35.9. EXERCISES 723

(b) Use continuity of w→ 12πi∫

C(z0,δ )φ ′(z)

φ(z)−w dz for |w|< δr2 and Problem 15 to con-

clude that there exists ε < rδ such that if |w|< ε there is one zero for φ (z)−win B(z0,δ ). In other words, φ (B(z0,δ ))⊇ B(0,ε) . Then also φ

m (B(z0,δ ))⊇B(0,εm). Hint: If you have w ∈ B(0,εm) , then there are m mth roots of w

equally spaced around B(

0, |w|1/m)

. Thus these roots are on a circle of ra-dius less than ε . Pick one. Call it ŵ. Then there exists z ∈ B(z0,δ ) such thatφ (z) = ŵ. Then φ

m (z) = w. Fill in details.(c) Explain why f (B(z0,δ ))⊇ f (z0)+B(0,εm) and why for w∈ f (z0)+B(0,εm)

there are m different points in B(z0,δ ) ,z1, · · · ,zm such that f (z j) = w.

20. ↑Let Ω be an open connected set. Let f : Ω→ C be analytic. Suppose f (Ω) is nota single point. Then pick z0 ∈Ω. Explain why f (z) = f (z0)+(z− z0)

m g(z) for allz ∈V an open ball contained in Ω which contains z0 and g(z) ̸= 0 in V,g(z) analytic.If this were not so, then z0 would be a zero of infinite order and by the theorem onzeros, Theorem 35.6.9, f (z) = f (z0) for all z ∈ Ω which is assumed not to happen.Thus, every z0 in Ω has this property that near z0, f (z) = f (z0)+ (z− z0)

m g(z) fornonzero g(z). Now explain why f (z) = f (z0)+φ

m (z) where φ (z0) = 0 but φ′ (z0) ̸=

0 and φ (z) is some analytic function. Thus from Problem 19 above, there is δ suchthat f (Ω) ⊇ f (z0)+B(0,εm). Hence f (Ω) is open since each f (z0) is an interiorpoint of f (Ω). You only need to show that there is G(z) such that G(z)m = g(z) andthen φ (z)≡ (z− z0)G(z) will work fine. When you have done this, Problem 19 willyield a proof of the open mapping theorem which says that if f is analytic on Ω aconnected open set, then f (Ω) is either an open set or a single point. So here aresome steps for doing this.

(a) Consider z→ g′(z)g(z) . It is analytic on the open ball V and so it has a primitive on

V . In fact, you could take h(z)≡∫

γ(z0,z)g′(w)g(w) dw.

(b) Let the primitive be h(z) . Then consider(

g(z)e−h(z))′. Show this equals 0.

Then explain why this requires it to be constant. Explain why there is a+ ibsuch that g(z) = eh(z)+a+ib. Then use the primitive h(z)+a+ ib instead of theoriginal one. Call it h(z). Then

g(z) = eh(z)

You can then complete the argument by letting g(z)1/m ≡ eh(z)/m and

G(z)≡ (z− z0)g(z)1/m

(c) Show that this theorem is certainly not true when considering functions of areal variable by considering f (x) = x2.

21. If you have an open set U in C show that for all z ∈U, |z| < sup{|w| : w ∈U}. Inother words, z→ |z| never achieves its maximum on any open set U ∈ C.

22. Let f be analytic on U and let B(z,r)⊆U . Let γr be the positively oriented boundaryof B(z,r). Explain, using the Cauchy integral formula why

| f (z)| ≤max{| f (w)| : w ∈ γ∗r} ≡ mr

Show that if equality is achieved, then | f (w)| must be constantly equal to mr on γ∗r .