The Laurent series is like a power series except it allows for negative exponents.
First here is a definition of what is meant by the convergence of such a series.
Definition 52.2.4∑n=−∞∞an
(z − a)
nconverges if both the series,
∑∞ n ∑∞ −n
an(z − a) and a−n (z − a)
n=0 n=1
converge. When this is the case, the symbol,∑n=−∞∞an
1 [∫ f (w) ∫ f (w) ]
f (z) = 2πi w-− z-dw+ w−-z-dw
⌊ −γ1 γ2 ⌋
1 ∫ f (w ) ∫ f (w )
= 2πi⌈ ------[---w−-a]dw + ------[----z−a]dw ⌉
γ1 (z − a) 1− z−a γ2 (w − a) 1 − w−a
( )
-1-∫ f-(w) ∞∑ -z −-a n
= 2πi γ w − a w − a dw +
2 n=0
∫ ∞ ( )n
-1- f-(w)-∑ w-−-a dw. (52.2.7)
2πi γ1 (z − a)n=0 z − a
(52.2.7)
From the formula 52.2.7, it follows that for z ∈ann
(a,r1,r2)
, the terms in the first
sum are bounded by an expression of the form C
(-r2-)
r2+ε
n while those in the
second are bounded by one of the form C
( )
r1−r1ε-
n and so by the Weierstrass M
test, the convergence is uniform and so the integrals and the sums in the above
formula may be interchanged and after renaming the variable of summation, this
yields
∞ ( ∫ )
f (z) = ∑ -1- --f-(w-)--dw (z − a)n +
n=0 2πi γ2 (w − a)n+1
( )
−∑ 1 -1-∫ ---f (w)-- n
2πi γ (w − a)n+1 (z − a) . (52.2.8)
n=−∞ 1