2.5 Isolated Singularities 149
1
1_a
z
weobtain
1 1 1 i 2]"
="' = k+1'
Z—ZJ Z k=oZ
z
In an entirely similiar manner, for j = m + 1, ., N,
1 —1 1 °° z"
——' = _ k+1‘
Z—Zj Zj 1‘:on
Z.
J
This gives (12) and (13) after summing on j.
Example 12 Fi
2.5 Isolated Singularities 145
Z3 + ZS
. sin 2 3! 5'
f(z) = 3 = * Z3
1 1 22
‘Z—z—§+§—"', Z¢0.
The principal part is 1/22, and the residue at zero is 0. El
Example 9 Find the ﬁrst ﬁve terms of the Laurent series about 20 = 0 for the
function g(z) = cot 2
2.5 Isolated Singularities 143
maywrite
1 1 _ 1 1
w—z_(w—zo)—(z—zo)_w—z0 1—2—20
W—Zo
w (z—Zo)k
=2 k+1
Z — 20 < 1
W — 20
In an entirely similar way, we obtain
1 = _ i _ 20)]
C' Z i=0 (Z ‘ 20y“
for each C e y; here again, the series is absolut
2.5 Isolated Singularities 141
24—7i
500 '
Res(r; — 21') =
The pole at 1 is of order 3, so by (7), the residue of r at 1 is the value at 1 of one-half
of the second derivative of (z + 1)/(z2 + 4). This works out to be
— 12
R ; 1 = —.
es(r ) 125
2.5 Isolated Singularities 139
That is, to ﬁnd the residue of a function f which has a pole at the point zo, write the
expansion of f in positive and negative powers of z —— 20 in the form (6). This may
always be done according to the presentation under t
2.4 Consequences of Cauchy's Formula 129
Morera's* Theorem
Cauchy’s Theorem, on which so much of the development of complex variables
depends, has a converse. This theorem, given precisely below, states that if the
integral of a continuous function f over
2.5
2.5 Isolated Singularities 135
28. Use the result of Exercise 27 to give the solution to each of these differential
equations:
(a) f ’(Z) - 22f(2) = 0
(b) f '(z) + e7(z) = 0
= 0
1
(c) f’(z) + Ere)
Bessel Functions*
29. Let v be an integer, v 2 0. Show
2.4 Consequences of Cauchy's Formula 133
Again using (12) of Section 2.2, we obtain f (“(0): (ak)k! and g‘"“"(0)= (bn_k)(n— k)!.
When we substitute these into the formula above for h‘"’(0) and cancel the factorials,
we get
h("’(0)= n! 2 akbrk,
k=0
which i
2.4 Consequences of Cauchy's Formula 131
1 2M 21rR
Ing <2——
—~>0 asR—+oo.
n R R—ICI
Consequently, g = 0. Thus, F (2) = F (0) for all z. I
Application 2 Analytic Logarithms
Suppose f is analytic and zero-free in a simply-connected domain D. Fix 20 e D
an
1) MASER Microwave Amplica3on by S3mulated Emission of Radia3on
2) A) The laser color is green.
power
beamarea
2Js 1
8Js 1
I=
=
= 2.55x108 Wm 2
(10 4 m)2 3.14x10 8 m 2
4
N
hc 2 N
N
I
B) I=h
c=
= 2
V
V
V
hc
The intensity is I =
N
2.55 108
1
dU
L2
Ze 2
F =
= ( 3 +
)=0
2
dr
r 4 0 r
L2
Ze 2
n22
=
=
3
2
3
r 4 0 rn r n
n2
Which gives the energy with rn = a0
Z
Solve for rn
n 2 2 4 0 n 2 0 h 2
n2
rn =
= ( 2 ) = a0
2
Ze
Z e
Z
Ze 2
e2
Z2
En =
=
( 2)
8 0 rn
8 0 a0 n
Solve for
A z I z ),
I H : (i .9! .39. a1 9; L
zwaxm, as! 337; W-fjd. 9 Pei a?
2 (H axb 31 1 a;
a) a; a; 5 3
an? r 3?; f 92:
2. E
i T , (1H) (:21) (H 2)
M
" (H I)
# 1,3 z 2 2 g?
t ' / + + :
H gap/ML U I I > W
Em / EL; r Em ; Ll (21+ )1+ [1) r
S/VV'ZLL $4451
j/f mf
2.5 Isolated Singularities 147
with
—1, k=—1,—2,—3,.
We return to our discussion of the principal part of a function with poles.
Suppose that f is analytic on some domain D except for a pole of order m at a point
20 of D. Then f may be written as
f(Z)