Complex Analysis - Maths KU

y
–(n + ��) + ni (n + ��) + ni
–n
–1
–(n + ��) – ni (n + ��) – ni
Figure 6.17 Rectangular contour C
enclosing poles of (37)
1
n
6.6 Some Consequences of the Residue Theorem 367
since � �
3 9
2 ≈ 38.44. We conclude from (34) that because f has a zero of order
, all nine zeros of g lie within the same disk.
9 within the disk |z| < 3
2
By slightly subtler reasoning, we can demonstrate that the function g in
Example 7 has some zeros inside the unit disk |z| < 1.To see this suppose
we choose f(z) =−8z 2 + 5.Then for all z on |z| =1,
|f(z) − g(z)| = � �(−8z 2 +5)− (z 9 − 8z 2 +5) � � = � �−z 9� � = |z| 9 = (1) 9 =1. (35)
But from (10) of Section 1.2 we have, for all z on |z| =1,
|f(z)| = |−f(z)| = � �8z 2 − 5 � �
� �
≥ �8|z| 2 �
�
−|−5| � = |8 − 5| =3. (36)
The values in (35) and (36) show, for all z on |z| = 1, that
|f(z)
�
− g(z)| < |f(z)|.Because f has two zeros within |z| < 1 (namely,
5 ± 8 ≈±0.79), we can conclude from Theorem 6.21 that two zeros of g also
lie within this disk.
We can continue the reasoning of the previous paragraph.Suppose now
we choose f(z) = 5 and |z| = 1
1
2 .Then for all z on |z| = 2 ,
|f(z) − g(z)| = � � �
� 9 2
5 − (z − 8z +5) � = �−z9 2
+8z � � 9 2 � �
1 9
≤|z| +8|z| = 2 +2≈ 2.002.
C
x
We now have |f(z) − g(z)| < |f(z)| = 5 for all z on |z| = 1
2 .Since f has
no zeros within the disk |z| < 1,
neither does g.At this point we are able
2
to conclude that all nine zeros of g(z) =z9 − 8z2 + 5 lie within the annular
region 1
3
1
2 < |z| < 2 ; two of these zeros lie within 2 < |z| < 1.
6.6.5 Summing Infinite Series
Using cot πz In some specialized circumstances, the residues at the
simple poles of the trigonometric function cot πz enable us to find the sum of
an infinite series.
In Section 4.3 we saw that the zeros of sin z were the real numbers z = kπ,
k =0,±1, ±2, ... .Thus the function cot πz has simple poles at the zeros
of sin πz, which are πz = kπ or z = k, k =0,±1, ±2, ... .If a polynomial
function p(z) has (i) real coefficients, (ii) degree n ≥ 2, and (iii) nointeger
zeros, then the function
f(z) =
π cot πz
p(z)
(37)
has an infinite number of simple poles z = 0, ±1, ±2, ... from cot πz
and a finite number of poles zp1 , zp2 , ... , zprfrom the zeros of p(z).The
closed rectangular contour C shown in Figure 6.17 has vertices � n + 1
�
� 2 + ni,
− ni, where n is taken large
− � n + 1
2
� + ni, − � n + 1
2
� − ni, and � n + 1
2