Kenneth Kuttler

1782 CHAPTER 56. APPROXIMATION BY RATIONAL FUNCTIONS

This is well defined by 6 and is easily seen to be a primitive. You just write the differencequotient and take a limit using 6.

limw→0

F (z+w)−F (z)w

= limw→0

(∫γ(z0,z+w)

f (u)du−∫

γ(z0,z)f (u)du

)= lim

w→0

∫γ(z,z+w)

f (u)du

= limw→0

∫ 1

0f (z+ tw)wdt = f (z) .

7⇒8 Suppose then that f ,1/ f are both analytic. Then f ′/ f is analytic and so it has aprimitive by 7. Let this primitive be g1. Then(

e−g1 f)′

= e−g1(−g′1

)f + e−g1 f ′

= −e−g1

(f ′

)f + e−g1 f ′ = 0.

Therefore, since Ω is connected, it follows e−g1 f must equal a constant. (Why?) Let theconstant be ea+ibi. Then f (z) = eg1(z)ea+ib. Therefore, you let g(z) = g1 (z)+a+ ib.

8⇒9 Suppose then that f ,1/ f are both analytic on Ω. Then by 8 f (z) = eg(z). Letφ (z)≡ eg(z)/2.

9⇒1 There are two cases. First suppose Ω = C. This satisfies condition 9 because iff ,1/ f are both analytic, then the same argument involved in 8⇒9 gives the existence ofa square root. A homeomorphism is h(z) ≡ z√

1+|z|2. It obviously maps onto B(0,1) and

is continuous. To see it is 1 - 1 consider the case of z1 and z2 having different arguments.Then h(z1) ̸= h(z2) . If z2 = tz1 for a positive t ̸= 1, then it is also clear h(z1) ̸= h(z2) .To show h−1 is continuous, note that if you have an open set in C and a point in this openset, you can get a small open set containing this point by allowing the modulus and theargument to lie in some open interval. Reasoning this way, you can verify h maps open setsto open sets. In the case where Ω ̸= C, there exists a one to one analytic map which mapsΩ onto B(0,1) by the Riemann mapping theorem. This proves the theorem.

56.3 Exercises1. Let a ∈ C. Show there exists a sequence of polynomials, {pn} such that pn (a) = 1

but pn (z)→ 0 for all z ̸= a.

2. Let l be a line in C. Show there exists a sequence of polynomials {pn} such thatpn (z)→ 1 on one side of this line and pn (z)→ −1 on the other side of the line.Hint: The complement of this line is simply connected.

3. Suppose Ω is a simply connected region, f is analytic on Ω, f ̸= 0 on Ω, and n ∈ N.Show that there exists an analytic function, g such that g(z)n = f (z) for all z ∈ Ω.That is, you can take the nth root of f (z) . If Ω is a region which contains 0, is itpossible to find g(z) such that g is analytic on Ω and g(z)2 = z?

1782 CHAPTER 56. APPROXIMATION BY RATIONAL FUNCTIONSThis is well defined by 6 and is easily seen to be a primitive. You just write the differencequotient and take a limit using 6.lim F(etw) =F) = lim I (fre auf or) au)w—0 Ww w—0 W= lim— dw>0W Mezaw)) "_ 1 /!= lim — | f (z+tw) wdt = f (z).w0 Ww JO7=>8 Suppose then that f,1/f are both analytic. Then f’/f is analytic and so it has aprimitive by 7. Let this primitive be g;. Then(es! f)' = eS (—s) fte sf"= —e 81 (4) fte sf —0.Therefore, since Q is connected, it follows e~*! f must equal a constant. (Why?) Let theconstant be e*+#, Then f (z) = e8!e“t, Therefore, you let g(z) = gi (z) +a+ib.8=>9 Suppose then that f,1/f are both analytic on Q. Then by 8 f(z) = e&©). Let@ (z) = es /2,9=1 There are two cases. First suppose Q = C. This satisfies condition 9 because iff,1/f are both analytic, then the same argument involved in 8=-9 gives the existence ofa square root. A homeomorphism is h(z) = Joe It obviously maps onto B (0,1) andis continuous. To see it is | - 1 consider the case of z; and zz having different arguments.Then h(z1) # h(z2). If z2 = tz for a positive t £ 1, then it is also clear h(z1) #4 h(z2).To show h7! is continuous, note that if you have an open set in C and a point in this openset, you can get a small open set containing this point by allowing the modulus and theargument to lie in some open interval. Reasoning this way, you can verify 4 maps open setsto open sets. In the case where Q F C, there exists a one to one analytic map which mapsQ onto B (0,1) by the Riemann mapping theorem. This proves the theorem.56.3. Exercises1. Let a € C. Show there exists a sequence of polynomials, {p,} such that p, (a) = 1but pn (z) + 0 for all za.2. Let J be a line in C. Show there exists a sequence of polynomials {p,} such thatPn(z) — 1 on one side of this line and p,(z) + —1 on the other side of the line.Hint: The complement of this line is simply connected.3. Suppose Q is a simply connected region, f is analytic on Q, f #0 on Q, andn EN.Show that there exists an analytic function, g such that g(z)" = f(z) for all z € Q.That is, you can take the n™ root of f (z). If Q is a region which contains 0, is itpossible to find g(z) such that g is analytic on Q and g(z)* =z?