Appendix I

(To Chapters III, IV, V)

The Proof that every Equation has a Root

LET

Z = P(z) = α0zn + α 1zn1 + + α n
be a polynomial in z, with real or complex coefficients. We can represent the values of z and Z by points in two planes, which we may call the z-plane and the Z-plane respectively. It is evident that if z describes a closed path γ in the z-plane, then Z describes a corresponding closed path Γ in the Z-plane. We shall assume for the present that the path Γ does not pass through the origin.

To any value of Z correspond an infinity of values of  am Z, differing by multiples of 2π, and each of these values varies continuously as Z describes Γ.112 We can select a particular value of  am Z corresponding to each point


pict
Fig. A.
pict
Fig. B.

of Γ, by first selecting a particular value corresponding to the initial value of Z, and then following the continuous variation of this value as Z moves along Γ. We shall, in the argument which follows, use the phrase ‘the amplitude of Z’ and the formula  am Z to denote the particular value of the amplitude of Z thus selected. Thus am Z denotes a one-valued and continuous function of X and Y , the real and imaginary parts of Z.

When Z, after describing Γ, returns to its original position, its amplitude may be the same as before, as will certainly be the case if Γ does not enclose the origin, like path (a) in Fig. B, or it may differ from its original value by any multiple of 2π. Thus if its path is like (b) in Fig. B, winding once round the origin in the positive direction, then its amplitude will have increased by 2π. These remarks apply, not merely to Γ, but to any closed contour in the Z-plane which does not pass through the origin. Associated with any such contour there is a number which we may call ‘the increment of  am Z when Z describes the contour’, a number independent of the initial choice of a particular value of the amplitude of Z.

We shall now prove that if the amplitude of Z is not the same when Z returns to its original position, then the path of z must contain inside or on it at least one point at which Z = 0.

We can divide γ into a number of smaller contours by drawing parallels to the axes at a distance δ1 from one another, as in Fig. C.113 If there is, on the boundary of any one of these contours, a point at which Z = 0, what we wish to prove is already established. We may therefore suppose


pict
Fig. C.
pict
Fig. D.

that this is not the case. Then the increment of  am Z, when z describes γ, is equal to the sum of all the increments of  am Z obtained by supposing z to describe each of these smaller contours separately in the same sense as γ. For if z describes each of the smaller contours in turn, in the same sense, it will ultimately (see Fig. D) have described the boundary of γ once, and each part of each of the dividing parallels twice and in opposite directions. Thus PQ will have been described twice, once from P to Q and once from Q to P. As z moves from P to Q, am Z varies continuously, since Z does not pass through the origin; and if the increment of  am Z is in this case θ, then its increment when z moves from Q to P is  θ; so that, when we add up the increments of  am Z due to the description of the various parts of the smaller contours, all cancel one another, save the increments due to the description of parts of γ itself.

Hence, if am Z is changed when z describes γ, there must be at least one of the smaller contours, say γ1, such that am Z is changed when z describes γ1. This contour may be a square whose sides are parts of the auxiliary parallels, or may be composed of parts of these parallels and parts of the boundary of γ. In any case every point of the contour lies in or on the boundary of a square Δ1 whose sides are parts of the auxiliary parallels and of length δ1.

We can now further subdivide γ1 by the help of parallels to the axes at a smaller distance δ2 from one another, and we can find a contour γ2, entirely included in a square Δ2, of side δ2 and itself included in Δ1 such that am Z is changed when z describes the contour.

Now let us take an infinite sequence of decreasing numbers δ1, δ2, …, δm, …, whose limit is zero.114 By repeating the argument used above, we can determine a series of squares Δ1, Δ2, …, Δm, … and a series of contours γ1, γ2, …, γm, … such that (i) Δm+1 lies entirely inside Δm, (ii) γm lies entirely inside Δm, (iii)  am Z is changed when z describes γm.

If (xm,ym) and (xm + δm,ym + δm) are the lower left-hand and upper right-hand corners of Δm, it is clear that x1, x2, …, xm, … is an increasing and x1 + δ1, x2 + δ2, …, xm + δm, … a decreasing sequence, and that they have a common limit x0. Similarly ym and ym + δm have a common limit y0, and (x0,y0) is the one and only point situated inside every square Δm. However small δ may be, we can draw a square which includes (x0,y0), and whose sides are parallel to the axes and of length δ, and inside this square a closed contour such that am Z is changed when z describes the contour.

It can now be shown that

P(x0 + iy0) = 0.
For suppose that P(x0 + iy0) = a, where a = ρ > 0. Since P(x + iy) is a continuous function of x and y, we can draw a square whose centre is (x0,y0) and whose sides are parallel to the axes, and which is such that
P(x + iy) P(x0 + iy0) < 1 2ρ
at all points x + iy inside the square or on its boundary. At all such points
P(x + iy) = a + φ,
where φ < 1 2ρ. Now let us take any closed contour lying entirely inside this square. As z describes this contour, Z = a + φ also describes a closed contour. But the latter contour evidently lies inside the circle whose centre is a and whose radius is 1 2ρ, and this circle does not include the origin. Hence the amplitude of Z is unchanged.

But this contradicts what was proved above, viz. that inside each square Δm we can find a closed contour the description of which by z changes  am Z. Hence P(x0 + iy0) = 0.

All that remains is to show that we can always find some contour such that am Z is changed when z describes γ. Now

Z = a0zn 1 + a1 a0z + a2 a0z2 + + an a0zn .
We can choose R so that
a1 a0R + a2 a0R2 + + an a0Rn < δ,
where δ is any positive number, however small; and then, if γ is the circle whose centre is the origin and whose radius is R, we have
Z = a0zn(1 + ρ),
where ρ < δ, at all points on γ. We can then show, by an argument similar to that used above, that am(1 + ρ) is unchanged as z describes γ in the positive sense, while am zn on the other hand is increased by 2nπ. Hence am Z is increased by 2nπ, and the proof that Z = 0 has a root is completed.

We have assumed throughout the argument that neither Γ, nor any of the smaller contours into which it is resolved, passes through the origin. This assumption is obviously legitimate, for to suppose the contrary, at any stage of the argument, is to admit the truth of the theorem.

We leave it as an exercise to the reader to infer, from the discussion which precedes and that of § 43, that when z describes any contour γ in the positive sense the increment of  am Z is 2kπ, where k is the number of roots of Z = 0 inside γ, multiple roots being counted multiply.

There is another proof, proceeding on different lines, which is often given. It depends, however, on an extension to functions of two or more variables of the results of §§ 102 et seq.

We define, precisely on the lines of § 102, the upper and lower bounds of a function f(x,y), for all pairs of values of x and y corresponding to any point of any region in the plane of (x,y) bounded by a closed curve. And we can prove, much as in § 102, that a continuous function f(x,y) attains its upper and lower bounds in any such region.

Now

Z = P(x + iy)
is a positive and continuous function of x and y. If m is its lower bound for points on and inside γ, then there must be a point z0 for which Z = m, and this must be the least value assumed by Z. If m = 0, then P(z0) = 0, and we have proved what we want. We may therefore suppose that m > 0.

The point z0 must lie either inside or on the boundary of γ: but if γ is a circle whose centre is the origin, and whose radius R is large enough, then the last hypothesis is untenable, since P(z) as z. We may therefore suppose that z0 lies inside γ.

If we put z = z0 + ζ, and rearrange P(z) according to powers of ζ, we obtain

P(z) = P(z0) + A1ζ + A2ζ2 + + A nζn,
say. Let Ak be the first of the coefficients which does not vanish, and let Ak = μ, ζ = ρ. We can choose ρ so small that
Ak+1ρ + Ak+2ρ2 + + A nρnk < 1 2μ.
Then
P(z) P(z0) Akζk < 1 2μρk,
and
P(z) < P(z0 + Akζk + 1 2μρk.

Now suppose that z moves round the circle whose centre is z0 and radius ρ. Then

P(z0) + Akζk
moves k times round the circle whose centre is P(z0) and radius Akζk = μρk, and passes k times through the point in which this circle is intersected by the line joining P(z0) to the origin. Hence there are k points on the circle described by z at which P(z0) + Akζk = P(z 0) μρk and so
P(z) < P(z0) μρk + 1 2μρk = m 1 2μρk < m;
and this contradicts the hypothesis that m is the lower bound of P(z).

It follows that m must be zero and that P(z0) = 0.

Examples on Appendix I

1. Show that the number of roots of f(z) = 0 which lie within a closed contour which does not pass through any root is equal to the increment of

{logf(z)}/2πi
when z describes the contour.

2. Show that if R is any number such that

a1 R + a2 R2 + + an Rn < 1,
then all the roots of zn + a1zn1 + + an = 0 are in absolute value less than R. In particular show that all the roots of z5 13z 7 = 0 are in absolute value less than 2 1 67.

3. Determine the numbers of the roots of the equation z2p + az + b = 0 where a and b are real and p odd, which have their real parts positive and negative. Show that if a > 0, b > 0 then the numbers are p 1 and p + 1; if a < 0, b > 0 they are p + 1 and p 1; and if b < 0 they are p and p. Discuss the particular cases in which a = 0 or b = 0. Verify the results when p = 1.

[Trace the variation of am(z2p + az + b) as z describes the contour formed by a large semicircle whose centre is the origin and whose radius is R, and the part of the imaginary axis intercepted by the semicircle.]

4. Consider similarly the equations

z4q + az + b = 0,z4q1 + az + b = 0,z4q+1 + az + b = 0.

5. Show that if α and β are real then the numbers of the roots of the equation z2n + α2z2n1 + β2 = 0 which have their real parts positive and negative are n 1 and n + 1, or n and n, according as n is odd or even.

(Math. Trip. 1891.)

6. Show that when z moves along the straight line joining the points z = z1, z = z2, from a point near z1 to a point near z2, the increment of

am 1 z z1 + 1 z z2
is nearly equal to π.

7. A contour enclosing the three points z = z1, z = z2, z = z3 is defined by parts of the sides of the triangle formed by z1z2z3, and the parts exterior to the triangle of three small circles with their centres at those points. Show that when z describes the contour the increment of

am 1 z z1 + 1 z z2 + 1 z z3
is equal to  2π.

8. Prove that a closed oval path which surrounds all the roots of a cubic equation f(z) = 0 also surrounds those of the derived equation f(z) = 0. [Use the equation

f(z) = f(z) 1 z z1 + 1 z z2 + 1 z z3 ,
where z1z2z3 are the roots of f(z) = 0, and the result of Ex. 7.]

9. Show that the roots of f(z) = 0 are the foci of the ellipse which touches the sides of the triangle (z1,z2,z3) at their middle points. [For a proof see Cesàro’s Elementares Lehrbuch der algebraischen Analysis, p. 352.]

10. Extend the result of Ex. 8 to equations of any degree.

11. If f(z) and φ(z) are two polynomials in z, and γ is a contour which does not pass through any root of f(z), and φ(z)< f(z)at all points on γ, then the numbers of the roots of the equations

f(z) = 0,f(z) + φ(z) = 0
which lie inside γ are the same.

12. Show that the equations

ez = az,ez = az2,ez = az3,
where a > e, have respectively (i) one positive root (ii) one positive and one negative root and (iii) one positive and two complex roots within the circle z= 1.
(Math. Trip. 1910.)