Solutions Functions of One Complex Variable I, Second Edition by John B. Conway

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Solutions Manual for

Functions of One Complex Variable I, Second Edition

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©

Copyright by Andreas Kleefeld, 2009

All Rights Reserved

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1_{by John B. Conway}

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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 intented for all students taking a graduate level Complex Analysis course. Students can check their answers to homework problems assigned from the excellent book “Functions of One Com-plex 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 [email protected] (LA_TEXpreferred

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 [email protected]. 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

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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 + 5i 7i + 1;       −1 + i√3 2       3 ;       −1 − i√3 2       6 ; in; 1 + i_√ 2 !n for 2_{≤ n ≤ 8.}

Solution. Let z = x + iy. Then a) Re 1 z ! = x x2_{+ y}2 = Re(z) |z|2 Im 1 z ! = ₋ y x2_{+ y}2 =− Im(z) |z|2 b) Re z − a z + a = x 2_{+ y}2 − a2 x2_{+ y}2_{+ 2ax + a}2 = |z|2 − a2 |z|2_{+ 2aRe(z) + a}2 Im z − a z + a = 2ya x2_{+ y}2_{+ 2xa + a}2 = 2Im(z)a |z|2_{+ 2aRe(z) + a}2 c)

Rez3 = _x3_{− 3xy}2_{= Re(z)}3_{− 3Re(z)Im(z)}2

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d) Re 3 + 5i 7i + 1 ! = 19 25 Im 3 + 5i 7i + 1 ! = ₋8 25 e) Re               −1 + i√3 2       3        = ₁ Im               −1 + i√3 2       3        = ₀ f) Re               −1 − i√3 2       6        = ₁ Im               −1 − i√3 2       6        = ₀ g) Re (in) =              0, n is odd 1, n_{∈ {4k : k ∈ Z}} −1, n ∈ {2 + 4k : k ∈ Z} Im (in) =              0, n is even 1, n_{∈ {1 + 4k : k ∈ Z}} −1, n ∈ {3 + 4k : k ∈ Z}

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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 + 3i); √3− i

2 + 3i ; 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 + 3i) = 5 + 10i, _{|z| = 5}√5, ¯z = 5_{− 10i}

d z = _√3− i 2 + 3i, |z| = 1 11 √ 110, ¯z = _√3 + i 2_{− 3i} e) z = i i + 3 = 1 10+ 3 10i, |z| = 1 10 √ 10, ¯z = 1 10− 3 10i f) z = (1 + i)6=_−8i, _{|z| = 8,} _{¯z = 8i} g) i17 = i, _{|z| = 1,} ¯z =_−i

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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_{+ 2Re(z ¯}_{w) +}_|w|2_. |z − w|2 ₌ |z|2 − 2Re(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) = z¯z + 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_{+ 2Re(z ¯}_{w) =}_|z|2_{+ 2Re(z ¯}_{w) +}_|w|2_. |z − w|2 ₌ _(z − w)(z − w) = (z − w)(¯z − ¯w) = z¯z − 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_{− 2Re(z ¯w) = |z|}2_{− 2Re(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 = z1+. . . + zn; w = w1w2. . . wn:

|w| = |w1| . . . |wn|; ¯z = ¯z1+. . . + ¯zn; ¯w = ¯w1. . . ¯wn.

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) = anz

n_{+ a}

n₋₁zn−1+. . . a0

bmzm+ bm−1zm−1+. . . b0

where n, m are nonnegative integers. Let all coefficients of R(z) be real, that is a0, a1, . . . , an, b0, b1, . . . , bm∈ R. Then R(z) = anz n_{+ a} n−1zn−1+. . . a0 bmzm+ bm₋₁zm−1+. . . b0 = anz n_{+ a} n−1zn−1+. . . a0 bmzm+ bm−1zm−1+. . . b0 = anz n_{+ a} n₋₁zn−1+. . . a0 bmzm+ bm−1zm−1+. . . b0 = an¯z n_{+ a} n₋₁¯zn−1+. . . a0 bm¯zm+ bm−1¯zm−1+. . . b0 = R(¯z).

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1.3 The complex plane

Exercise 1. Prove (3.4) and give necessary and sufficient conditions for equality. Solution. Let z and w be complex numbers. Then

||z| − |w|| = ||z − w + w| − |w|| ≤ ||z − w| + |w| − |w||

= _{||z − w||}

= _{|z − w|}

Notice that_{|z| and |w| is the distance from z and w, respectively, to the origin while |z − w| is the distance} between z and w. Considering the construction of the implied triangle, in order to guarantee equality, it is necessary and sufficient that

||z| − |w|| = |z − w| ⇐⇒ (_{|z| − |w|)}2=_{|z − w|}2 ⇐⇒ (_{|z| − |w|)}2=_|z|2_{− 2Re(z ¯w) + |w|}2 ⇐⇒ |z|2 − 2|z||w| + |w|2₌ |z|2 − 2Re(z ¯w) + |w|2 ⇐⇒ |z||w| = Re(z ¯w) ⇐⇒ |z ¯w| = Re(z ¯w)

Equivalently, this is z ¯w _{≥ 0. Multiplying this by} w_w, we get z ¯w_· w_w = _|w|2 _· z

w ≥ 0 if w , 0. If t = _wz = 1 |w|2 · |w|2_· z w. Then t≥ 0 and z = tw.

Exercise 2. Show that equality occurs in (3.3) if and only if zk/zl≥ 0 for any integers k and l, 1 ≤ k, l ≤ n,

for which zl,0.

Exercise 3. Let a_{∈ R and c > 0 be fixed. Describe the set of points z satisfying}

|z − a| − |z + a| = 2c

for every possible choice of a and c. Now let a be any complex number and, using a rotation of the plane, describe the locus of points satisfying the above equation.

1.4 Polar representation and roots of complex numbers

Exercise 1. Find the sixth roots of unity.

Solution. Start with z6 _{= 1 and z = rcis(θ), therefore r}6_{cis(6θ) = 1. Hence r = 1 and θ =} 2kπ

6 with k ∈ {−3, −2, −1, 0, 1, 2}. The following table gives a list of principle values of arguments and the corresponding

value of the root of the equation z6= 1.

θ0= 0 z0= 1 θ1= π 3 z1= cis( π 3) θ2= 2π 3 z2= cis( 2π 3) θ3=_π _z₃_{= cis(π) =}₋₁ θ4= −2π₃ z4= cis(−2π₃ ) θ5= −π₃ z5= cis(−π₃ )

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Exercise 2. Calculate the following: a) the square roots of i

b) the cube roots of i

c) the square roots of √3 + 3i

Solution. c) The square roots of √3 + 3i.

Let z = √3 + 3i. Then r =_{|z| =}

q_√ 32+ 32= √_{12 and α = tan}−1 3 √ 3 = π

3. So, the 2 distinct roots of z

are given by√2_r_cosα+2kπ

n + i sin α+2kπ n where k = 0, 1. Specifically, √ z = √412 cos π 3+ 2kπ 2 + i sin π 3 + 2kπ 2 ! .

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

z0= 4 √ 12cosπ₆+ i sinπ₆=√4₁₂ √ 3 2 + 1 2i z1= 4 √ 12cos7π₆ + i sin7π₆=√4₁₂ −√3 2 − 1 2i .

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

√4 108 2 , 4 √ 12 2 and −√4108 2 ,− 4 √ 12 2 .

Exercise 3. A primitive nth root of unity is a complex number a such that 1, a, a2, ..., an−1are 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?

Exercise 4. Use the binomial equation

(a + b)n= n X 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−2θ sin2θ + n 4 ! cosn−4θ sin4θ_{− . . .} sin nθ = n 1 ! cosn−1θ sin θ− n₃ ! cosn−3θ sin3θ + . . .

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Solution. The summation of the finite geometric sequence 1, z, z2_{, . . . , z}n−1_{can be calculated as}Pn

j=1zj−1= zn₋₁

z₋₁. We want to show that z

n _{is an n}th _{root of unity. So, using de Moivre’s formula, z}n ₌_cis2π

n n = cisn_·2π n

= cis(2π) = 1. It follows that 1 + z + z2₊_{... + z}n₋₁₌ zn₋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}.}

Exercise 7. If z_{∈ C and Re(z}n₎_{≤ 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(z}n₎ _{≥ 0. Since z}n ₌

rn_{(cos(nθ) + i sin(nθ)), we have}

Re(zn) = rncos(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) = rncos(nθ)_{≥ 0}

implies cos(nθ) _{≥ 0 for all n. This implies θ = 0 as we will show next. Clearly, θ < [π/2, 3π/2]. If}

θ_{∈ (0, π/2), then there exists a k ∈ {2, 3, . . .} such that}_k+1π _{≤ 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, 2π).} Then 2π_{− π/k ≤ θ < 2π − 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 = 0

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

1.6 The extended plane and its spherical representation

Exercise 1. Give the details in the derivation of (6.7) and (6.8). Solution. Not available.

Exercise 2. For each of the following points in C, give the corresponding point of S : 0, 1 + i, 3 + 2i. Solution. We have Φ(z) = 2x |z|2_{+ 1}, 2y |z|2_{+ 1}, |z|2_{− 1} |z|2_{+ 1} ! .

If z1= 0, then|z1| = 0 and therefore

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Thus, z1= 0 corresponds to the point (0, 0,−1) on the sphere S . If z2= 1 + i, then|z2| = √ 2 and therefore Φ(z2) = 2 3, 2 3, 1 3 ! .

Thus, z2= 1 + i corresponds to the point ₂ 3, 2 3, 1 3 on the sphere S . If z3= 3 + 2i, then|z3| = √ 13 and therefore Φ(z3) = 3 7, 2 7, 6 7 ! .

Thus, z3= 3 + 2i corresponds to the point ₃ 7, 2 7, 6 7 on the sphere S .

Exercise 3. Which subsets of S correspond to the real and imaginary axes in C. Solution. If z is on the real axes, then z = x which implies_|z|2_{= x}2_{. Thus}

Φ(z) = 2x x2_{+ 1}, 0, x2_{− 1} x2_{+ 1} ! .

Therefore the set

( 2x x2_{+ 1}, 0, x2_{− 1} x2_{+ 1} ! |x ∈ R ) ⊂ S

corresponds to the real axes in C. That means the unit circle x2+ z2= 1 lying in the xz_{−plane corresponds}

to the real axes in C.

If z is on the imaginary axes, then z = iy which implies_|z|2= y2. Thus

Φ(z) = 0, 2y y2_{+ 1}, y2_{− 1} y2_{+ 1} ! .

Therefore the set

( 0, 2y y2_{+ 1}, y2_{− 1} y2_{+ 1} ! |y ∈ R ) ⊂ S

corresponds to the imaginary axes in C. That means the unit circle y2 _{+ z}2 _{= 1 lying in the yz}_−plane

corresponds to the imaginary axes in C.

Exercise 4. Let Λ be a circle lying in S . Then there is a unique plane P in R3_{such that P}_{∩ S = Λ. Recall}

from analytic geometry that

P =_{(x1, x2, x3) : x1β1+ x2β2+ x3β3= l}

where (β1, β2, β3) is a vector orthogonal to P and l is some real number. It can be assumed that β2 1+β

2 2+β

2

3=

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 Z0be points on S corresponding to z and z0respectively. Let W be the point on S corresponding to z + z0. Find the coordinates of W in terms of the coordinates of Z and Z0.

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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 : z}n_{= 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 ε}z= 1−|z|₂ , then B(z, εz)⊂ A is open and therefore A =Sz_∈AB(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 ε}z = |Imz|₂ , 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 + iε₂ _{∈ B(x, ε) but x + i}₂ε _{∈ C − R. Thus B is not open.} (c) C :=_{{z ∈ C : z}n_{= 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 z0:= (1 +ε₄)cis(θ).

To show that C cannot be closed, note that for p_q _{∈ Q with p ∈ Z and q ∈ N the number z :=}

cisp_q2π_{∈ C since}

zq = cis(p2π) = cos(p2π) + i sin(p2π) = 1.

Now fix x _{∈ R − Q and let {x}n}n be a rational sequence that converges to x. Let m be any natural

number. Now zm_{= 1 implies that sin(mx2π) = 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.

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(d) D :=_{{z ∈ C : z is real and 0 ≤ z < 1}}

The set cannot be open by the argument given in2.1) and it is not closed because zn := 1−1_n is a

sequence in D that converges to a point outside of D. (e) E :=_{{z ∈ C : z is real and 0 ≤ z ≤ 1}}

This set E is not open by the observation in2.1) and it is closed because its complement is open: If z , E and z real, then B(z,min{|x|,|x−1|}₂ ) is contained in the complement of E, if z is imaginary then

B(z,|Imy|₂ ) is completely contained in the complement of E.

Exercise 3. If (X, d) is any metric space show that every open ball is, in fact, an open set. Also, show that every closed ball is a closed set.

Exercise 4. Give the details of the proof of (1.9c). Solution. Not available.

Exercise 5. Prove Proposition 1.11. Solution. Not available.

Exercise 6. Prove that a set G_{⊂ X is open if and only if X − G is closed.} Solution. Not available.

Exercise 7. Show that (C_∞, d) where d is given (I. 6.7) and (I. 6.8) is a metric space.

Solution. To show (C_∞, d) with d(z, z0) = _[(1+ 2|z−z0|

|z|2₎₍₁₊_|z0_|2_)]1/2 for z, z0 ∈ C and d(z, ∞) = 2 (1+_|z|2₎1/2, z ∈ C is a metric space. i) d(z, z0)_{≥ 0.} Since_|z−z0_{| ≥ 0, we have d(z, z}0) = _[(1+ 2|z−z0|

|z|2₎₍₁₊_|z0_|2_)]1/2 ≥ 0 for all z, z0∈ C (the denominator is always positive).

Obviously, d(z,_{∞) =} 2

(1+|z|2₎1/2 ≥ 0 for all z ∈ C.

ii) d(z, z0) = 0 iff z = z0.

We have 2_{|z − z}0_{| = 0 iff z = z}0. Therefore, d(z, z0) = _[(1+ 2|z−z0|

|z|2₎₍₁₊_|z0_|2_)]1/2 = 0 iff z = z0. d(∞, ∞) =

limz_→∞₍₁₊_|z|22₎1/2 = 0. Thus, d(∞, 0) = 0 iff z = ∞.

iii) d(z, z0) = d(z0, z).

We have d(z, z0) = _[(1+ 2|z−z0|

|z|2₎₍₁₊_|z0_|2_)]1/2 =

2|z0_−z|

[(1+_|z0_|2₎₍₁₊_|z|2_)]1/2 = d(z0, z) for all z, z0 ∈ C. Also d(z, ∞) = 2 (1+_|z|2₎1/2 =

d(_{∞, z) by the symmetry of d(z, z}0) in general.

iv) d(z, z0)_{≤ d(z, x) + d(x, z}0). We have d(z, z0) = 2|z − z 0_| [(1 +_|z|2_{)(1 +}_|z0_|2_)]1/2 ≤ _{[(1 +} 2|z − x| |z|2_{)(1 +}_|x|2_)]1/2 + 2_{|x − z}0_| [(1 +_|x|2_{)(1 +}_|z₀_|2_)]1/2 = _{d(z, x) + d(x, z}0₎

for all x, z, z0_{∈ C. The first inequality will be shown next.} First, we need the following equation

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which follows from a simple computation

(z_{− x)(1 + ¯xz}0) + (x_{− z}0)(1 + ¯xz) = _z_{− x + ¯xz}0_{z + x ¯xz}0_{+ x}_{− z}0_{+ x ¯xz}_{− ¯xz}0_z = _{z + x ¯xz}_{− z}0_{− x ¯xz}0

= _(z_{− z}0_{)(1 + ¯xx).}

Using (2.1) and taking norms gives

|(z − z0)(1 + ¯xx)_{| = |(z − z}0)(1 +_|x|2)_{| = |(z − z}0)_{|(1 + |x|}2) = _{|(z − x)(1 + ¯xz}0_{) + (x}_{− z}0_{)(1 + ¯xz)}_| ≤ |z − x|(1 + |x|2₎1/2_{(1 +} |z0|2₎1/2₊ |x − z0|(1 + |x|2₎1/2_{(1 +} |z|2₎1/2 _(2.2)

where the last inequality follows by

|z + ¯xz0| ≤ (1 + |x|2₎1/2_{(1 +} |z0|2₎1/2 ⇔ (1 + ¯xz0_{)(1 + x¯z}0₎_{≤ (1 + |x|}2_{)(1 +} |z0_|2₎ and |z + ¯xz| ≤ (1 + |x|2)1/2(1 +_|z|2)1/2 ⇔ (1 + ¯xz)(1 + x¯z) ≤ (1 + |x|2_{)(1 +} |z|2₎ since (1 + ¯xz)(1 + x¯z)_{≤ (1 + |x|}2)(1 +_|z|2) ⇔ ¯xz + x¯z_{≤ |x|}2+_|y|2 ⇔ 2Re(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)1/2yields

|z − z0_{|(1 + |x|}2₎1/2 ≤ |z − x|(1 + |z0_|2₎1/2₊ |x − z0_{|(1 + |z|}2₎1/2_. Multiplying this by 2 (1 +_|x|2₎1/2_{(1 +}_|z0_|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 G1 ⊂ Y is open in (Y, d), there is an open set G ⊂ X such that G1 = G∩ Y.

Solution. Set G1 = G∩ Y, let G be open in (X, d) and Y ⊂ X. In order to show that G1 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 G1in the metric d. Thus G1is open in Y (Proposition 1.13a).

Let G1be an open set in Y. Then, for every p∈ G1, there exists an −ball

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Thus

G1 = [

p∈G1

BY(p; ).

Since we can write

BY(p; ) = BX(p; )_{∩ Y,} we get G1= [ p∈G1 BY(p; ) = [ p∈G1 BX(p; )_{∩ Y}= [ p∈G1 BX(p; )_{∩ Y = G ∩ Y} where G =Sp∈G1B

X_{(p; ) is open in (X, d). (Proposition 1.9c since each B}X_{(p; ) is open).}

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

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) ∩}n1 + 1

n : n≥ 1

o

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

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This function is obviously continuous, since f (1/3) = lim_t_→1 3 − f (t) = lim_t →1 3 + f (t) = Re(x) ∈ X and f (2/3) = lim_t →2 3 − f (t) = lim_t →2 3

+ f (t) = Re(y)∈ X. In addition, we have f (0) = x and f (1) = y. Therefore

X is path-connected and hence X is connected.

b) There is no way to connect_{{2} and {}3₂_{}. Therefore X is not connected. The components are [0, 1), {2},}

{3 2}, { 4 3}, . . ., {1 + 1 n}.

c) X is not connected, since there is no way to connect (2, 1) and (1,_{−2). The k − th component is given by}

C_{− {A ∩ B} where}

A = [2πk_{− 2π, 2πk)} and

B =_{{z = r cis θ : r = θ, 2πk − 2π ≤ θ < 2πk},} k_{∈ {1, 2, 3, . . .}.}

Exercise 4. Prove the following generalization of Lemma 2.6. If_{Dj : j∈ J} is a collection of connected

subsets of X and if for each j and k in J we have Dj∩ Dk, then D =S{Dj: j∈ J} is connected.

Solution. Let D =Sj∈JDjandC = {Dj: j∈ J}. If D is connected, we could write D as the disjoint union

A_{∪ B where A and B are nonempty subsets of X. Thus, for each C ∈ C either C ⊂ A or C ⊂ B.}

We have C _{⊂ A ∀C ∈ C or C ⊂ B ∀C ∈ C. If not, then there exist E, F ∈ C such that E ⊂ A and F ⊂ B.} But, we assume A_{∪ B is disjoint and thus E ∪ F is disjoint which contradicts the assumption E ∩ F , ∅.} Therefore, all members of_{C are contained in either A or all B. Thus, either D = A and B = ∅ or D = B and}

A =_{∅. Both contradicting the fact that A, B are assumed to be nonempty. Hence,}

D =[

j_∈J

Dj

is connected.

Exercise 5. Show that if F _{⊂ X is closed and connected then for every pair of points a, b in F and each}

 > 0 there are points z0, z1, . . . , zn in F with z0 = a, zn = b and d(zk−1, zk) < for 1 ≤ k ≤ n. Is the

hypothesis that F be closed needed? If F is a set which satisfies this property then F is not necessarily connected, even if F is closed. Give an example to illustrate this.

2.3 Sequences and completeness

Exercise 1. Prove Proposition 3.4.

Solution. a) A set is closed iff it contains all its limit points. Let S _{⊂ X be a set.}

“_{⇐”: Assume S contains all its limit points. We have to show that S is closed or that S}c is open. Let

x_{∈ S}c. By assumption x is not a limit point and hence there exists an open _{−ball around x, B(x; ) such} that B(x; )_{∩ S = ∅ (negation of Proposition 1.13f). So B(x; ) ⊂ S}cand therefore Scis open.

“_{⇒”: Let S be closed and x be a limit point. We claim x ∈ S . If not, S}c (open) would be an open

neighborhood of x, that does not intersect S . (Proposition 1.13f again) which contradicts the fact that x is a limit point of S .

b) If A_{⊂ X, then A}−= A_{∪ {x : x is a limit point of A} := A ∪ A}0. Let A_{⊂ X be a set.}

“_{⊆”: Let x ∈ ¯}A. We want to show x_{∈ A ∪ A}0. If x_{∈ A, then obviously x ∈ A ∪ A}0. Suppose x < A. Since

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intersect A in a point different from x. In particular, for every integer n there is a point xnin B(x;1_n)∩ A.

Thus d(x, xn) < 1_n which implies xn→ x (see Proof of Proposition 3.2). Then x ∈ A0, so x∈ A ∪ A0.

“_{⊇”: To show A ∪ A}0 _{⊆ A}−. Let x_{∈ A ∪ A}0. If x_{∈ A, then x ∈ A}−(A_{⊂ A}−). Now, assume x_{∈ A}0but 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) =

2

p

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

|zn− z| → 0

since the denominator p(1 +_|zn|2)(1 +|z|2) ≥ 1 is bounded below away from 0. To see the converse, let

d(zn, z) → 0 or equivalently d2(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

d2(zn, z) = 4_|zn|2− zn¯z− ¯znz +|z|2 1 +_|zn|2 1 + |z|2 =4 |zn|2− 2Re(zn¯z) +|z|2 1 +_|z|2 |z n|2+ 1 +|z|2 = 41₋2Re(zn¯z) |zn|2 + |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 z}n, z ∈ C. Next assume that |zn| → ∞. Then clearly also

p

1 +_|zn|2 → ∞ and

therefore d(zn,∞) = √2

1+_|zn|2 → 0. The last thing to show is that if z

n→ ∞ in (C∞, d), also|zn| → ∞. But

d(zn,∞) → 0 implies

p

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) < 2 + 2 =, ∀n, m ≥ N

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

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

q R1

−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= x₂n+_x1_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_∞.)

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(x}n, 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 nk0> M + N, then

d(xn, x)≤ d(xn, xn_k0) + d(xn_k0, x) <

2+

2 =, ∀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 = (p1, . . . , pn) and q = (q1, . . . , qn) be points in Rn with pk < qk for each k. Let

R = [p1, q1]_{× . . . × [p}n, qn] and show that

diam R = d(p, q) =        n X k=1 (qk− pk)2        1 2 . Solution. By definition diam(R) = sup x∈R,y∈R d(x, y).

Obviously, R is compact, so we have

diam(R) = max

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Let x = (x1, . . . , xn) ∈ R and y = (y1, . . . , yn) ∈ R. Then clearly, pi ≤ xi ≤ qiand pi ≤ yi ≤ qi for all i = 1, . . . , n. We also have (yi− xi)2≤ (qi− pi)2, ∀i = 1, . . . , n. (2.3) Therefore, by (2.3) we obtain d(x, y) = v t n X i=1 (yi− xi)2 ≤ v t n X i=1 (qi− pi)2= d(p, q).

Note that we get equality if for example xi= piand yi= qifor all i = 1, . . . , n. Thus the maximum distance

is obtained, so diam(R) = sup x∈R,y∈R d(x, y) = v t n X i=1 (qi− pi)2= d(p, q).

Exercise 3. Let F = [a1, b1]_{× . . . × [a}n, bn] ⊂ Rn and let > 0; use Exercise 2 to show that there are

rectangles R1, . . . , Rmsuch that F = Smk=1Rkand diam Rk < for each k. If xk ∈ Rkthen it follows that

Rk⊂ B(xk; ).

Exercise 4. Show that the union of a finite number of compact sets is compact.

Solution. Let K =Sn_i=1Kibe a finite union of compact sets. Let{Gλ}λ∈Γbe an open cover of K, that is

K_⊂[

λ_∈Γ

Gλ.

Of course,_{Gλ}λ_∈Γis an open cover of each Ki, i = 1, . . . , n, that is

Ki⊂

[

λ∈Γ

Gλ, ∀i = 1, . . . , n.

Since each Kiis compact,

Ki⊂ ki [

j=1

Gi, j

for each i = 1, . . . , n. Thus

K = n [ j=1 Ki⊂ n [ i=1 ki [ j=1 Gi, j

which is a finite union and therefore K is compact.

Exercise 5. Let X be the set of all bounded sequences of complex numbers. That is,_{xn} ∈ X iff sup{|xn| :

n_{≥ 1} < ∞. If x = {x}n} and y = {yn}, define d(x, y) = sup{|xn− yn| : n ≥ 1}. Show that for each x in X and

 > 0, ¯B(x; ) is not totally bounded although it is complete. (Hint: you might have an easier time of it if

you first show that you can assume x = (0, 0, . . .).) Solution. Not available.

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Solution. Let (X, d) be a given metric space and let S _{⊂ X be totally bounded. Let > 0. Since S is totally} bounded, we have by Theorem 4.9 d) p. 22 that there exist a finite number of points x1, . . . , xn∈ S such that

S _⊆

n

[

k=1

B(xk; /2).

Taking the closure, gives the desired result

¯ S _⊆ n [ k=1 B(xk; ).

2.5 Continuity

Exercise 2. Show that if f and g are uniformly continuous (Lipschitz) functions from X into C then so is f + g.

Solution. The distance in C is ρ(x, y) =_{|x − y|. The distance in X is d(x, y). Let f : X → C be Lipschitz,} that is, there is a constant M > 0 such that

ρ( f (x), f (y)) =| f (x) − f (y)| ≤ Md(x, y), ∀x, y ∈ X

and g : X_{→ C be Lipschitz, that is, there is a constant N > 0 such that}

ρ(g(x), g(y)) =_{|g(x) − g(y)| ≤ Nd(x, y),} _{∀x, y ∈ X.}

We have

ρ( f (x) + g(x), f (y) + g(y)) = _{| f (x) + g(x) − f (y) − g(y)| = | f (x) − f (y) + g(x) − g(y)|}

≤ | f (x) − f (y)| + |g(x) − g(y)|

≤ Md(x, y) + Nd(x, y) = (M + N)d(x, y), ∀x, y ∈ X.

Since there is a constant K = M + N > 0 such that

ρ( f (x) + g(x), f (y) + g(y))≤ Kd(x, y), ∀x, y ∈ X,

we have that f + g is Lipschitz.

Now, let f , g be both uniformly continuous, that is,

∀1> 0∃δ1> 0 such that ρ( f (x), f (y)) =| f (x) − f (y)| < 1whenever d(x, y) < δ1

and

∀2> 0∃δ2> 0 such that ρ( f (x), f (y)) =| f (x) − f (y)| < 2whenever d(x, y) < δ2.

We have

ρ( f (x) + g(x), f (y) + g(y)) = _{| f (x) + g(x) − f (y) − g(y)| = | f (x) − f (y) + g(x) − g(y)|}

≤ | f (x) − f (y)| + |g(x) − g(y)|

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whenever d(x, y) < δ1and d(x, y) < δ2, that is, whenever d(x, y) < min(δ1, δ2). So, choosing = 1+2and

δ = min(δ1, δ2), we have shown that

∀ > 0 ∃δ > 0 such that ρ( f (x), f (y)) = | f (x) − f (y)| < whenever d(x, y) < δ.

Thus f + g is uniformly continuous.

Exercise 3. We say that f : X_{→ C is bounded if there is a constant M > 0 with | f (x)| ≤ M for all x in X.} Show that if f and g are bounded uniformly continuous (Lipschitz) functions from X into C then so is f g. Solution. Let f be bounded, that is, there exists a constant M1> 0 with| f (x)| < M1for all x∈ X and let g

be bounded, that is, there exists a constant M2> 0 with|g(x)| < M2for all x∈ X. Obviously f g is bounded,

because

| f (x)g(x)| ≤ | f (x)| |g(x)| ≤ M1· M2 ∀x ∈ X.

So, there exists a constant M = M1M2with

| f (x)g(x)| ≤ M ∀x ∈ X.

Let f be Lipschitz, that is, there exists a constant N1> 0 such that

ρ( f (x), f (y))≤ N1d(x, y) ∀x, y ∈ X

and let g be Lipschitz, that is, there exists a constant N2> 0 such that

ρ(g(x), g(y))_{≤ N}1d(x, y) ∀x, y ∈ X.

Now,

ρ( f (x)g(x), f (y)g(y)) = _{| f (x)g(x) − f (y)g(y)| =}

= _{| f (x)g(x) − f (x)g(y) + f (x)g(y) − f (y)g(y)|}

≤ | f (x)g(x) − f (x)g(y)| + | f (x)g(y) − f (y)g(y)| ≤ | f (x)| |g(x) − g(y)| + | f (x)| |g(y) − g(y)|

≤ M1N2d(x, y) + M2N1d(x, y) = (M1N2+ M2N1)d(x, y) ∀x, y ∈ X.

Thus, there exists a constant K = M1N2+ M2N1 > 0 with

ρ( f (x)g(x), f (y)g(y))≤ Kd(x, y) ∀x, y ∈ X,

so f g is Lipschitz and bounded.

Now, let f and g be both bounded and uniformly continuous, that is

∃ M1such that_{| f (x)| ≤ M1} _{∀x ∈ X}

∃ M2such that|g(x)| ≤ M2 ∀x ∈ X

∀ 1> 0∃δ1> 0 such that ρ( f (x), f (y)) =| f (x) − f (y)| < 1whenever d(x, y) < δ1 ∀ 2> 0∃δ2> 0 such that ρ( f (x), f (y)) =| f (x) − f (y)| < 2whenever d(x, y) < δ2.

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It remains to verify that f g is uniformly continuous, since we have already shown that f g is bounded. We have

ρ( f (x)g(x), f (y)g(y)) = _{| f (x)g(x) − f (y)g(y)| =}

= _{| f (x)g(x) − f (x)g(y) + f (x)g(y) − f (y)g(y)|}

≤ | f (x)g(x) − f (x)g(y)| + | f (x)g(y) − f (y)g(y)| ≤ | f (x)| |g(x) − g(y)| + | f (x)| |g(y) − g(y)|

≤ M12+ M21,

whenever d(x, y) < min(δ1, δ2). So choosing = M12+ M21and δ = min(δ1, δ2), we have∀ > 0 ∃δ > 0

such that

| f (x)g(x) − f (y)g(y)| < 

whenever d(x, y) < δ. Thus, f g is uniformly continuous and bounded.

Exercise 4. Is the composition of two uniformly continuous (Lipschitz) functions again uniformly continu-ous (Lipschitz)?

Exercise 5. Suppose f : X_{→ Ω is uniformly continuous; show that if {x}n} is a Cauchy sequence in X then

{ f (xn)} is a Cauchy sequence in Ω. Is this still true if we only assume that f is continuous? (Prove or give

a counterexample.)

Solution. Assume f : X_{→ Ω is uniformly continuous, that is, for every > 0 there exists δ > 0 such that}

ρ( f (x), f (y)) < if d(x, y) < δ. If{xn} is a Cauchy sequence in X, we have, for every 1 > 0 there exists

N_{∈ N such that d(x}n, xm) < 1for all n, m≥ N. But then, by the uniform continuity, we have that

ρ( f (xn), f (xm)) < ∀n, m ≥ N

whenever d(xn, xm) < δ which tells us that{ f (xn)} is a Cauchy sequence in Ω.

If f is continuous, the statement is not true. Here is a counterexample: Let f (x) = 1_xwhich is continuous on (0, 1). The sequence xn = 1_n is apparently convergent and therefore a Cauchy sequence in X. But

{ f (xn)} = { f (_n1)} = {n} is obviously not Cauchy. Note that f (x) = 1_x is not uniformly continuous on (0, 1).

To see that pick = 1. Then there is no δ > 0 such that_{| f (x) − f (y)| < 1 whenever |x − y| < δ. Assume there} exists such a δ. WLOG assume δ < 1 since the interval (0, 1) is considered. Let y = x + δ/2 and set x = δ/2, then | f (x) − f (y)| = 1 x− 1 y =y− x xy = δ/2 δ/2_{· δ} = 1 δ > 1,

that is no matter what δ < 1 we choose, we always obtain_{| f (x) − f (y)| > 1. Therefore f (x) =} 1_xcannot be uniformly continuous.

Exercise 6. Recall the definition of a dense set (1.14). Suppose that Ω is a complete metric space and that f : (D, d)_{→ (Ω; ρ) is uniformly continuous, where D is dense in (X, d). Use Exercise 5 to show that there} is a uniformly continuous function g : X_{→ Ω with g(x) = f (x) for every x in D.}

Exercise 7. Let G be an open subset of C and let P be a polygon in G from a to b. Use Theorems 5.15 and 5.17 to show that there is a polygon Q_{⊂ G from a to b which is composed of line segments which are} parallel to either the real or imaginary axes.

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Exercise 8. Use Lebesgue’s Covering Lemma (4.8) to give another proof of Theorem 5.15.

Solution. Suppose f : X _{→ Ω is continuous and X is compact. To show f is uniformly continuous. Let}

 > 0. Since f is continuous, we have for all x∈ X there is a δx> 0 such that ρ( f (x), f (y)) < /2 whenever

d(x, y) < δx. In addition,

X =[

x∈X

B(x; δx)

is an open cover of X. Since X is by assumption compact (it is also sequentially compact as stated in Theorem 4.9 p. 22), we can use Lebesgue’s Covering Lemma 4.8 p. 21 to obtain a δ > 0 such that x_{∈ X} implies that B(x, δ)_{⊂ B(z; δ}z) for some z∈ X. More precisely, x, y ∈ B(z; δz) and therefore

ρ( f (x), f (z))≤ ρ( f (x), f (z)) + ρ( f (z), f (y)) < ₂ +

2 =

and hence f is uniformly continuous on X.

Exercise 9. Prove the following converse to Exercise 2.5. Suppose (X, d) is a compact metric space having the property that for every > 0 and for any points a, b in X, there are points z0, z1, . . . , znin X with z0= a,

zn = b, and d(zk_−l, zk) < for 1≤ k ≤ n. Then (X, d) is connected. (Hint: Use Theorem 5.17.)

Exercise 10. Let f and g be continuous functions from (X, d) to (Ω, p) and let D be a dense subset of X. Prove that if f (x) = g(x) for x in D then f = g. Use this to show that the function g obtained in Exercise 6 is unique.

2.6 Uniform convergence

Exercise 1. Let_{{ f}n} be a sequence of uniformly continuous functions from (X, d) into (Ω, ρ) and suppose

that f = u_{− − lim f}n exists. Prove that f is uniformly continuous. If each fn is a Lipschitz function with

constant Mnand sup Mn <∞, show that f is a Lipschitz function. If sup Mn =∞, show that f may fail to

be Lipschitz.

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Chapter 3

Elementary Properties and Examples of

Analytic Functions

3.1 Power series

Exercise 2. Give the details of the proof of Proposition 1.6. Solution. Not available.

Exercise 3. Prove that lim sup(an+ bn)≤ lim sup an+ lim sup bnand lim inf(an+ bn)≥ lim inf an+ lim inf bn

for bounded sequences of real numbers_{an} and {bn}.

Solution. Let r > lim sup_n_→∞an(we know there are only finitely many by definition) and let s > lim supn→∞

(same here, there are only finitely many by definition). Then r + s > an+ bnfor all but finitely many n’s.

This however, implies that

r + s_{≥ lim sup}

n_→∞

(an+ bn).

Since this holds for any r > lim sup_n_→∞anand s > lim supn_→∞bn, we have

lim sup n→∞ (an+ bn)≤ lim sup n→∞ an+ lim sup n→∞ bn.

Let r < lim infn→∞an (we know there are only finitely many by definition) and let s < lim infn→∞(same

here, there are only finitely many by definition). Then r + s < an+ bnfor all but finitely many n’s. This

however, implies that

r + s_{≤ lim inf}

n→∞ (an+ bn).

Since this holds for any r < lim infn→∞anand s < lim infn→∞bn, we have

lim inf

n→∞ (an+ bn)≥ lim infn→∞ an+ lim infn→∞ bn.

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Solution. Let m = lim infn_→∞an and bn = inf{an, an+1, . . .}. Let M = lim supn→∞an. Take any s > M.

Then, by definition of the lim sup_n_→∞an = M, we obtain that an < s for infinitely many n’s which implies

that bn < s for all n and hence lim supn→∞bn = m < s. This holds for all s > M. But the infimum of all

these s’s is M. Therefore m_{≤ M which is}

lim inf

n→∞ an≤ lim supn→∞

an.

Exercise 5. If_{an} is a convergent sequence in R and a = lim an, show that a = lim inf an= lim sup an.

Solution. Suppose that_{an} is a convergent sequence in R with limit a = limn_→∞an. Then by definition, we

have:_{∀ > 0 ∃N > 0 such that ∀n ≥ N, we have |a}n− a| ≤ , that is a − ≤ an≤ a + . This means that all

but finitely many an’s are≤ a + and ≥ a − . This shows that

a_{− ≤ lim inf} n→∞ an | {z } =:m ≤ a +  and a_{− ≤ lim sup} n_→∞ an | {z } =:M ≤ a + .

By the previous Exercise 4, we also have

a_{− ≤ m ≤ M ≤ a + .}

Hence,

0_{≤ M − m ≤ 2.}

Since > 0 is arbitrary, we obtain m = M and further

a_{− ≤ lim inf} n→∞ an≤ lim supn→∞ an≤ a + , we obtain lim inf n_→∞ an= lim sup_n_→∞ an= a.

Exercise 6. Find the radius of convergence for each of the following power series: (a)P∞_n=0an_zn_{, a}_{∈ C;}

(b)P∞_n=0an2 zn_{, a}_{∈ C; (c)}P∞ n=0knzn, k an integer , 0; (d) P∞ n=0zn!. Solution. a) We haveP∞_n=0anzn= P∞

k=0bkznwith bk= ak, a∈ C. We also have,

lim sup k→∞ |b k|1/k = lim sup k→∞ |a k |1/k_{= lim sup} k→∞ |a| = |a|. Therefore, R = 1/_{|a|, so} R =        1 |a|, a , 0 ∞, a = 0. b) In this case, bn= an 2 where a_{∈ C.} R = _lim n_→∞ bn bn+1 = limn_→∞ an2 a(n+1)2 = lim n_→∞ an2 an2_+2n+1 = lim n_→∞ 1 a2n+1 = _lim n_→∞ 1 |a|2n+1 =              0, _{|a| > 1} 1, _{|a| = 1} ∞, |a| < 1 .

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c) Now, bn= kn, k is an integer , 0. We have R = lim sup n→∞ |bn| 1/n_{= lim sup} n→∞ |k n |1/n= lim sup n→∞ |k| = |k|. So R = 1 |k|=        1 k, k > 0, k integer −1 k, k < 0, k integer . d) We can writeP∞_n=0zn!_{= P}∞ k=0akzkwhere ak=                    0, k = 0 2, k = 1 1, k = n!, n_{∈ N, n > 1} 0, otherwise Thus, lim sup k_→∞ |a k|1/k = lim sup k_→∞ |1| 1/k! = 1.

Therefore 1/R = 1 which implies R = 1.

Exercise 7. Show that the radius of convergence of the power series

∞ X n=1 (₋₁₎n n z n(n+1)

is 1, and discuss convergence for z = 1,_{−1, and i. (Hint: The nth coefficient of this series is not (−1)}n/n.)

Solution. Rewrite the power series in standard form, then

∞ X n=1 (₋₁₎n n z n(n+1)₌ ∞ X n=1 akzkwith ak=        (−1)n n if∃ n ∈ N s.t. k = n(n + 1) 0 else .

To find the radius of convergence we use the root criterion and therefore need the estimates

1_≤ n(n+1)√

n_≤√n

n for n_{∈ N.}

The first inequality is immediate from the fact that n_{≥ 1 and hence n}n(n+1)1 ≥ 1. For the second inequality

note that n_{≤ n}n+1 ⇔ nn(n+1)1 ≤ n1n ⇔ n(n+1)√ n_≤√n n .

Using this one obtains

n(n+1) r |(−1) n n | = 1 n(n+1)√_n ≤ 1 and n(n+1) r |(−1) n n | = 1 n(n+1)√_n ≥ 1 n √ n.

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Vague memories of calculus classes tell me that√n_n

→ 1, thus 1

R = lim sup

n(n+1)√_a

n = 1, i.e. R = 1.

If z = 1 the series reduces toP∞_n=1 (−1)_nn which converges with the Leibniz Criterion.

If z =_{−1 we note that the exponents n(n + 1) are always even integers and therefore the series is the} same as in the previous case of z = 1.

Now let z = i. The expression in(n+1)will always be real, so if the series converges at z = i, it converges to a real number. We also note that formally

∞ X n=1 (₋₁₎n n i n(n+1) ₌ ∞ X n=1 cnwith        1 n ifn mod 4∈ {0, 1} −1 n if n mod 4∈ {2, 3} .

Define the partial sums Sk:=Pkn=0ck. We claim that the following chain of inequalities holds

0_{≤ S}4k+3 a) < S4k b) < S4k+4 c) < S4k+2 d) < S4k+1≤ 1.

To verify this, note that

S4k+3− S4k= c4k+1+ c4k+2+ c4k+3=− 16k2+ 8k_{− 1} (4k + 1)(4k + 2)(4k + 3) < 0, hence a) S4k+4− S4k= c4k+3+ c4k+4=− 1 4k + 3+ 1 4k + 4 < 0, hence c) S4k+2− S4k+1= c4k+2< 0, hence d).

Relation b) is obvious and so are the upper bound c1= 1 and the non-negativity constraint. We remark that

{S4k+l}k_≥1, l∈ {0, 1, 2, 3} describe bounded and monotone subsequences that converge to some point. Now

that_|cn| & 0 the difference between S4k+land S4k+m, l, m ∈ 0, 1, 2, 3 tends to zero, i.e. all subsequences

converge to the same limit. Therefore the power series converges also in the case of z = i.

3.2 Analytic functions

Exercise 1. Show that f (z) =_|z|2_{= x}2_{+ y}2_{has a derivative only at the origin.}

Solution. The derivative of f at z is given by

f0(z) = lim

h→0

f (z + h)_{− f (z)}

h , h∈ C provided the limit exist. We have

f (z + h)_{− f (z)} h = |z + h| 2_{− |z|}2 h = (z + h)(¯z + ¯h)− z¯z h = z¯z + h¯z + z¯h + h¯h− z¯z h = _{¯z + ¯h + z}¯h h =: D.

If the limit of D exists, it may be found by letting the point h = (x, y) approach the origin (0,0) in the complex plane C in any manner.

1.) Take the path along the real axes, that is y = 0. Then ¯h = h and thus

D = ¯z + h + zh

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and therefore, if the limit of D exists, its value has to be ¯z + z.

2.) Take the path along the imaginary axes, that is x = 0. Then ¯h =_{−h and thus} D = ¯z_{− h − z}h

h = ¯z− h − z

and therefore, if the limit of D exists, its value has to be ¯z_{− z.} Because of the uniqueness of the limit of D, we must have

¯z + z = ¯z_{− z ⇐⇒ z = −z ⇐⇒ z = 0,}

if the limit of D exists. It remains to show that the limit of D exists at z = 0. Since z = 0, we have that D = ¯h and thus the limit of D is 0.

In summary, the function f (z) =_|z|2_{= x}2_{+ y}2_{has a derivative only at the origin with value 0.}

Exercise 2. Prove that if bn, an are real and positive and 0 < b = lim bn < ∞, a = lim sup an then

ab = lim sup(anbn). Docs this remain true if the requirement of positivity is dropped?

Solution. Let a = lim sup

n→∞

an<∞. Then there exists a monotonic subsequence {ank} of {an} that converges

to a. Since lim

k→∞bnk = b, limn_→∞ankbnk = ab. Hence,{ankbnk} is a subsequence of anbnthat converges to ab. So

ab_{≤ lim sup}

n_→∞

anbn. Hence, lim sup n_→∞

anbn ≥ b lim sup n_→∞

an.

Now, let a = lim sup

n→∞

an =∞. Then there exists a subsequence {ank} of {an} such that lim_k

→∞ank = a > 0.

And, since lim

n_→∞bn > 0, limk→∞ankbnk

= _{∞. Hence lim sup}

n_→∞

anbn = ∞. In this second case, lim sup

n_→∞

anbn ≥

b lim sup

n→∞

an.

In both cases, we have established that ab_{≤ lim sup}

n_→∞

anbn. Now, since for all n∈ N, an> 0 and bn> 0,

consider lim

n→∞

1

bn

= 1

b. Applying the inequality we have established replacing bnwith

1 bn and anreplaced with anbn: lim sup n→∞ an= lim sup n→∞ 1 bn (anbn)≥ 1 blim supn→∞ anbn. Rearranging, lim sup n→∞ anbn≤ b lim sup n→∞ an.

It follows that ab = lim sup

n_→∞

anbnas required.

Now, consider the case where we drop the positivity requirement. Let bn = 0,−1₂, 0,−1₃, 0,−1₄, . . . and

note 0 = b = lim

n_→∞bn <∞. Also let an = 0,−2, 0, −3, 0, −4 . . . and note lim sup_n_→∞ an = a = 0. In this case,

ab = 0 , 1 = lim sup

n→∞

anbn.

Solutions Functions of One Complex Variable I, Second Edition by John B. Conway

Solutions Manual for

Functions of One Complex Variable I, Second Edition

©

Copyright by Andreas Kleefeld, 2009

All Rights Reserved

PREFACE

CONTRIBUTION

Contents

Chapter 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

Chapter 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

Chapter 3

Elementary Properties and Examples of

Analytic Functions

3.1

Power series

3.2

Analytic functions