55.3. RIEMANN MAPPING THEOREM 1745

and so |ψ ′ (z0)| ≤ (1/2π)2πr(1/r2

)= 1/r. Therefore, η < ∞ as desired. For ψ defined

above in 55.3.7

ψ′ (z0) =

−rφ′ (z0)

(φ (z0)+a)2 =−r (1/2)

(√z0−ξ

)−1

(φ (z0)+a)2 ̸= 0.

Therefore, η > 0. It remains to verify the existence of the function, h.By Theorem 55.3.1, there exists a sequence, {ψn}, of functions in F and an analytic

function, h, such that ∣∣ψ ′n (z0)∣∣→ η (55.3.9)

andψn→ h,ψ ′n→ h′, (55.3.10)

uniformly on all compact subsets of Ω. It follows∣∣h′ (z0)∣∣= lim

n→∞

∣∣ψ ′n (z0)∣∣= η (55.3.11)

and for all z ∈Ω,|h(z)|= lim

n→∞|ψn (z)| ≤ 1. (55.3.12)

By 55.3.11, h is not a constant. Therefore, in fact, |h(z)|< 1 for all z ∈Ω in 55.3.12 bythe open mapping theorem.

Next it must be shown that h is one to one in order to conclude h ∈F . Pick z1 ∈ Ω

and suppose z2 is another point of Ω. Since the zeros of h−h(z1) have no limit point, thereexists a circular contour bounding a circle which contains z2 but not z1 such that γ∗ containsno zeros of h−h(z1).

z1

γ

z2

Using the theorem on counting zeros, Theorem 52.6.1, and the fact that ψn is one toone,

0 = limn→∞

12πi

∫γ

ψ ′n (w)ψn (w)−ψn (z1)

dw

=1

2πi

∫γ

h′ (w)h(w)−h(z1)

dw,

which shows that h− h(z1) has no zeros in B(z2,r) . In particular z2 is not a zero of h−h(z1) . This shows that h is one to one since z2 ̸= z1 was arbitrary. Therefore, h ∈F . Itonly remains to verify that h(z0) = 0.