Exercise 1. Let G be a region and let{an} and {bm} be two sequences of distinct points in G without limit points in G such that an an , bmfor all n, m. Let Sn(z) be a singular part at an and let pmbe a positive integer. Show that there is a meromorphic function f on G whose only poles and zeros are{an} and {bm} respectively, the singular part at z = anis Sn(z), and z = bmis a zero of multiplicity pm.
Solution. Let G be a region and let{bm} be a sequence of distinct points in G with no limit point in G;
and let{pm} be a sequence of integers. By Theorem 5.15 p.170 there is an analytic function g defined on G whose only zeros are at the points bm; furthermore, bmis a zero of g of multiplicity pm.
Since g∈ H(g) and {an} ∈ G, g has a Taylor series in a neighborhood B(an; Rn) of each an, that is
Proof of the claim:
X∞ where the last step follows by choosing
X∞ k=0
αkBjk= Ajn.
Since G is a region, G is open. Let{an} be a sequence of distinct points without a limit point in G and such that an,bmfor all n, m. Let{rn(z)} be the sequence of rational functions given by
(see (8.1)). By Mittag-Leffler’s Theorem, there is a meromorphic function h on G whose poles are exactly the points{an} and such that the singular part of h at anis rn(z).
Set f = g· h. Then by construction f is the meromorphic function on G whose only poles and zeros are {an} and{bm} respectively, the singular part at z = anis Sn(z), and z = bmis a zero of multiplicity pm. (Note that the zeros do not cancel the poles since by assumption an,bm∀ n, m).
Exercise 2. Let{an} be a sequence of points in the plane such that |an| → ∞, and let {bn} be an arbitrary sequence of complex numbers.
(a) Show that if integers{kn} can be chosen such that X∞
converges absolutely for all r > 0 then X∞
(c) Show that if there is an integer k such that the series X∞ n=1
bn
ak+1n (3.6)
converges absolutely, then (3.4) converges absolutely if kn = k for all n.
(d) Suppose there is an r > 0 such that|an− am| ≥ r for all n , m. Show thatP |an|−3<∞. In particular, if the sequence{bn} is bounded then the series (3.6) with k = 2 converges absolutely. (This is somewhat involved and the reader may prefer to prove part (f) directly since this is the only application.)
(e) Show that if the series (3.5) converges in M(C) to a meromorphic function f then
f (z) =
(f) Let ω and ω0be two complex numbers such that Im(ω0/ω) , 0. Using the previous parts of this exercise show that the series
ζ(z) = 1 meromorphic function ζ with simple poles at the points 2nω + 2n0ω0. This function is called the Weierstrass zeta function.
(g) LetP(z) = −ζ0(z);P is called the Weierstrass pe function. Show that P(z) = 1
where the sum is over the same w as in part (f). Also show that P(z) = P(z + 2nω + 2n0ω0)
for all integers n and n0. That is,P is doubly periodic with periods 2ω and 2ω0. Solution. Not available.
Exercise 3. This exercise shows how to deduce Weierstrass’s Theorem for the plane (Theorem VII. 5.12) from Mittag-Leffler’s Theorem.
(a) Deduce from Exercises 2(a) and 2(b) that for any sequence{an} in C with lim an=∞ and an,0 there is a sequence of integers{kn} such that
h(z) =
is a meromorphic function on C with simple poles at a1, a2, . . .. The remainder of the proof consists of showing that there is a function f such that h = f0/ f . This function f will then have the appropriate zeros.
(b) Let z be an arbitrary but fixed point in C− {a1, a2, . . .}. Show that if γ1and γ2are any rectifiable curves the curve γ and the resulting function f is analytic.
(d) Suppose that z ∈ {a1, a2, . . .}; show that z is a removable singularity of the function f defined in part (c). Furthermore, show that f (z) = 0 and that the multiplicity of this zero equals the number of times that z appears in the sequence{a1, a2, . . .}.
Remark. We could have skipped parts (b), (c), and (d) and gone directly from (a) to (e). However this would have meant that we must show that (3.7) converges in H(C) and it could hardly be classified as a new proof. The steps outlined in parts (a) through (d) give a proof of Weierstrass’s Theorem without introducing infinite products.
Solution. Not available.
Exercise 4. This exercise assumes a knowledge of the terminology and results of Exercise VII. 5.11.
(a) Define two functions f and g in H(G) to be relatively prime (in symbols, ( f , g) = 1) if the only common function f then ( f ) is called a principal ideal.
(e) Show that every finitely generated ideal in H(G) is a principal ideal.
(f) An idealJ is called a fixed ideal if Z(J) , ; otherwise it is called a free ideal. Prove that if J = (S) thenZ(J) = Z(S) and that a proper principal ideal is fixed.
(g) Let fn(z) = sin(2−nz) for all n≥ 0 and let J = ({ f1, f2, . . .}). Show that J is a fixed ideal in H(C) which is not a principal ideal.
(h) LetJ be a fixed ideal and prove that there is an f in H(G) with Z( f ) = Z(J) and J ⊂ ( f ). Also show
Exercise 5. Let G be a region and let{an} be a sequence of distinct points in G with no limit point in G.
For each integer n≥ 1 choose integers kn≥ 0 and constants A(k)n , 0≤ k ≤ kn. Show that there is an analytic function f on G such that f(k)(an) = k!A(k)n . (Hint: Let g be an analytic function on G with a zero at anof multiplicity kn+ 1. Let h be a meromorphic function on G with poles at each an of order kn+ 1 and with singular part Sn(z). Choose the Snso that f = gh has the desired property.)
Solution. Not available.
Exercise 6. Find a meromorphic function with poles of order 2 at 1,√ 2,√
3, . . . such that the residue at each pole is 0 and lim(z−√
n)2f (z) = 1 for all n.
Solution. We claim that the function
f (z) = for every z∈ C that is not one of the poles. Moreover f converges in M(C). To see this let r > 0 and choose N∈ N so large that √ This is the majorant independent of z for the Weierstrass’ Criterion.
Next fix m∈ N and consider the limit in equation limz→√
Lastly we have to check that the residuals are 0 at the poles. By Proposition V 2.4, p.113, for a function g(z) := (z−√
m)2f (z), it has to be verified that g0(z)|z=√ m= 0.
First find the derivative of f as
f0(z) = X∞ n=1
6√
nz2− 6nz − 2z3 n3/2(z−√
m)3 then compute the residuum at z = √
m Res( f ,√
m) = g0(z)|z=√m= 2(z−√
m) f (z) + (z− √
m)2f0(z) z=√m
=
X∞
n=1
2(z−√m)(3√nz2− 2z3) n3/2(z− √
n)2 +
X∞ n=1
(z− √ m)2(6√
nz2− 6nz − 2z3) n3/2(z−√m)3
z=√m
=
X∞
n=1
2(z−p (n))(3√
nz2− 2z3) + (z−√ m)(6√
nz2)− 6nz − 2z3)
· z−√ m n3/2(z− √n)3
#
z=√m
If n , m and z = √m, the last factor equals zero, in the case n = m, (z−√m)2cancels directly and we are left with
12√m− 2z3− 6mz n3/2(z− √m)
z=m
= −6z(z−√m) n3/2
z=m
= 0.
The residuum vanishes for all √
m, m∈ N and therefore the function considered in this exercise has all the required properties.
Chapter 9
Analytic Continuation and Riemann Surfaces
9.1 Schwarz Reflection Principle
Exercise 1. Let γ be a simple closed rectifiable curve with the property that there is a point a such that for all z on γ the line segment [a, z] intersects{γ} only at z; i.e. [a, z] ∩ {γ} = {z}. Define a point w to be inside γ if [a, w]∩ {γ} = and let G be the collection of all points that are inside γ.
(a) Show that G is a region and G−= G∪ {γ}.
(b) Let f : G−→ C be a continuous function such that f is analytic on G. Show thatR
γ f = 0.
(c) Show that n(γ; z) =±1 if z is inside γ and n(γ; z) = 0 if z < G−.
Remarks. It is not necessary to assume that γ has such a point a as above; each part of this exercise remains true if γ is only assumed to be a simple closed rectifiable curve. Of course, we must define what is meant by the inside of γ. This is difficult to obtain. The fact that a simple closed curve divides the plane into two pieces (an inside and an outside) is the content of the Jordan Curve Theorem. This is a very deep result of topology.
Solution. Not available.
Exercise 2. Let G be a region in the plane that does not contain zero and let G∗be the set of all points z such that there is a point w in G where z and w are symmetric with respect to the circle|ξ| = 1. (See III.
3.17.)
(a) Show that G∗={z : (1/¯z) ∈ G}.
(b) If f : G→ C is analytic, define f∗: G∗→ C by f∗(z) = f (1/¯z). Show that f∗is analytic.
(c) Suppose that G = G∗and f is an analytic function defined on G such that f (z) is real for z in G with
|z| = 1. Show that f = f∗.
(d) Formulate and prove a version of the Schwarz Reflection Principle where the circle|ξ| = 1 replaces R.
Do the same thing for an arbitrary circle.
Solution. Not available.
Exercise 3. Let G, G+, G−, G0 be as in the statement of the Schwarz Reflection Principle and let f : G+∪ G0→ C∞be a continuous function such that f is meromorphic on G+. Also suppose that for x in G0 f (x)∈ R. Show that there is a meromorphic function g : G → C∞such that g(z) = f (z) for z in G+∪ G0. Is it possible to allow f to assume the value∞ on G0?
Solution. Not available.