1764 CHAPTER 55. COMPLEX MAPPINGS

Corollary 55.5.18 Suppose f is entire and nonconstant and not a polynomial. Then fassumes every complex value infinitely many times with the possible exception of one.

Proof: Since f is entire, f (z) = ∑∞n=0 anzn. Define for z ̸= 0,

g(z)≡ f(

1z

)=

∞

∑n=0

an

(1z

)n

.

Thus 0 is an isolated essential singular point of g. By the big Picard theorem, Theorem55.5.17 it follows g takes every complex number but possibly one an infinite number oftimes. This proves the corollary.

Note the difference between this and the little Picard theorem which says that an entirefunction which is not constant must achieve every value but two.

55.6 Exercises1. Prove that in Theorem 55.3.1 it suffices to assume F is uniformly bounded on each

compact subset of Ω.

2. Find conditions on a,b,c,d such that the fractional linear transformation, az+bcz+d maps

the upper half plane onto the upper half plane.

3. Let D be a simply connected region which is a proper subset of C. Does there existan entire function, f which maps C onto D? Why or why not?

4. Verify the conclusion of Theorem 55.3.1 involving the higher order derivatives.

5. What if Ω = C? Does there exist an analytic function, f mapping Ω one to one andonto B(0,1)? Explain why or why not. Was Ω ̸=C used in the proof of the Riemannmapping theorem?

6. Verify that |φ α (z)|= 1 if |z|= 1. Apply the maximum modulus theorem to concludethat |φ α (z)| ≤ 1 for all |z|< 1.

7. Suppose that | f (z)| ≤ 1 for |z| = 1 and f (α) = 0 for |α| < 1. Show that | f (z)| ≤|φ α (z)| for all z ∈ B(0,1) . Hint: Consider f (z)(1−αz)

z−αwhich has a removable sin-

gularity at α. Show the modulus of this function is bounded by 1 on |z| = 1. Thenapply the maximum modulus theorem.

8. Let U and V be open subsets of C and suppose u : U → R is harmonic while h is ananalytic map which takes V one to one onto U . Show that u◦h is harmonic on V .

9. Show that for a harmonic function, u defined on B(0,R) , there exists an analyticfunction, h = u+ iv where

v(x,y)≡∫ y

0ux (x, t)dt−

∫ x

0uy (t,0)dt.