1652 CHAPTER 52. THE OPEN MAPPING THEOREM

Shrinking r if necessary you can assume φ′ (z) ̸= 0 on B(z0,r). Is there an open set V

contained in B(z0,r) such that φ maps V onto B(0,δ ) for some δ > 0?Let φ (z) = u(x,y)+ iv(x,y) where z = x+ iy. Consider the mapping(

xy

)→(

u(x,y)v(x,y)

)where u,v are C1 because φ is given to be analytic. The Jacobian of this map at (x,y) ∈B(z0,r) is ∣∣∣∣ ux (x,y) uy (x,y)

vx (x,y) vy (x,y)

∣∣∣∣= ∣∣∣∣ ux (x,y) −vx (x,y)vx (x,y) ux (x,y)

∣∣∣∣= ux (x,y)

2 + vx (x,y)2 =

∣∣φ ′ (z)∣∣2 ̸= 0.

This follows from a use of the Cauchy Riemann equations. Also(u(x0,y0)v(x0,y0)

)=

(00

)Therefore, by the inverse function theorem there exists an open set, V, containing z0 andδ > 0 such that (u,v)T maps V one to one onto B(0,δ ) . Thus φ is one to one onto B(0,δ )as claimed. Applying the same argument to other points, z of V and using the fact thatφ′ (z) ̸= 0 at these points, it follows φ maps open sets to open sets. In other words, φ

−1 iscontinuous.

It also follows that φm maps V onto B(0,δ m) . Therefore, the formula 52.1.1 implies

that f maps the open set, V, containing z0 to an open set. This shows f (Ω) is an openset because z0 was arbitrary. It is connected because f is continuous and Ω is connected.Thus f (Ω) is a region. It remains to verify that φ

−1 is analytic on B(0,δ ) . Since φ−1 is

continuous,

limφ(z1)→φ(z)

φ−1 (φ (z1))−φ

−1 (φ (z))φ (z1)−φ (z)

= limz1→z

z1− zφ (z1)−φ (z)

=1

φ′ (z)

.

Therefore, φ−1 is analytic as claimed.

It only remains to verify the assertion about the case where f is one to one. If m > 1,then e

2πim ̸= 1 and so for z1 ∈V,

e2πim φ (z1) ̸= φ (z1) . (52.1.2)

But e2πim φ (z1)∈ B(0,δ ) and so there exists z2 ̸= z1(since φ is one to one) such that φ (z2) =

e2πim φ (z1) . But then

φ (z2)m =

(e

2πim φ (z1)

)m= φ (z1)

m

implying f (z2) = f (z1) contradicting the assumption that f is one to one. Thus m = 1and f ′ (z) = φ

′ (z) ̸= 0 on V. Since f maps open sets to open sets, it follows that f−1 is