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 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
2.5 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 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 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 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
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)