the fact that
γ
is compact and does not pass through
z
k
, we know
∑
∞
l
=1
a
k

l
(
z

z
k
)

l
converges uniformly on
γ
, thus, using the fact that
H
γ
(
z

z
k
)

k
= 0 for
k
6
= 1,
I
γ
S
k
(
z
)
dz
=
∞
X
l
=1
I
γ
a
k

l
(
z

z
k
)

l
dz
=
I
γ
a
k

1
(
z

z
k
)

1
dz
= 2
πiRes
f
(
z
k
)
·
n
(
γ,z
k
)
Therefore, in sum, combine the above results, we conclude that
I
γ
f
(
z
)
dz
=
I
γ
(
F
(
z
) +
G
(
z
))
dz
=
I
γ
G
(
z
)
dz
=
n
X
k
=1
I
γ
S
k
dz
= 2
πi
n
X
k
=1
Res
f
(
z
k
)
·
n
(
k
)
.
This proved our desired Residue Formula.
2
Let
g
=
f
(
ζ
)
/
(
ζ

z
0
). Observe that
Res
g
(
z
0
) =
f
(
z
0
) by the Laurent expansion
of
g
=
f
(
ζ
)
/
(
ζ

z
0
) at the point
z
0
. Apply the above residue theorem to
g
to
yield that, for any (piecewise smooth) closed simple curve,
γ
with
z
0
∈
int
(
γ
),
since
n
(
0
) = 1,
f
(
z
0
) =
1
2
πi
I
γ
f
(
ζ
)
(
ζ

z
0
)
dζ.
Thus the above Cauchy’s integral formula is proved.
2
To prove the argument principle, we ﬁrst suppose
f
is a meromorphic function
on a open set
U
,
¯
D
(
P,r
)
⊆
U
, and
f
has no poles or zeros on
∂D
(
), then
we shall show that
1
2
πi
I
∂D
(
P,r
)
f
0
(
z
)
f
(
z
)
dz
=
p
X
j
=1
n
j

q
X
k
=1
m
k
where
n
1
,
···
,n
p
are the multiplicities of zeros
z
1
,
,z
p
of
f
in
D
(
), and
m
1
,
,m
p
are the poles
w
1
,
,w
q
of
f
in
D
(
).
To see this, for the function
h
(
z
) =
f
0
(
z
)
/f
(
z
), we know that the possible
singularities (actually poles) for it are
z
1
,
p
and
w
1
,
q
. Next, for any
2