f z
ð Þ ¼
1
ffiffiffiffiffiffiffiffi
r 1 r 2
p
e
Ài θ 1 þθ 2
ð
Þ =2
:
ð6:258Þ
Let C a and C b be a small circle of radius ε centered at z ¼ a and z ¼ b,
respectively. When we are evaluating the integral on C b , we can put
z À b ¼ εe
iθ 2 Àπ θ 2 π
ð
Þ ; i:e:, r 2 ¼ ε:
We also have r 1 % b À a; in Fig. 6.30 we assume b > a. Hence, we have
Z
C b
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
z À a
ð
Þ z À b
ð
Þ
p
dz
¼
Z π
Àπ
iε
ffiffiffiffiffiffi ffi
r 1 ε
p e
Ài θ 1 þθ 2
ð
Þ =2 dθ 2
Z π
Àπ
ffiffi ffi
ε
p
ffiffiffiffi
r 1
p dθ 2 :
Therefore, we get
lim
ε!0
Z
C b
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
z À a
ð
Þ z À b
ð
Þ
p
dz
¼ 0 or lim
ε!0
Z
C b
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
z À a
ð
Þ z À b
ð
Þ
p
dz ¼ 0:
In a similar manner, we have
lim
ε!0
Z
C a
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
z À a
ð
Þ z À b
ð
Þ
p
dz ¼ 0:
Meanwhile, on the line Q
0 P
0 we have
z À a ¼ r 1 e
iθ 1 and z À b ¼ r 2 e
iθ 2 :
ð6:259Þ
In (6.259) we have r 1 ¼ x À a, θ 1 ¼ 0 and r 2 ¼ b À x, θ 2 ¼ π; see Fig. 6.30. Also,
we have dz ¼ dx. Hence, we have
Fig. 6.30 Branch cut (shown with a doubled broken line) and contour for the integration of
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
zÀa
ð
Þ zÀb
ð
Þ
p
a < b
ð
Þ. Only the argument θ 1 is depicted. With θ 2 see text
260
6 Theory of Analytic Functions