53.2. SINGULARITIES AND THE LAURENT SERIES 1683

Lemma 53.2.3 Let g be analytic on ann(a,R1,R2) . Then if γr (t)≡ a+ reit for t ∈ [0,2π]and r ∈ (R1,R2) , then

∫γr

g(z)dz is independent of r.

Proof: Let R1 < r1 < r2 < R2 and denote by −γr (t) the curve, −γr (t)≡ a+ rei(2π−t)

for t ∈ [0,2π] . Then if z∈ B(a,R1), Lemma 53.2.2 implies both n(γr2

,z)

and n(γr1

,z)= 1

and son(−γr1

,z)+n(γr2

,z)=−1+1 = 0.

Also if z /∈ B(a,R2) , then Lemma 53.2.2 implies n(

γr j,z)= 0 for j = 1,2. Therefore,

whenever z /∈ ann(a,R1,R2) , the sum of the winding numbers equals zero. Therefore,by Theorem 51.7.19 applied to the function, f (w) = g(z)(w− z) and z ∈ ann(a,R1,R2)\∪2

j=1γr j([0,2π]) ,

f (z)(n(γr2

,z)+n(−γr1

,z))

= 0(n(γr2

,z)+n(−γr1

,z))

=

12πi

∫γr2

g(w)(w− z)w− z

dw− 12πi

∫γr1

g(w)(w− z)w− z

dw

=1

2πi

∫γr2

g(w)dw− 12πi

∫γr1

g(w)dw

which proves the desired result.

53.2.2 The Laurent SeriesThe Laurent series is like a power series except it allows for negative exponents. First hereis a definition of what is meant by the convergence of such a series.

Definition 53.2.4 ∑∞n=−∞ an (z−a)n converges if both the series,

∞

∑n=0

an (z−a)n and∞

∑n=1

a−n (z−a)−n

converge. When this is the case, the symbol, ∑∞n=−∞ an (z−a)n is defined as

∞

∑n=0

an (z−a)n +∞

∑n=1

a−n (z−a)−n .

Lemma 53.2.5 Suppose

f (z) =∞

∑n=−∞

an (z−a)n

for all |z−a| ∈ (R1,R2) . Then both ∑∞n=0 an (z−a)n and ∑

∞n=1 a−n (z−a)−n converge abso-

lutely and uniformly on {z : r1 ≤ |z−a| ≤ r2} for any r1 < r2 satisfying R1 < r1 < r2 < R2.