Math 411: Honours Complex Variables - University of Alberta

12.1. APPLICATIONS OF THE RESIDUE THEOREM TO REAL INTEGRALS77
so that
f(z) = 1
iz ·
=
=
� a+ 1
2
1
� z + 1
z
−4iz
(z 2 +2az +1) 2
−4iz
(z −z1) 2 (z −z2) 2,
where z1 and z2 are again given by Eq. 12.1.
At z1, the function f has a pole of order two. In order to calculate res(f,z1),
set
g(z) := (z −z1) 2 f(z) = −4iz
(z −z2) 2,
so that
it follows that
�� 2
g ′ �
1 2z
(z) = −4i −
(z −z2) 2 (z −z2) 3
�
= −4i
(z −z2) 3[(z
−z2)−2z]
= 4i(z +z2)
;
(z −z2) 3
res(f,z1) = g ′ (z1)
From Proposition 12.1, we conclude that
Problem 12.1.
� 2π
0
= −4i(−2a)
8 �√ a2 −1 �3 −ai
= �√ �3 .
a2 −1
dt
(a+cost) 2 = 2πi res(f,z1) =
2πa
�√ a 2 −1 � 3 .
Let D ⊂ C be open, let f: D → C be holomorphic, and let z0 ∈ D be a zero of order
one of f. Show that �
1
res
f ;z0
�
= 1
f ′ (z0) .