Location of zeros

25.3 LOCATION OF ZEROS
y
s
w0
φ=0
π/α
w=z
z0
φ=0
α
x
Φ=0
(a)
(b)
Φ=0
r
w0∗
Figure 25.4 (a) An infinite conducting wedge with interior angle π/α and a
line charge at z = z0 ; (b) after the transformation w = z α , with an additional
image charge placed at w = w0∗ .
Substituting w = z α into the above shows that the required complex potential in the
original z-plane is
α
q
z − z0∗α
.
f(z) =
ln
α
2π0
z α − z0
It should be noted that the appearance of a complex conjugate in the final
expression is not in conflict with the general requirement that the complex
potential be analytic. It is z ∗ that must not appear; here, z0∗α is no more than a
parameter of the problem.
25.3 Location of zeros
The residue theorem, relating the value of a closed contour integral to the sum of
the residues at the poles enclosed by the contour, was discussed in the previous
chapter. One important practical use of an extension to the theorem is that of
locating the zeros of functions of a complex variable. The location of such zeros
has a particular application in electrical network and general oscillation theory,
since the complex zeros of certain functions (usually polynomials) give the system
parameters (usually frequencies) at which system instabilities occur. As the basis
of a method for locating these zeros we next prove three important theorems.
(i) If f(z) has poles as its only singularities inside a closed contour C and is
not zero at any point on C then
0 f (z)
dz = 2πi
(Nj − Pj ).
(25.14)
C f(z)
j
Here Nj is the order of the jth zero of f(z) enclosed by C. Similarly Pj is the
order of the jth pole of f(z) inside C.
To prove this we note that, at each position zj , f(z) can be written as
f(z) = (z − zj )mj φ(z),
879
(25.15)
APPLICATIONS OF COMPLEX VARIABLES
where φ(z) is analytic and non-zero at z = zj and mj is positive for a zero and
negative for a pole. Then the integrand f (z)/f(z) takes the form
mj
φ (z)
f (z)
=
.
+
f(z)
z − zj
φ(z)
(25.16)
Since φ(zj ) = 0, the second term on the RHS is analytic; thus the integrand has a
simple pole at z = zj , with residue mj . For zeros mj = Nj and for poles mj = −Pj ,
and thus (25.14) follows from the residue theorem.
(ii) If f(z) is analytic inside C and not zero at any point on it then
Nj = ∆C [arg f(z)],
(25.17)
2π
j
where ∆C [x] denotes the variation in x around the contour C.
Since f is analytic, there are no Pj ; further, since
d
f (z)
= [Ln f(z)],
f(z)
dz
equation (25.14) can be written
0 f (z)
dz = ∆C [Ln f(z)].
2πi
Nj =
C f(z)
(25.18)
(25.19)
However,
∆C [Ln f(z)] = ∆C [ln |f(z)|] + i∆C [arg f(z)],
(25.20)
and, since C is a closed contour, ln |f(z)| must return to its original value; so the
real term on the RHS is zero. Comparison of (25.19) and (25.20) then establishes
(25.17), which is known as the principle of the argument.
(iii) If f(z) and g(z) are analytic within and on a closed contour C and
|g(z)| < |f(z)| on C then f(z) and f(z) + g(z) have the same number of zeros
inside C; this is Rouché’s theorem.
With the conditions given, neither f(z) nor f(z) + g(z) can have a zero on C.
So, applying theorem (ii) with an obvious notation,
2π j Nj (f + g) = ∆C [arg(f + g)]
= ∆C [arg f] + ∆C [arg(1 + g/f)]
k Nk (f) + ∆C [arg(1 + g/f)].
= 2π
(25.21)
Further, since |g| < |f| on C, 1 + g/f always lies within a unit circle centred
on z = 1; thus its argument always lies in the range −π/2 < arg(1 + g/f) < π/2
and cannot change by any multiple of 2π. It must therefore return to its
original value when z returns to its starting point having traversed C. Hence
the second term on the RHS of (25.21) is zero and the theorem is established.
The importance of Rouché’s theorem is that for some functions, in particular
880
25.3 LOCATION OF ZEROS
polynomials, only the behaviour of a single term in the function need be considered if the contour is chosen appropriately. For example, for a polynomial,
i
f(z) + g(z) = N
0 bi z , only the properties of its largest power, taken as f(z), need
be investigated if a circular contour is chosen with radius R sufficiently large that,
on the contour, the magnitude of the largest power term, |bN R N |, is greater than
the sum of the magnitudes of all other terms. It is obvious that f(z) = bN z N has
N zeros inside |z| = R (all at the origin); consequently, f + g also has N zeros
inside the same circle.
The corresponding situation, in which only the properties of the polynomial’s
smallest power, again taken as f(z), need be investigated is a circular contour
with a radius R chosen sufficiently small that, 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, since f(z) = b0 has no zeros inside |z| = R, neither
does f + g.
A weak form of the maximum-modulus theorem may also be deduced. This
states that if f(z) is analytic within and on a simple closed contour C then |f(z)|
attains its maximum value on the boundary of C. The proof is as follows.
Let |f(z)| ≤ M on C with equality at at least one point of C. Now suppose
that there is a point z = a inside C such that |f(a)| > M. Then the function
h(z) ≡ f(a) is such that |h(z)| > | − f(z)| on C, and thus, by Rouché’s theorem,
h(z) and h(z) − f(z) have the same number of zeros inside C. But h(z) (≡ f(a))
has no zeros inside C, and, again by Rouché’s theorem, this would imply that
f(a) − f(z) has no zeros in C. However, f(a) − f(z) clearly has a zero at z = a,
and so we have a contradiction; the assumption of the existence of a point z = a
inside C such that |f(a)| > M must 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 of f(z) is not attained at any interior
point except for the case where f(z) is a constant.
Show that the four zeros of h(z) = z 4 + z + 1 occur one in each quadrant of the Argand
diagram and that all four lie between the circles |z| = 2/3 and |z| = 3/2.
Putting z = x and z = iy shows 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
of h(z) = 0.
Now take C as the contour OXY O shown in figure 25.5 and consider the changes
∆[arg h] in the argument of h(z) as z traverses C.
(i) OX: arg h is everywhere zero, since h is real, and thus ∆OX [arg h] = 0.
(ii) XY : z = R exp iθ and so arg h changes by an amount
∆XY [arg h] = ∆XY [arg z 4 ] + ∆XY [arg(1 + z −3 + z −4 )]
2
1
= ∆XY [arg R 4 e4iθ ] + ∆XY arg[1 + O(R −3 )]
= 2π + O(R −3 ).
881
(25.22)