In other words, every polynomial has a root in C.

If the reader believes that this is obvious, then she should stop reading

at this point and write down the ˜obvious argument™. In fact, Leibniz and

other mathematicians doubted the truth of the result. Although d™Alembert,

10

Repeated in Martin Gardner™s Mathematical Diversions. See also Exercise K.89.

114 A COMPANION TO ANALYSIS

Euler and Lagrange o¬ered proofs of the result, they were unsatisfactory and

the ¬rst satisfactory discussion is due to Gauss11 .

The ¬rst point to realise is that the ˜fundamental theorem of algebra™ is

in fact a theorem of analysis!

Exercise 5.8.2. Suppose z = u + iv with u, v ∈ R. If z 2 ’ 2 = 0, show that

u2 ’ v 2 = 2

uv = 0

and deduce that v = 0, u2 = 2.

If we write

Q + iQ = {x + iy : x, y ∈ Q},

show that the equation

z2 ’ 2 = 0

has no solution with z ∈ Q + iQ.

Since Q + iQ and C = R + iR share the same algebraic structure, Exer-

cise 5.8.2 shows that the truth of Theorem 5.8.1 must depend in some way of

the fundamental axiom of analysis. We shall use Theorem 4.3.4, which states

that any continuous function on a closed bounded set in Rn has a minimum,

to establish the following key step of our proof.

Lemma 5.8.3. If P is a polynomial, then there exists a z0 ∈ C such that

|P (z)| ≥ |P (z0 )|

for all z ∈ C.

We then complete the proof by establishing the following lemma.

Lemma 5.8.4. If P is a non-constant polynomial and |P | attains a mini-

mum at z0 , then P (z0 ) = 0.

Clearly, Lemmas 5.8.3 and 5.8.4 together imply Theorem 5.8.1. Our

proofs of the two lemmas make use of simple results given in the next exercise.

11

See [29], Chapter 19, section 4 and Chapter 25 sections 1 and 2.

115

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise 5.8.5. (i) Let P (z) = n aj z j with n ≥ 1 and an = 0. Show

j=0

’1

that, if we set R0 = 2n|an | (1 + max0¤j¤n |aj |), then, whenever |z| ≥ R0 ,

|aj | |aj | |an |

¤ ¤

|z|n’j R0 2n

for all 0 ¤ j ¤ n ’ 1. Hence, or otherwise, show that

n’1

|an |

aj

≥

an +

z n’j 2

j=0

and so

n

|an ||z|n

j

≥

aj z

2

j=0

for all |z| ≥ R0 .

(ii) By using the result of (i), show that, given any real number K ≥ 0,

we can ¬nd an R(K) > 0 such that |P (z)| ≥ K whenever |z| ≥ R(K).

(iii) Let Q(z) = n bj z j with n ≥ k ≥ 1 and bk = 0. Show that there

j=k

exists a ·0 > 0 such that

n

|bk ||z|k

j

¤

bj z

2

j=k+1

for all |z| ¤ ·0 .

Proof of Lemma 5.8.3. We wish to show that, if P is any polynomial, then

|P | has a minimum. If P is a constant polynomial there is nothing to prove,

n j

j=0 aj z with n ≥ 1 and an = 0. By

so we may suppose that P (z) =

Exercise 5.8.5 (ii), we can ¬nd an R > 0 such that |P (z)| ≥ |P (0)| + 1

whenever |z| ≥ R.

Identifying C with R2 in the usual way, we observe that

¯

DR = {z ∈ C : |z| ¤ R}

is a closed bounded set and that the function |P | : C ’ R is continuous.

Thus we may use Theorem 4.3.4 which states that a continuous function on

¯

a closed bounded set attains its minimum to show the existence of a z0 ∈ DR

¯

with |P (z0 )| ¤ |P (z)| for all z ∈ DR .

We note, in particular, that |P (z0 )| ¤ |P (0)|. Thus, if |z| ≥ R, then

|P (z)| ≥ |P (0)| + 1 > |P (0)| ≥ |P (z0 )|.

It follows that |P (z0 )| ¤ |P (z)| for all z ∈ C as required.

116 A COMPANION TO ANALYSIS

Exercise 5.8.6. De¬ne f : C ’ R by f (z) = ’|z|2 . Show that f attains a

minimum on every set

¯

DR = {z ∈ C : |z| ¤ R}

but has no minimum on C. Explain brie¬‚y why the proof above works for |P |

but not for f .

We must now show that, if z0 gives the minimum value of the modulus |P |

of a non-constant polynomial P , then P (z0 ) = 0. We start with a collection

of remarks intended to simplify the algebra.

Exercise 5.8.7. (i) Let P be a non-constant polynomial whose modulus |P |

has a minimum at z0 . Show that if Q(z) = P (z + z0 ), then Q is a non-

constant polynomial whose modulus |Q| has a minimum at 0. Show further

that, if Q(0) = 0, then P (z0 ) = 0.

(ii) Let Q be a non-constant polynomial whose modulus |Q| has a mini-

mum at 0. Show that, for an appropriate φ ∈ R, to be de¬ned, the function

R(z) = eiφ Q(z) has R(0) real and positive12 . Show that R is a non-constant

polynomial whose modulus |R| has a minimum at 0 and that, if R(0) = 0,

then Q(0) = 0.

(iii) Let R be a non-constant polynomial whose modulus |R| has a mini-

mum at 0 and such that R(0) is real and positive. Explain why we have

n

aj z j

R(z) = a0 +

j=k

where a0 is real and positive, k ≥ 1 and ak = 0. Set S(z) = R(eiψ z). Show

that, for an appropriate ψ ∈ R, to be de¬ned,

n

bj z j

S(z) = b0 +

j=k

where b0 is real and positive, k ≥ 1 and bk is real and strictly negative (that

is bk < 0).

Most mathematicians would consider the results of Exercise 5.8.7 to be

trivial and use a phrase like ˜Without loss of generality we may suppose that

z0 = 0 and P (z) = a0 + n aj z j where a0 is real and positive, k ≥ 1 and

j=k

ak is real and strictly negative™ or (better) ˜By considering eiφ P (eiψ (z ’ z0 ))

we may suppose that z0 = 0 and P (z) = a0 + n aj z j where a0 is real and

j=k

positive, k ≥ 1 and ak is real and strictly negative™.

12

That is to say, non-negative.

117

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Proof of Lemma 5.8.4. We want to show that if P is a non-constant poly-

nomial and z0 gives a minimum of |P |, then P (z0 ) = 0. Without loss of

generality we may suppose that z0 = 0 and P (z) = a0 + n aj z j where a0

j=k

is real and positive, k ≥ 1 and ak is real and strictly negative. If a0 = 0 then

P (0) = 0 and we are done. We suppose that a0 is strictly positive and seek

a contradiction.

By Exercise 5.8.5, we can ¬nd an ·0 > 0 such that

n

|ak z k |

j

¤

aj z

2

j=k+1

for all |z| ¤ ·0 . Now choose ·1 , a real number with 0 < ·1 ¤ ·0 and

k

a0 > |ak |·1 /2 (·1 = min(·0 , 1, ’a0 /(2ak )) will do). Remembering that a0 is

real and strictly positive and ak is real and strictly negative, we see that,

whenever · is real and 0 < · < ·1 , we have

n n

j k

aj · j

|P (·)| = a0 + ¤ |a0 + ak · | +

aj ·

j=k j=k+1

¤ |a0 + ak · k | + |ak · k |/2 = a0 + ak · k ’ ak · k /2 = a0 + ak · k /2 < P (0),

contradicting the statement that 0 is a minimum for P . The result follows

by reductio ad absurdum.

The proof of Theorem 5.8.1 may look a little complicated but really it

only amounts to a ¬‚eshing out of the following sketch argument.

Outline proof of Theorem 5.8.1. Let P be a non constant polynomial. Since

|P (z)| ’ ∞ as |z| ’ ∞, P must attain a minimum. By translation, we may

suppose that the minimum occurs at 0. If P (0) = 0, then

n

aj z j

P (z) = a0 +

j=k

with k ≥ 1 and a0 , ak = 0. Close to zero,

P (z) ≈ a0 + ak z k .

Choosing an appropriate φ, we have |a0 + ak (eiφ ·)k | < |a0 | whenever · is

small and strictly positive, contradicting the statement that |P | attains a

minimum, at 0. The result follows by reductio ad absurdum.

Exercise 5.8.8. Give an explicit value for φ in the outline proof just sketched.

118 A COMPANION TO ANALYSIS

Exercise 5.8.9. We say that z0 is a local minimum of a function G : C ’ R

if we can ¬nd a δ > 0 such that G(z) ≥ G(z0 ) for all z with |z ’ z0 | < δ.

Show that if P is a non-constant polynomial and z0 is a local minimum of

|P |, then P (z0 ) = 0.

We have already used the strategy of looking for a minimum (or maxi-

mum) and then considering the behaviour of the function near that ˜extreme™

point in our proof of Rolle™s theorem (Theorem 4.4.4). Another example oc-

curs in Exercise K.30 if the reader wishes to try it and other examples will

crop up in this book. The method is very powerful but we must be careful to

establish that an extreme point actually exists (see, as a warning example,

the discussion beginning on page 199 of a counterexample due, essentially,

to Weierstrass). Notice that our proof required the ability to ˜look in all

directions™. The minimum had to be in the open set

DR = {z ∈ C : |z| < R}

and not merely in the set

¯

DR = {z ∈ C : |z| ¤ R}.

Exercise 5.8.10. This exercise recalls material that is probably familiar from

algebra. We work in C.

(i) Show, by induction on the degree of P , or otherwise, that if P is a

non-constant polynomial and » ∈ C, then there exists a polynomial Q and

an r ∈ C such that

P (z) = (z ’ »)Q(z) + r.

(ii) If P is a non-constant polynomial and » ∈ C is such that P (») = 0,

then there is a polynomial Q such that

P (z) = (z ’ »)Q(z).

(iii) Use the fundamental theorem of algebra and induction on the degree

of n to show that any polynomial P of degree n can be written in the form

n

(z ’ »j ).

P (z) = a

j=1

(iv) Show that a polynomial of degree n can have at most n distinct roots.

What is the minimum number of distinct roots it can have?

119

Please send corrections however trivial to twk@dpmms.cam.ac.uk

(v) If P has real coe¬cients show13 that P (z)— = P (z — ) and deduce that,

if » is a root of P , so is »— .

(vi) Use part (v) and induction to show that, if P is a polynomial with

real coe¬cients, then P can be written in the form

m

P (z) = a Qj (z)

j=1

where a ∈ R and, for each j, either Qj (z) = z + aj with aj ∈ R, or Qj =

z 2 + aj z + bj with aj , bj ∈ R.

In the days before mathematicians acquired our present con¬dence with

complex numbers, the fundamental theorem of algebra was given the less gen-

eral statement that any polynomial with real coe¬cients could be written as

the product of linear and quadratic terms with real coe¬cients.

It is natural to ask if this restricted result which does not mention complex

numbers can be proved without using complex numbers. Gauss™s ¬rst proof of

the restricted result used complex numbers but he later gave a second proof

without using complex numbers which depends only on the fact that a real

polynomial of odd degree must have a root (Exercise 1.6.4) and so uses the

fundamental axiom in the form of the intermediate value theorem. As might

be expected, his proof and its modern sucessors are rather subtle. The reader

is advised to wait until she has studied the rudiments of Galois theory before

pursuing these ideas further.

Exercise 5.8.11. Let P (z) = n aj z j be a non-constant polynomial with

j=0

a root at z0 .

(i) Explain why we can ¬nd an ·0 > 0 such that P (z) = 0 for all z with

0 < |z ’ z0 | < ·0 .

(ii) If 0 < · < ·0 , use the fact that a continuous function on a closed

bounded set is bounded and attains its bounds to show that there is a δ(·) > 0

such that |P (z)| ≥ δ(·) > 0 for all z with |z ’ z0 | = ·.

(iii) Continuing with the notations and assumptions of (ii), show that if

Q(z) is a polynomial with |P (z) ’ Q(z)| < δ(·)/2 for all z with |z ’ z 0 | ¤ ·,

then |Q| has a local minimum (and so Q has a root) z1 with |z1 ’ z0 | < ·.

(iv) Show that given any δ > 0, we can ¬nd an > 0 (depending on δ, n,

a0 , a1 , . . . , an ) such that, if |aj ’ bj | < , for 0 ¤ j ¤ n then n bj z j has

j=0

at least one root z1 with |z0 ’ z1 | < δ.

[Note that this result is not true if we work over R. The equation x2 = 0

has a real root at 0 but x2 + = 0 has no real roots if > 0 however small

may be.]

We write z — for the complex conjugate of z. Thus, if x and y are real (x+iy)— = x’iy.

13

Some authors use z .

¯

120 A COMPANION TO ANALYSIS

Exercise 5.8.12. (This exercise requires countability and a certain willing-

ness to think like an algebraist.)

It is sometimes said that we have to introduce R in order to provide

equations like x2 ’ 2 = 0 with a root. A little thought shows that this is too

simple a view of the matter. Recall that a system (F, +, —) satisfying all the

axioms set out in Axioms A except axioms P1 to P4 (the axioms of order) is

called a ¬eld. If (F, +, —) is a ¬eld and and G ⊆ F is such that

(a) 0, 1, ’1 ∈ G,

(b) if x, y ∈ G, then x + y, xy ∈ G,

(c) if x ∈ G and x = 0, then x’1 ∈ G,

then we say that G is a sub¬eld of F. It easy to see that a sub¬eld is itself

a ¬eld. In this exercise we show that there is a countable sub¬eld L of C

containing Q and such that, if a0 , a1 , . . . , an ∈ L, with an = 0, then we can

¬nd a, »1 , . . . , »n ∈ L such that

n n

j

(z ’ »j )

aj z = a

j=0 k=1

for all z ∈ L. In other words, every polynomial with coe¬cients in L has all

its roots in L. Here are the steps in the proof.

(i) If K is a countable sub¬eld of C, show that the set of polynomials

with degree n with coe¬cients in K is countable. Deduce that the set of

polynomials P(K) with coe¬cients in K is countable. Show also that the set

Z(K) of roots in C of polynomials in P(K) is countable.

(ii) If K is a sub¬eld of C and ω ∈ C, we write K(ω) for the set of

numbers P (ω)/Q(ω) with P , Q ∈ P(K) and Q(ω) = 0. Show that K(ω) is a

sub¬eld of C containing K and ω. If K is countable, show that K(ω) is.

(iii) Let K be a sub¬eld of C and ω = (ω1 , ω2 , . . . ) where ωj ∈ C. Set

K0 = K and de¬ne Kn = Kn’1 (ωn ) for all n ≥ 1. If we set K(ω) = ∞ Kn ,n=0

show that K(ω) is a sub¬eld of C containing K and ωj for each j ≥ 1. If K

is countable, show that K(ω) is.

(iv) Let K be a countable sub¬eld of C (we could take K = Q). Set

K0 = K. Show by induction, using part (iii), that we may de¬ne inductively

a sequence Kn of countable sub¬elds of C such that Kn contains Z(Kn’1 ) for

each n ≥ 1. If we set L = ∞ Kn , show that L is a countable sub¬eld of C

n=0

such that every polynomial with coe¬cients in L has all its roots in L.

[We say that ¬elds like L are ˜algebraically closed™. The work we have

had to do to obtain an ˜algebraically closed™ L from K shows the fundamental

theorem of algebra in a remarkable light. Although R is not algebraically

closed, adjoining a single root i of a single equation z 2 + 1 = 0 to form

R(i) = C produces an algebraically closed ¬eld!]

Chapter 6

Di¬erentiation

6.1 Preliminaries

This section is as much propaganda as technical mathematics and, as with

much propaganda, most points are made more than once.

We have already looked brie¬‚y at di¬erentiation of functions f : R ’ R.

Unfortunately, nature is not one dimensional and we must consider the more

general case of a function f : Rm ’ Rp . The de¬nition of the derivative in

terms of the limit of some ratio is not available since we cannot divide by

vectors.

The ¬rst solution that mathematicians found to this problem is via ˜di-

rectional derivatives™ or, essentially equivalently, via ˜partial derivatives™. We

shall give formal de¬nitions later but the idea is to reduce a many dimen-

sional problem to a collection of one dimensional problems by only examining

changes in one direction at at time. Suppose, for example, that f : Rm ’ R

is well behaved. If we wish to examine how f behaves near x we choose a unit

vector u and look at fu (t) = f (x + tu) with t ∈ R. The function fu : R ’ R

is ˜one dimensional™ and we may look at its derivative

f (x + hu) ’ f (x)

fu (x) = lim .

h

h’0

By choosing m unit vectors uj at right angles and looking at the associated

˜directional derivatives™ fuj (x) we can obtain a picture of the way in which

f changes.

But to echo Maxwell

. . . the doctrine of Vectors . . . is a method of thinking and not,

at least for the present generation, a method of saving thought.

121

122 A COMPANION TO ANALYSIS

It does not, like some more popular mathematical methods, en-

courage the hope that mathematicians may give their minds a

holiday, by transferring all their work to their pens. It calls on us

at every step to form a mental image of the geometrical features

represented by the symbols, so that in studying geometry by this

method we have our minds engaged with geometrical ideas, and

are not permitted to call ourselves geometers when we are only

arithmeticians. (Page 951, [38])

Is there a less ˜coordinate bound™ and more ˜geometric™ way of looking at

di¬erentiation in many dimensions? If we are prepared to spend a little time

and e¬ort acquiring new habits of thought, the answer is yes.

The original discoverers of the calculus thought of di¬erentiation as the

process of ¬nding a tangent. If f : R ’ R is well behaved then the tangent

at x is the line y = b + a(t ’ x) which touches the curve y = f (t) at (x, f (x))

that is the ˜line which most resembles f close to x™. In other words