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Solutions Manual for

Functions of One Complex Variable I, Second Edition^1

© Copyright by Andreas Kleefeld, 2009

All Rights Reserved^2

(^1) by John B. Conway (^2) Last updated on April, 15th 2016

PREFACE

Most of the exercises I solved were assigned homeworks in the graduate courses Math 713 and Math 714 ”Complex Analysis I and II” at the University of Wisconsin – Milwaukee taught by Professor Dashan Fan in Fall 2008 and Spring 2009. The solutions manual is intended for all students taking a graduate level Complex Analysis course. Stu- dents can check their answers to homework problems assigned from the excellent book “Functions of One Complex Variable I”, Second Edition by John B. Conway. Furthermore students can prepare for quizzes, tests, exams and final exams by solving additional exercises and check their results. Maybe students even study for preliminary exams for their doctoral studies. However, I have to warn you not to copy straight of this book and turn in your homework, because this would violate the purpose of homeworks. Of course, that is up to you. I strongly encourage you to send me solutions that are still missing to a.kleefeld@fz-juelich.de (LATEXpreferred but not mandatory) in order to complete this solutions manual. Think about the contribution you will give to other students. If you find typing errors or mathematical mistakes pop an email to a.kleefeld@fz-juelich.de. The recent version of this solutions manual can be found at http://www.math.tu-cottbus.de/INSTITUT/kleefeld/Files/Solution.html. The goal of this project is to give solutions to all of the 452 exercises.

CONTRIBUTION

I thank (without special order)

Christopher T. Alvin Martin J. Michael David Perkins Robert Browning for contributions to this book.

• 1 The Complex Number System
  • 1.1 The real numbers
  • 1.2 The field of complex numbers
  • 1.3 The complex plane
  • 1.4 Polar representation and roots of complex numbers
  • 1.5 Lines and half planes in the complex plane
  • 1.6 The extended plane and its spherical representation
• 2 Metric Spaces and the Topology of C
  • 2.1 Definitions and examples of metric spaces
  • 2.2 Connectedness
  • 2.3 Sequences and completeness
  • 2.4 Compactness
  • 2.5 Continuity
  • 2.6 Uniform convergence
• 3 Elementary Properties and Examples of Analytic Functions
  • 3.1 Power series
  • 3.2 Analytic functions
  • 3.3 Analytic functions as mappings. Möbius transformations
• 4 Complex Integration
  • 4.1 Riemann-Stieltjes integrals
  • 4.2 Power series representation of analytic functions
  • 4.3 Zeros of an analytic function
  • 4.4 The index of a closed curve
  • 4.5 Cauchy’s Theorem and Integral Formula
  • 4.6 The homotopic version of Cauchy’s Theorem and simple connectivity
  • 4.7 Counting zeros; the Open Mapping Theorem
  • 4.8 Goursat’s Theorem
• 5 Singularities
  • 5.1 Classification of singularities
  • 5.2 Residues
  • 5.3 The Argument Principle
• 6 The Maximum Modulus Theorem
  • 6.1 The Maximum Principle
  • 6.2 Schwarz’s Lemma
  • 6.3 Convex functions and Hadamard’s Three Circles Theorem
  • 6.4 The Phragmén-Lindelöf Theorem
• 7 Compactness and Convergence in the Space of Analytic Functions
  • 7.1 The space of continuous functions C(G, Ω)
  • 7.2 Spaces of analytic functions
  • 7.3 Spaces of meromorphic functions
  • 7.4 The Riemann Mapping Theorem
  • 7.5 The Weierstrass Factorization Theorem
  • 7.6 Factorization of the sine function
  • 7.7 The gamma function
  • 7.8 The Riemann zeta function
• 8 Runge’s Theorem
  • 8.1 Runge’s Theorem
  • 8.2 Simple connectedness
  • 8.3 Mittag-Leffler’s Theorem
• 9 Analytic Continuation and Riemann Surfaces
  • 9.1 Schwarz Reflection Principle
  • 9.2 Analytic Continuation Along a Path
  • 9.3 Monodromy Theorem
  • 9.4 Topological Spaces and Neighborhood Systems
  • 9.5 The Sheaf of Germs of Analytic Functions on an Open Set
  • 9.6 Analytic Manifolds
  • 9.7 Covering spaces
• 10 Harmonic Functions
  • 10.1 Basic properties of harmonic functions
  • 10.2 Harmonic functions on a disk
  • 10.3 Subharmonic and superharmonic functions
  • 10.4 The Dirichlet Problem
  • 10.5 Green’s Function
• 11 Entire Functions
  • 11.1 Jensen’s Formula
  • 11.2 The genus and order of an entire function
  • 11.3 Hadamard Factorization Theorem
• 12 The Range of an Analytic Function
  • 12.1 Bloch’s Theorem
  • 12.2 The Little Picard Theorem
  • 12.3 Schottky’s Theorem
  • 12.4 The Great Picard Theorem

Chapter 1

The Complex Number System

1.1 The real numbers

No exercises are assigned in this section.

1.2 The field of complex numbers

Exercise 1. Find the real and imaginary parts of the following:

1 z

z − a z + a

(a ∈ R); z^3 ; 3 + 5 i 7 i + 1

 −^1 +^ i^

3 ;

   − 1 − i

6 ; in;

1 + i √ 2

)n for 2 ≤ n ≤ 8.

Solution. Let z = x + iy. Then a)

Re

z

x x^2 + y^2

Re(z) |z|^2

Im

z

y x^2 + y^2

Im(z) |z|^2

b)

Re

( (^) z − a

z + a

x^2 + y^2 − a^2 x^2 + y^2 + 2 ax + a^2

|z|^2 − a^2 |z|^2 + 2 aRe(z) + a^2

Im

( (^) z − a z + a

2 ya x^2 + y^2 + 2 xa + a^2

2 Im(z)a |z|^2 + 2 aRe(z) + a^2

c)

Re

z^3

= x^3 − 3 xy^2 = Re(z)^3 − 3 Re(z)Im(z)^2

Im

z^3

= 3 x^2 y − y^3 = 3 Re(z)^2 Im(z) − Im(z)^3

h)

Re

1 + i √ 2

)n)

0 , n = 2 −

√ 2 2 ,^ n^ =^3 − 1 , n = 4 −

√ 2 2 ,^ n^ =^5 (^0) √, n = 6 2 2 ,^ n^ =^7 1 , n = 8

Im

1 + i √ 2

)n)

(^1) √, n = 2 2 2 ,^ n^ =^3 0 , n = 4 −

√ 2 2 ,^ n^ =^5 − 1 , n = 6 −

√ 2 2 ,^ n^ =^7 0 , n = 8

Exercise 2. Find the absolute value and conjugate of each of the following:

− 2 + i; −3; (2 + i)(4 + 3 i);

3 − i √ 2 + 3 i

i i + 3

; (1 + i)^6 ; i^17.

Solution. It is easy to calculate: a) z = − 2 + i, |z| =

5 , ¯z = − 2 − i

b) z = − 3 , |z| = 3 , ¯z = − 3

c) z = (2 + i)(4 + 3 i) = 5 + 10 i, |z| = 5

5 , ¯z = 5 − 10 i

d

z =

3 − i √ 2 + 3 i

, |z| =

110 , z¯ =

3 + i √ 2 − 3 i

e)

z = i i + 3

i, |z| =

10 , ¯z =

i

f) z = (1 + i)^6 = − 8 i, |z| = 8 , z¯ = 8 i

g) i^17 = i, |z| = 1 , ¯z = −i

Exercise 3. Show that z is a real number if and only if z = ¯z.

Solution. Let z = x + iy. ⇒: If z is a real number, then z = x (y = 0 ). This implies ¯z = x and therefore z = ¯z. ⇔: If z = ¯z, then we must have x + iy = x − iy for all x, y ∈ R. This implies y = −y which is true if y = 0 and therefore z = x. This means that z is a real number.

Exercise 4. If z and w are complex numbers, prove the following equations:

|z + w|^2 = |z|^2 + 2 Re(z w¯) + |w|^2. |z − w|^2 = |z|^2 − 2 Re(z w¯) + |w|^2. |z + w|^2 + |z − w|^2 = 2

|z|^2 + |w|^2

Solution. We can easily verify that ¯¯z = z. Thus

|z + w|^2 = (z + w)(z + w) = (z + w)(¯z + w¯) = zz¯ + z w¯ + w¯z + w w¯ = |z|^2 + |w|^2 + z w¯ + zw¯ = |z|^2 + |w|^2 + z w¯ + ¯z w¯¯

= |z|^2 + |w|^2 + z w¯ + z w¯ = |z|^2 + |w|^2 + 2

z w¯ + z w¯ 2 = |z|^2 + |w|^2 + 2 Re(z w¯) = |z|^2 + 2 Re(z w¯) + |w|^2.

|z − w|^2 = (z − w)(z − w) = (z − w)(¯z − w¯) = zz¯ − z w¯ − w¯z + w w¯ = |z|^2 + |w|^2 − z w¯ − zw¯ = |z|^2 + |w|^2 − z w¯ − ¯z w¯¯

= |z|^2 + |w|^2 − z w¯ − z w¯ = |z|^2 + |w|^2 − 2

z w¯ + z w¯ 2 = |z|^2 + |w|^2 − 2 Re(z w¯) = |z|^2 − 2 Re(z w¯) + |w|^2.

|z + w|^2 + |z − w|^2 = |z|^2 + Re(z w¯) + |w|^2 + |z|^2 − Re(z w¯) + |w|^2 = |z|^2 + |w|^2 + |z|^2 + |w|^2 = 2 |z|^2 + 2 |w|^2 = 2

|z|^2 + |w|^2

Exercise 5. Use induction to prove that for z = z 1 +... + zn; w = w 1 w 2... wn:

|w| = |w 1 |... |wn|; ¯z = ¯z 1 +... + ¯zn; ¯w = w¯ 1... w¯n.

Solution. Not available.

Exercise 6. Let R(z) be a rational function of z. Show that R(z) = R(¯z) if all the coefficients in R(z) are real.

Solution. Let R(z) be a rational function of z, that is

R(z) =

anzn^ + an− 1 zn−^1 +... a 0 bmzm^ + bm− 1 zm−^1 +... b 0

where n, m are nonnegative integers. Let all coefficients of R(z) be real, that is

a 0 , a 1 ,... , an, b 0 , b 1 ,... , bm ∈ R.

Then

R(z) =

anzn^ + an− 1 zn−^1 +... a 0 bmzm^ + bm− 1 zm−^1 +... b 0

anzn^ + an− 1 zn−^1 +... a 0 bmzm^ + bm− 1 zm−^1 +... b 0

= anzn^ + an− 1 zn−^1 +... a 0 bmzm^ + bm− 1 zm−^1 +... b 0

an z¯n^ + an− 1 ¯zn−^1 +... a 0 bm ¯zm^ + bm− 1 z¯m−^1 +... b 0

= R(¯z).

Exercise 2. Calculate the following:

a) the square roots of i

b) the cube roots of i

c) the square roots of

3 + 3 i

Solution. c) The square roots of

3 + 3 i.

Let z =

3 + 3 i. Then r = |z| =

12 and α = tan−^1

√^3 3

= π 3. So, the 2 distinct roots of z

are given by 2

r

cos α+ n^2 k π+ i sin α+ n^2 kπ

where k = 0 , 1. Specifically,

√ z =

cos

π 3 +^2 kπ 2

• i sin

π 3 +^2 kπ 2

Therefore, the square roots of z, zk, are given by

z 0 =

cos π 6 + i sin π 6

3 2 +^

1 2 i

z 1 = 4

cos 76 π + i sin 76 π

√ 3 2 −^

1 2 i

So, in rectangular form, the second roots of z are given by

108 2 ,^

√ (^412) 2

and

√ (^4108) 2 ,^ −^

√ (^412) 2

Exercise 3. A primitive nth root of unity is a complex number a such that 1 , a, a^2 , ..., an−^1 are distinct nth roots of unity. Show that if a and b are primitive nth and mth roots of unity, respectively, then ab is a kth root of unity for some integer k. What is the smallest value of k? What can be said if a and b are nonprimitive roots of unity?

Solution. Not available.

Exercise 4. Use the binomial equation

(a + b)n^ =

∑^ n

k= 0

n k

an−kbk,

where (^) ( n k

n! k!(n − k)!

and compare the real and imaginary parts of each side of de Moivre’s formula to obtain the formulas:

cos nθ = cosn^ θ −

n 2

cosn−^2 θ sin^2 θ +

n 4

cosn−^4 θ sin^4 θ −...

sin nθ =

n 1

cosn−^1 θ sin θ −

n 3

cosn−^3 θ sin^3 θ +...

Solution. Not available.

Exercise 5. Let z = cis (^2) nπ for an integer n ≥ 2. Show that 1 + z +... + zn−^1 = 0.

Solution. The summation of the finite geometric sequence 1 , z, z^2 ,... , zn−^1 can be calculated as

∑n j= 1 z^ j− (^1) = zn− 1 z− 1. We want to show that z

n (^) is an nth (^) root of unity. So, using de Moivre’s formula, zn (^) = (cis (^2 π n

))n

cis

n · (^2) nπ

= cis(2π) = 1. It follows that 1 + z + z^2 + ... + zn−^1 = z n− 1 z− 1 =^

1 − 1 z− 1 =^0 as required.

Exercise 6. Show that ϕ(t) = cis t is a group homomorphism of the additive group R onto the multiplicative group T = {z : |z| = 1 }.

Solution. We have

ϕ(s + t) = cos(s + t) + i sin(s + t) = [cos(s) cos(t) − sin(s) sin(t)] + i [sin(s) cos(t) + cos(s) sin(t)] = cos(s) cos(t) + i cos(s) sin(t) + i sin(s) cos(t) − sin(s) sin(t) = (cos(s) + i sin(s)) (cos(t) + i sin(t)) = ϕ(s)· ϕ(t) , ∀ s, t ∈ R

and |ϕ(t)| = 1 for any t ∈ R.

Exercise 7. If z ∈ C and Re(zn) ≤ 0 for every positive integer n, show that z is a non-negative real number.

Solution. Let n be an arbitrary but fixed positive integer and let z ∈ C and Re(zn) ≥ 0. Since zn^ = rn(cos(nθ) + i sin(nθ)), we have Re(zn) = rn^ cos(nθ) ≥ 0.

If z = 0 , then we are done, since r = 0 and Re(zn) = 0. Therefore, assume z , 0 , then r > 0. Thus

Re(zn) = rn^ cos(nθ) ≥ 0

implies cos(nθ) ≥ 0 for all n. This implies θ = 0 as we will show next. Clearly, θ < [π/ 2 , 3 π/2]. If θ ∈ (0, π/2), then there exists a k ∈ { 2 , 3 ,.. .} such that (^) k+π 1 ≤ θ < π k. If we choose n = k + 1 , we have

π ≤ nθ < (k + 1)π k

which is impossible since cos(nθ) ≥ 0. Similarly, we can derive a contradiction if we assume θ ∈ (3π/ 2 , 2 π). Then 2 π − π/k ≤ θ < 2 π − π/(k + 1) for some k ∈ { 2 , 3 ,.. .}.

1.5 Lines and half planes in the complex plane

Exercise 1. Let C be the circle {z : |z − c| = r}, r > 0 ; let a = c + rcis α and put

Lβ =

z : Im

( (^) z − a

b

where b = cisβ. Find necessary and sufficient conditions in terms of β that Lβ be tangent to C at a.

Solution. Not available.

Exercise 4. Let Λ be a circle lying in S. Then there is a unique plane P in R^3 such that P ∩ S = Λ. Recall from analytic geometry that

P = {(x 1 , x 2 , x 3 ) : x 1 β 1 + x 2 β 2 + x 3 β 3 = l}

where (β 1 , β 2 , β 3 ) is a vector orthogonal to P and l is some real number. It can be assumed that β^21 +β^22 +β^23 =

1. Use this information to show that if Λ contains the point N then its projection on C is a straight line. Otherwise, Λ projects onto a circle in C.

Solution. Not available.

Exercise 5. Let Z and Z′^ be points on S corresponding to z and z′^ respectively. Let W be the point on S corresponding to z + z′. Find the coordinates of W in terms of the coordinates of Z and Z′.

Solution. Not available.

Chapter 2

Metric Spaces and the Topology of C

2.1 Definitions and examples of metric spaces

Exercise 1. Show that each of the examples of metric spaces given in (1.2)-(1.6) is, indeed, a metric space. Example (1.6) is the only one likely to give any difficulty. Also, describe B(x; r) for each of these examples.

Solution. Not available.

Exercise 2. Which of the following subsets of C are open and which are closed: (a) {z : |z| < 1 }; (b) the real axis; (c) {z : zn^ = 1 for some integer n ≥ 1 }; (d) {z ∈ C is real and 0 ≤ z < 1 }; (e) {z ∈ C : z is real and 0 ≤ z ≤ 1 }?

Solution. We have

(a) A := {z ∈ C : |z| < 1 } Let z ∈ A and set εz = 1 −| 2 z|, then B(z, εz) ⊂ A is open and therefore A =

z∈A B(z, εz)^ is open also. A cannot be closed, otherwise A and C − A were both closed and open sets yet C is connected.

(b) B := {z ∈ C : z = x + iy, y = 0 } (the real axis) Let z ∈ C − B, then Imz , 0. Set εz = |Im 2 z |, then B(z, εz) ⊂ C − B.Hence B is closed since its complement is open. For any real x and any ε > 0 the point x + i ε 2 ∈ B(x, ε) but x + i ε 2 ∈ C − R. Thus B is not open.

(c) C := {z ∈ C : zn^ = 1 for some integer n ≥ 1 } Claim: C is neither closed nor open. C is not open because if zn^ = 1 then z = rcis(θ) with r = 1 and any ε-ball around z would contain an element z′^ := (1 + ε 4 )cis(θ). To show that C cannot be closed, note that for (^) qp ∈ Q with p ∈ Z and q ∈ N the number z := cis

( (^) p q 2 π

∈ C since zq^ = cis(p 2 π) = cos(p 2 π) + i sin(p 2 π) = 1. Now fix x ∈ R − Q and let {xn}n be a rational sequence that converges to x. Let m be any natural number. Now zm^ = 1 implies that sin(mx 2 π) = 0 which in turn means that mx ∈ Z contradicting the choice x ∈ R − Q. We have constructed a sequence of elements of C that converges to a point that is not element of C. Hence C is not closed.

which follows from a simple computation

(z − x)(1 + xz¯′) + (x − z′)(1 + xz¯) = z − x + xz¯′z + x xz¯′^ + x − z′^ + x xz¯ − xz¯′z = z + x xz¯ − z′^ − x xz¯′ = (z − z′)(1 + xx¯).

Using (2.1) and taking norms gives

|(z − z′)(1 + xx¯)| = |(z − z′)(1 + |x|^2 )| = |(z − z′)|(1 + |x|^2 ) = |(z − x)(1 + xz¯′) + (x − z′)(1 + xz¯)| ≤ |z − x|(1 + |x|^2 )^1 /^2 (1 + |z′|^2 )^1 /^2 + |x − z′|(1 + |x|^2 )^1 /^2 (1 + |z|^2 )^1 /^2 (2.2)

where the last inequality follows by

|z + xz¯′| ≤ (1 + |x|^2 )^1 /^2 (1 + |z′|^2 )^1 /^2 ⇔ (1 + xz¯′)(1 + x¯z′) ≤ (1 + |x|^2 )(1 + |z′|^2 )

and

|z + xz¯| ≤ (1 + |x|^2 )^1 /^2 (1 + |z|^2 )^1 /^2 ⇔ (1 + xz¯)(1 + xz¯) ≤ (1 + |x|^2 )(1 + |z|^2 )

since

(1 + xz¯)(1 + xz¯) ≤ (1 + |x|^2 )(1 + |z|^2 ) ⇔ ¯xz + x¯z ≤ |x|^2 + |y|^2 ⇔ 2 Re(x¯z) ≤ |x|^2 + |y|^2

which is true by Exercise 4 part 2 on page 3. Thus, dividing (2.2) by (1 + |x|^2 )^1 /^2 yields

|z − z′|(1 + |x|^2 )^1 /^2 ≤ |z − x|(1 + |z′|^2 )^1 /^2 + |x − z′|(1 + |z|^2 )^1 /^2.

Multiplying this by 2 (1 + |x|^2 )^1 /^2 (1 + |z′|^2 )^1 /^2 (1 + |z|^2 )^1 /^2

gives the assertion above.

Exercise 8. Let (X, d) be a metric space and Y ⊂ X. Suppose G ⊂ X is open; show that G ∩ Y is open in (Y, d). Conversely, show that if G 1 ⊂ Y is open in (Y, d), there is an open set G ⊂ X such that G 1 = G ∩ Y.

Solution. Set G 1 = G ∩ Y, let G be open in (X, d) and Y ⊂ X. In order to show that G 1 is open in (Y, d), pick an arbitrary point p ∈ G 1. Then p ∈ G and since G is open, there exists an  > 0 such that

BX^ (p; ) ⊂ G.

But then BY^ (p; ) = BX^ (p; ) ∩ Y ⊂ G ∩ Y = G 1

which proves that p is an interior point of G 1 in the metric d. Thus G 1 is open in Y (Proposition 1.13a).

Let G 1 be an open set in Y. Then, for every p ∈ G 1 , there exists an −ball

BY^ (p : ) ⊂ G 1.

Thus G 1 =

p∈G 1

BY^ (p; ).

Since we can write BY^ (p; ) = BX^ (p; ) ∩ Y,

we get G 1 =

p∈G 1

BY^ (p; ) =

p∈G 1

BX^ (p; ) ∩ Y

p∈G 1

BX^ (p; ) ∩ Y = G ∩ Y

where G =

p∈G 1 B X (^) (p; ) is open in (X, d). (Proposition 1.9c since each BX (^) (p; ) is open).

Exercise 9. Do Exercise 8 with “closed” in place of “open.”

Solution. Not available.

Exercise 10. Prove Proposition 1.13.

Solution. Not available.

Exercise 11. Show that {cis k : k a non-negative integer } is dense in T = {z ∈ C : |z| = 1 }. For which values of θ is {cis(kθ) : k a non-negative integer } dense in T?

Solution. Not available.

2.2 Connectedness

Exercise 1. The purpose of this exercise is to show that a connected subset of R is an interval.

(a) Show that a set A ⊂ R is an interval iff for any two points a and b in A with a < b, the interval [a, b] ⊂ A.

(b) Use part (a) to show that if a set A ⊂ R is connected then it is an interval.

Solution. Not available.

Exercise 2. Show that the sets S and T in the proof of Theorem 2.3 are open.

Solution. Not available.

Exercise 3. Which of the following subsets X of C are connected; if X is not connected, what are its components: (a) X = {z : |z| ≤ 1 } ∪ {z : |z − 2 | < 1 }. (b) X = [0, 1) ∩

1 + (^1) n : n ≥ 1

. (c) X = C − (A ∩ B) where A = [0, ∞) and B = {z = r cis θ : r = θ, 0 ≤ θ ≤ ∞}?

Solution. a) Define X = {z : |z| ≤ 1 } ∪ {z : |z − 2 | < 1 } := A ∪ B. It suffices to show that X is path-connected. Obviously A is path-connected and B is path-connected. Next, we will show that X is path-connected. Recall that a space is path-connected if for any two points x and y there exists a continuous function f from the interval [0, 1] to X with f (0) = x and f (1) = y (this function f is called the path from x to y). Let x ∈ A and y ∈ B and define the function f (t) : [0, 1] → X by

f (t) =

(1 − 3 t)x + 3 tRe(x), 0 ≤ t ≤ (^13) (2 − 3 t)Re(x) + (3t − 1)Re(y), 13 < t ≤ (^23) (3 − 3 t)Re(y) + (3t − 2)y, 23 < t ≤ 1.

intersect A in a point different from x. In particular, for every integer n there is a point xn in B(x; (^1) n ) ∩ A. Thus d(x, xn) < (^1) n which implies xn → x (see Proof of Proposition 3.2). Then x ∈ A′, so x ∈ A ∪ A′. “⊇”: To show A ∪ A′^ ⊆ A−. Let x ∈ A ∪ A′. If x ∈ A, then x ∈ A−^ (A ⊂ A−). Now, assume x ∈ A′^ but not in A. Then there exists {xn} ⊂ A with limn→∞ xn = x. It follows, ∀ > 0 B(x; ) ∩ A , ∅ since {xn} ⊂ A. By Proposition 1.13f we get x ∈ A−.

Exercise 2. Furnish the details of the proof of Proposition 3.8.

Solution. Not available.

Exercise 3. Show that diam A = diam A−.

Solution. Not available.

Exercise 4. Let zn, z be points in C and let d be the metric on C∞. Show that |zn − z| → 0 if and only if d(zn, z) → 0. Also show that if |zn| → ∞ then {zn} is Cauchy in C∞. (Must {zn} converge in C∞?)

Solution. First assume that |zn − z| → 0 , then

d(zn, z) =

(1 + |zn|^2 )(1 + |z|^2 )

|zn − z| → 0

since the denominator

(1 + |zn|^2 )(1 + |z|^2 ) ≥ 1 is bounded below away from 0. To see the converse, let d(zn, z) → 0 or equivalently d^2 (zn, z) → 0. We need to show that if zn → z in the d-norm, then |zn| 9 ∞ because otherwise the denominator grows without bounds. In fact we will show that if |zn| → ∞, then d(zn, z) 9 0 for any z ∈ C. Then

d^2 (zn, z) =

|zn|^2 − zn ¯z − ¯znz + |z|^2

1 + |zn|^2

1 + |z|^2

|zn|^2 − 2 Re(zn ¯z) + |z|^2

1 + |z|^2

|zn|^2 + 1 + |z|^2

1 − 2 Re |z(nz| 2 n z¯)+ |z|

2 |zn|^2

1 + |z|^2 + 1 +|z|

2 |zn|^2

if |zn| , 0

in particular if |zn| → ∞, d(zn, z) → (^1) +^4 |z| 2 , 0. This shows that the denominator remains bounded as d(zn, z) → 0 and therefore the numerator 2 |zn − z| → 0. Hence convergence in d-norm implies convergence in |·|-norm for numbers zn, z ∈ C. Next assume that |zn| → ∞. Then clearly also

1 + |zn|^2 → ∞ and therefore d(zn, ∞) = √^2 1 +|zn|^2 → 0. The last thing to show is that if zn → ∞ in (C∞, d), also |zn| → ∞. But

d(zn, ∞) → 0 implies

1 + |zn|^2 → ∞ which is equivalent to |zn| → ∞.

Exercise 5. Show that every convergent sequence in (X, d) is a Cauchy sequence.

Solution. Let {xn} be a convergent sequence with limit x. That is, given  > 0 ∃N such that d(xn, x) <  2 if n > N and d(x, xm) <  2 if m > N. Thus

d(xn, xm) ≤ d(xn, x) + d(x, xm) <

= , ∀n, m ≥ N

and therefore {xn} is a Cauchy sequence.

Exercise 6. Give three examples of non complete metric spaces.

Solution. Example 1: Let X = C[− 1 , 1] and the metric d( f, g) =

1 − 1 [^ f^ (x)^ −^ g(x)]

(^2) dx, f, g ∈ X. Consider

the Cauchy sequence

fn(x) =

0 , − 1 ≤ x ≤ 0 nx, 0 < x ≤ (^1) n 1 , (^1) n < x ≤ 1

It is obvious that the limit function f is discontinuous. Hence, the metric space (X, d) is not complete. Example 2: Let X = (0, 1] with metric d(x, y) = |x − y|, x, y ∈ X. The sequence { (^1) n } is Cauchy, but converges to 0 , which is not in the space. Thus, the metric space is not complete. Example 3: Let X = Q, the rationals, with metric d(x, y) = |x − y|, x, y ∈ X. The sequence defined by x 1 = 1 , xn+ 1 = x 2 n + (^) x^1 n is a Cauchy sequence of rational numbers. The limit

2 is not a rational number. Therefore, the metric space is not complete.

Exercise 7. Put a metric d on R such that |xn − x| → 0 if and only if d(xn, x) → 0 , but that {xn} is a Cauchy sequence in (R, d) when |xn| → ∞. (Hint: Take inspiration from C∞.)

Solution. Not available.

Exercise 8. Suppose {xn} is a Cauchy sequence and {xnk } is a subsequence that is convergent. Show that {xn} must be convergent.

Solution. Since {xnk } is convergent, there is a x such that xnk → x as k → ∞. We have to show that xn → x as n → x. Let  > 0. Then we have ∃N ∈ N such that d(xn, xm) <  2 ∀n, m ≥ N since {xn} is Cauchy and ∃M ∈ N such that d(xnk , x) <  2 ∀nk ≥ M since xnk → x as k → ∞. Now, fix nk 0 > M + N, then

d(xn, x) ≤ d(xn, xnk 0 ) + d(xnk 0 , x) <

= , ∀n ≥ N.

Thus, xn → x as n → ∞ and therefore {xn} is convergent.

2.4 Compactness

Exercise 1. Finish the proof of Proposition 4.4.

Solution. Not available.

Exercise 2. Let p = (p 1 ,... , pn) and q = (q 1 ,... , qn) be points in Rn^ with pk < qk for each k. Let R = [p 1 , q 1 ] ×... × [pn, qn] and show that

diam R = d(p, q) =

∑^ n

k= 1

(qk − pk)^2

(^12) .

Solution. By definition diam(R) = sup x∈R,y∈R

d(x, y).

Obviously, R is compact, so we have

diam(R) = max x∈R,y∈R d(x, y).