Complex Analysis Summer 2001

88 SINGULARITIESLemma 9.3 Let g be analytic on ann (a, R 1 , R 2 ) . Then if γ r (t) ≡ a + re it for t ∈ [0, 2π] and r ∈ (R 1 , R 2 ) ,then ∫ γ rg (z) dz is independent of r.Proof: Let R 1 < r 1 < r 2 < R 2 and denote by −γ r (t) the curve, −γ r (t) ≡ a + re i(2π−t) for t ∈ [0, 2π] .Then if z ∈ B (a, R 1 ), Lemma 9.2 implies both n (γ r2 , z) and n (γ r1 , z) = 1 and son (−γ r1 , z) + n (γ r2 , z) = −1 + 1 = 0.Also if z /∈ B (a, R 2 ) , then Lemma 9.2 implies n ( γ rj , z ) = 0 for j = 1, 2. Therefore, whenever z /∈ann (a, R 1 , R 2 ) , the sum of the winding numbers equals zero. Therefore, by Theorem 7.8 applied to thefunction, f (z) = g (z) (w − z) and z ∈ ann (a, R 1 , R 2 ) \ ∪ 2 j=1 γ r j([0, 2π]) ,∫12πiwhich proves the desired result.9.2 The Laurent Series0 (n (γ r2 , z) + n (−γ r1 , z)) =g (w) (w − z)γ r2w − zdw − 1 ∫2πig (w) (w − z)γ r1w − z= 1 ∫g (w) dw − 1 ∫g (w) dw2πi γ r22πi γ r1The Laurent series is like a power series except it allows for negative exponents. First here is a definition ofwhat is meant by the convergence of such a series.Definition 9.4 ∑ ∞n=−∞ a n (z − a) n converges if both the series, ∑ ∞n=0 a n (z − a) n and ∑ ∞n=1 a −n (z − a) −nconverge. When this is the case, the symbol, ∑ ∞n=−∞ a n (z − a) n is defined as∞∑∞∑a n (z − a) n + a −n (z − a) −n .n=0n=1Lemma 9.5 Suppose f (z) = ∑ ∞n=−∞ a n (z − a) n for all |z − a| ∈ (R 1 , R 2 ) . Then both ∑ ∞n=0 a n (z − a) nand ∑ ∞n=1 a −n (z − a) −n converge absolutely and uniformly on {z : r 1 ≤ |z − a| ≤ r 2 } for any r 1 < r 2 satisfyingR 1 < r 1 < r 2 < R 2 .Proof: Let R 1 < |w − a| = r 1 − δ < r 1 . Then ∑ ∞n=1 a −n (w − a) −n converges and sowhich implies that for all n sufficiently large,Therefore,n=1lim |a −n| |w − a| −n = lim |a −n| (r 1 − δ) −n = 0n→∞ n→∞|a −n | (r 1 − δ) −n < 1.∞∑∞∑|a −n | |z − a| −n = |a −n | (r 1 − δ) −n (r 1 − δ) n |z − a| −n .n=1dw