and gz analytic on B(z,δ z)such that gz = f on B(z,δ z)∩B(a,r) . Since ∂B(a,r) is compact, there exist z1, · · · ,zn,points in ∂B(a,r) such that

{B(zk,δ zk

)}nk=1 covers ∂B(a,r) . Now define

g(z)≡{

f (z) if z ∈ B(a,r)gzk (z) if z ∈ B

(zk,δ zk

)Is this well defined? If z∈B(zi,δ zi)∩B

(z j,δ z j

), is gzi (z)= gz j (z)? Consider the following

picture representing this situation.

You see that if z ∈ B(zi,δ zi)∩B(z j,δ z j

)then I ≡ B(zi,δ zi)∩B

(z j,δ z j

)∩B(a,r) is

a nonempty open set. Both gzi and gz j equal f on I. Therefore, they must be equal onB(zi,δ zi)∩B

(z j,δ z j

)because I has a limit point. Therefore, g is well defined and analytic

on an open set containing B(a,r). Since g agrees with f on B(a,r) , the power series for gis the same as the power series for f and converges on a ball which is larger than B(a,r)contrary to the assumption that the radius of convergence of the above power series equalsr. This proves the theorem.