270 CHAPTER 11. FUNCTIONS OF ONE COMPLEX VARIABLE

Lemma 11.6.5 Suppose U is a region and g : U → N is continuous. Then g is constanton U.

Proof: Let z,w ∈U . Let h(t) = g(γ (t)) for t ∈ [0,1] where γ is a smooth curve withγ (0) = z and γ (1) = z. Then by the intermediate value theorem, h can have only one value.Thus g(z) = g(w).

Theorem 11.6.6 (Open mapping theorem) Let Ω be a region (open connected set)in C and suppose f : Ω→ C is analytic. Then f (Ω) is either a point or a region. In thecase where f (Ω) is a region, it follows that for each z0 ∈ Ω, there exists an open set Vcontaining z0 and m ∈ N such that for all z ∈V,

f (z) = f (z0)+φ (z)m (11.9)

where φ : V → B(0,δ ) is one to one, analytic and onto, φ (z0) = 0, φ′ (z) ̸= 0 on V.

Proof: Suppose f (Ω) is not a point. Then for z0 ∈Ω it follows there exists r > 0 suchthat f (z) ̸= f (z0) for all z ∈ B(z0,r)\{z0} . Otherwise, z0 would be a limit point of the set,

{z ∈Ω : f (z)− f (z0) = 0}

which would imply from Theorem 11.6.3 that f (z) = f (z0) for all z ∈Ω. Therefore, mak-ing r smaller if necessary, and using the power series of f ,

f (z) = f (z0)+(z− z0)m g(z) ?

= ( f (z0)+((z− z0)g(z)1/m

)m)

where g is analytic near z0 and g(z0) ̸= 0. Does an analytic g(z)1/m exist? By continuity,g(B(z0,r)) ⊆ B(g(z0) ,ε) where ε is small enough that 0 /∈ B(g(z0) ,ε), so there exists abranch of the logarithm on C⧹B(g(z0) ,ε). Call it log even though it might not be theprinciple branch. Then consider e(1/m) log(g(z)) ≡ g(z)1/m and so we can obtain an analyticfunction denoted by g(z)1/m as in the above formula. Let φ (z) = (z− z0)g(z)1/m . Thenφ (z0) = 0 and

φ′ (z) = e(1/m) log(g(z))+(z− z0)e(1/m) log(g(z)) 1

g(z)g′ (z)

so φ′ (z0) = e(1/m) log(g(z0)) ̸= 0. Shrinking r some more if necessary, assume φ

′ (z) ̸= 0 forall z ∈ B(z0,r). The representation

f (z) = f (z0)+φ (z)m ,z ∈ B(z0,r)

where φ′ (z) ̸= 0 for all z ∈ B(z0,r) and φ (z0) = 0 has been obtained.

Let δ be small enough that the only zero of φ (z)−φ (z0) is z0 in B(z0,δ ). If no suchsmall positive δ exists, then the zeroes of φ (z)−φ (z0) would have a limit point and so φ

would be a constant. This would force f to be constant also. Then φ (z0) /∈ φ (C (z0,δ ))

and so if |w−φ (z0)| is small enough, then w /∈ φ (C (z0,δ )) either. Thus there is ε > 0with B(φ (z0) ,ε)∩φ (C (z0,δ )) = /0. Consider for w ∈ B(φ (z0) ,ε) = B(0,ε) the formulafor counting zeroes.

12πi

∫C(z0,δ )

φ′ (z)

φ (z)−wdz