Mathematical Methods for Physics and Engineering : A Comprehensive Guide

25.3 LOCATION OF ZEROS

polynomials, only the behaviour of a single term in the function need be con-

sidered if the contour is chosen appropriately. For example, for a polynomial,

f(z)+g(z)=

∑N

0 biz

i, only the properties of its largest power, taken asf(z), need

be investigated if a circular contour is chosen with radiusRsufficiently large that,

on the contour, the magnitude of the largest power term,|bNRN|, is greater than

the sum of the magnitudes of all other terms. It is obvious thatf(z)=bNzNhas

Nzeros inside|z|=R(all at the origin); consequently,f+galso hasNzeros

inside the same circle.

The corresponding situation, in which only the properties of the polynomial’s

smallest power, again taken asf(z), need be investigated is a circular contour

with a radiusRchosen sufficientlysmallthat, on the contour, the magnitude of

the smallest power term (usually the constant term in a polynomial) is greater

than the sum of the magnitudes of all other terms. Then, a similar argument to

that given above shows that, sincef(z)=b 0 has no zeros inside|z|=R,neither

doesf+g.

Aweakformofthemaximum-modulus theoremmay also be deduced. This

states that iff(z) is analytic within and on a simple closed contourCthen|f(z)|

attains its maximum value on the boundary ofC. The proof is as follows.

Let|f(z)|≤MonCwith equality at at least one point ofC. Now suppose

that there is a pointz=ainsideCsuch that|f(a)|>M. Then the function

h(z)≡f(a) is such that|h(z)|>|−f(z)|onC, and thus, by Rouche’s theorem, ́

h(z)andh(z)−f(z) have the same number of zeros insideC.Buth(z)(≡f(a))

has no zeros insideC, and, again by Rouch ́e’s theorem, this would imply that

f(a)−f(z) has no zeros inC. However,f(a)−f(z) clearly has a zero atz=a,

and so we have a contradiction; the assumption of the existence of a pointz=a

insideCsuch that|f(a)|>Mmust be invalid. This establishes the theorem.

The stronger form of the maximum-modulus theorem, which we do not prove,

states, in addition, that the maximum value off(z) is not attained at any interior

point except for the case wheref(z) is a constant.

Show that the four zeros ofh(z)=z^4 +z+1occur one in each quadrant of the Argand

diagram and that all four lie between the circles|z|=2/ 3 and|z|=3/ 2.

Puttingz=xandz=iyshows that no zeros occur on the real or imaginary axes. They
must therefore occur in conjugate pairs, as can be shown by taking the complex conjugate
ofh(z)=0.
Now takeCas the contourOXY Oshown in figure 25.5 and consider the changes
∆[argh] in the argument ofh(z)asztraversesC.

(i)OX:arghis everywhere zero, sincehis real, and thus ∆OX[argh]=0.

(ii)XY:z=Rexpiθandsoarghchanges by an amount

∆XY[argh]=∆XY[argz^4 ]+∆XY[arg(1 +z−^3 +z−^4 )]

=∆XY[argR^4 e^4 iθ]+∆XY

{

arg[1+O(R−^3 )]

}

=2π+O(R−^3 ). (25.22)