Complex Analysis - Maths KU

6.6 Some Consequences of the Residue Theorem 365
To illustrate Theorem 6.20, suppose the simple closed contour is |z| =2
and the function f is the one given in (27).The result in (28) indicates that
in the evaluation of �
C f ′ (z) dz/f(z), each zero of f within C contributes 2πi
times the order of multiplicity of the zero and each pole contributes 2πi times
the negative of the order of the pole:
�
C
f ′ contribution
of zeros of f
(z) � �� �
dz = [2πi(1)+2πi(2)] + [
f(z)
contribution
of poles of f
� �� �
2πi(−2)+2πi(−2)+2πi(−6)] = −14πi.
Why the Name? Why is Theorem 6.20 called the argument principle?
This question may have occurred to you since no reference is made in the
proof of the theorem to any arguments of complex quantities.But in point
of fact there is a relation between the number N0 − Np in Theorem 6.20 and
arg (f(z)).More precisely,
N0 − Np = 1
2π
[change in arg (f(z)) as z traverses C once in the positive direction].
This principle can be easily verified using the simple function f(z) =z 2 and
the unit circle |z| = 1 as the simple closed contour C in the z-plane.Because
the function f has a zero of multiplicity 2 within C and no poles, we have
N0−Np = 2.Now, if C is parametrized by z = e iθ , 0 ≤ θ ≤ 2π, then its image
C ′ in the w-plane under the mapping w = z 2 is w = e i2θ , 0 ≤ θ ≤ 2π, which
is the unit circle |w| =1.Asz traverses C once starting at z =1(θ = 0) and
finishing at z =1(θ =2π), we see arg (f(z)) = arg (w) =2θ increases from 0
to 4π.Put another way, w traverses or winds around the circle |w| = 1 twice.
Thus,
1
1
[change in arg(f(z)) as z traverses C once in the positive direction] = [4π − 0] = 2.
2π 2π
Rouché’s Theorem The next result follows as a consequence of the
argument principle.The theorem is helpful in determining the number of
zeros of an analytic function.
Theorem 6.21 Rouché’s Theorem
Let C be a simple closed contour lying entirely within a domain D.
Suppose f and g are analytic in D.If the strict inequality
|f(z) − g(z)| < |f(z)| holds for all z on C, then f and g have the
same number of zeros (counted according to their order or multiplicities)
inside C.
Proof We start with the observation that the hypothesis “the inequality
|f(z) − g(z)| < |f(z)| holds for all z on C” indicates that both f and g have