Page 116 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 11: Rouche’s Theorem

But notice that ’(t) = f’((t))’(t). Hence, Notes

f'(z)  f'( (t))   '(t)

 dz    '(t)dt   dt

C f(z)  f( (t))   (t)
where |n| is the number of times  ”winds around” the origin. The integer n is positive in case
 is traversed in the positive direction, and negative in case the traversal is in the negative
direction.
f'(z)
Next, we shall use the Residue Theorem to evaluate the integral  dz. The singularities of
C f(z)
f'(z)
the integrand are the poles of f together with the zeros of f. Let’s find the residues at these
f(z)
points. First, let Z = {z , z , ..., z } be set of all zeros of f. Suppose the order of the zero z is n. Then
1
2
j
j
K
f(z) = (z – z) j h(z) and h(z)  0. Thus,
n
j
j
m
f'(z) (z p ) h'(z) m (z p ) m j 1 h(z) (z p )m j




j
j
j
j
j
f(z) = (z pj) 2m j . h(z)

h'(z) m
=  j .
h(z) (z p ) m j

j
Now then,
(z) = (z p ) m f'(z)  (z p ) m h'(z)  m ,

j
j

j
h(z)
j
f(z)
j
and so
Res f'   (p )   m .
z j p f j j
The sum of all these residues is
–P = –m – m – ... – m j
2
1
Then,
f'(z)
 dz  2 i(N P);


C f(z)
and we already found that
f'(z)
 f(z) dz  n2 i,

C
where n is the ”winding number”, or the number of times  winds around the origin—n > 0
means  winds in the positive sense, and n negative means it winds in the negative sense.
Finally, we have
n = N – P,
where N = n + n + ... + n is the number of zeros inside C, counting multiplicity, or the order of
K
1
2
the zeros, and P = m + m + ... + m is the number of poles, counting the order. This result is the
1
J
2
celebrated argument principle.
LOVELY PROFESSIONAL UNIVERSITY 109