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Applications of the Cauchy Theory Chapter 4 Applications Of The Cauchy Theory This chapter contains several applications of the material developed in Chapter 3. In the first section, we will describe the possible behavior of an analytic function near a singularity of that function. 4.1 Singularities We will say that f has an isolated singularity at z0 if f is analytic on D(z0,r) \{z0} for some r. What, if anything, can be said about the behavior of f near z0? The basic tool needed to answer this question is the Laurent series, an expansion of f(z)in powers of z − z0 in which negative as well as positive powers of z − z0 may appear. In fact, the number of negative powers in this expansion is the key to determining how f behaves near z0. From now on, the punctured disk D(z0,r) \{z0} will be denoted by D (z0,r). We will need a consequence of Cauchy’s integral formula. 4.1.1 Theorem Let f be analytic on an open set Ω containing the annulus {z : r1 ≤|z − z0|≤r2}, 0 <r1 <r2 < ∞, and let γ1 and γ2 denote the positively oriented inner and outer boundaries of the annulus. Then for r1 < |z − z0| <r2, we have 1 f(w) − 1 f(w) f(z)= − dw − dw. 2πi γ2 w z 2πi γ1 w z Proof. Apply Cauchy’s integral formula [part (ii)of (3.3.1)]to the cycle γ2 − γ1. ♣ 1 2 CHAPTER 4. APPLICATIONS OF THE CAUCHY THEORY 4.1.2 Definition For 0 ≤ s1 <s2 ≤ +∞ and z0 ∈ C, we will denote the open annulus {z : s1 < |z−z0| <s2} by A(z0,s1,s2). 4.1.3 Laurent Series Representation { }∞ If f is analytic on Ω = A(z0,s1,s2), then there is a unique two-tailed sequence an n=−∞ such that ∞ n f(z)= an(z − z0) ,z ∈ Ω. n=−∞ In fact, if r is such that s1 <r<s2, then the coefficients an are given by 1 f(w) a = dw, n =0, ±1, ±2,... n − n+1 2πi C(z0,r) (w z0) Also, the above series converges absolutely on Ω and uniformly on compact subsets of Ω. Proof. Choose r1 and r2 such that s1 <r1 <r2 <s2 and consider the Cauchy type integral 1 f(w) ∈ − dw, z D(z0,r2). 2πi C(z0,r2) w z Then proceeding just as we did in the proof of Theorem 2.2.16, we obtain ∞ 1 f(w) dw = a (z − z )n 2πi w − z n 0 C(z0,r2) n=0 where 1 f(w) a = dw. n − n+1 2πi C(z0,r2) (w z0) The series converges absolutely on D(z0,r2), and uniformly on compact subsets of D(z0,r). Next, consider the Cauchy type integral − 1 f(w) | − | − dw, z z0 >r1. 2πi C(z0,r1) w z This can be written as ∞ n−1 1 f(w) 1 (w − z0) − dw = f(w) dw. w z0 − n 2πi (z − z0)[1 − ] 2πi (z z0) C(z0,r1) z−z0 C(z0,r1) n=1 By the Weierstrass M-test, the series converges absolutely and uniformly for w ∈ C(z0,r1). Consequently, we may integrate term by term to obtain the series ∞ 1 f(w) b (z − z )−n, where b = dw. n 0 n 2πi (w − z )−n+1 n=1 C(z0,r1) 0 4.1. SINGULARITIES 3 This is a power series in 1/(z −z0), and it converges for |z −z0| >r1, and hence uniformly on sets of the form {z : |z − z0|≥1/ρ} where (1/ρ) >r1. It follows that the convergence is uniform on compact (indeed on closed)subsets of {z : z − z0| >r1. The existence part of the theorem now follows from (4.1.1)and the above computa- tions, if we note two facts. First, if s1 <r<s2 and k =0, ±1, ±2,..., f(w) f(w) f(w) dw = dw = dw. − k+1 − k+1 − k+1 C(z0,r) (w z0) C(z0,r1) (w z0) C(z0,r2) (w z0) Second, any compact subset of A(z0,s1,s2)is contained in {z : ρ1 ≤|z − z0|≤ρ2} for some ρ1 and ρ2 with s1 <ρ1 <ρ2 <s2. { } We turn now to the question of uniqueness. Let bn be a sequence such that f(z)= ∞ − n ∈ n=−∞ bn(z z0) for z A(z0,s1,s2). As in the above argument, this series must converge uniformly on compact subsets of A(z0,s1,s2). Therefore if k is any integer and s1 <r<s2, then ∞ 1 f(w) 1 dw = b (w − z )n−k−1 dw 2πi (w − z )k+1 2πi n 0 C(z0,r) 0 C(z0,r) n=−∞ ∞ 1 = b (w − z )n−k−1 dw n 2πi 0 n=−∞ C(z0,r) = bk, because − − − 1 n−k−1 1ifn k 1= 1 (w − z0) dw = 2πi C(z0,r) 0 otherwise. The theorem is completely proved. ♣ We are now in a position to analyze the behavior of f near an isolated singularity. As the preceding discussion shows, if f has an isolated singularity at z0, then f can be represented uniquely by ∞ n f(z)= an(z − z0) n=−∞ in some deleted neighborhood of z0. 4.1.4 Definition ∞ − n Suppose f has an isolated singularity at z0, and let n=−∞ an(z z0) be the Laurent expansion of f about z0, that is, the series given in (4.1.3). We say that f has a removable singularity at z0 if an = 0 for all n<0; f has a pole of order m at z0 if m is the largest positive integer such that a−m = 0. (A pole of order 1 is called a simple pole.)Finally, if an = 0 for infinitely many n<0, we say that f has an essential singularity at z0. The next theorem relates the behavior of f(z)for z near z0 to the type of singularity that f has at z0. 4 CHAPTER 4. APPLICATIONS OF THE CAUCHY THEORY 4.1.5 Theorem Suppose that f has an isolated singularity at z0. Then (a) f has a removable singularity at z0 iff f(z)approaches a finite limit as z → z0 iff f(z) is bounded on the punctured disk D (z0,δ)for some δ>0. m (b)For a given positive integer m, f has a pole of order m at z0 iff (z−z0) f(z)approaches a finite nonzero limit as z → z0. Also, f has a pole at z0 iff |f(z)|→+∞ as z → z0. (c) f has an essential singularity at z0 iff f(z)does not approach a finite or infinite limit as z → z , that is, f(z)has no limit in Cˆ as z → z . 0 0 { }∞ ∞ − n | − | Proof. Let an n=−∞ and r>0 be such that f(z)= −∞ an(z z0) for 0 < z z0 <r. (a)If an = 0 for all n<0, then limz→z0 f(z)=a0. Conversely, if limz→z0 f(z)exists (in C ), then f can be defined (or redefined)at z0 so that f is analytic on D(z0,r). It follows { }∞ ∞ − n ∈ that there is a sequence bn n=0 such that f(z)= n=0 bn(z z0) for z D (z0,r).By uniqueness of the Laurent expansion, we conclude that an = 0 for n<0 and an = bn for n ≥ 0. (Thus in this case, the Laurent and Taylor expansions coincide.)The remaining equivalence stated in (a)is left as an exercise (Problem 1). (b)If f has a pole of order m at z0, then for 0 < |z − z0| <r, ∞ −m −1 n f(z)=a−m(z − z0) + ···+ a−1(z − z0) + an(z − z0) n=0 m where a−m = 0. Consequently, (z − z0) f(z) → a−m =0as z → z0. Conversely, if − m − m limz→z0 (z z0) f(z) = 0, then by (a)applied to ( z z0) f(z), there is a sequence { }∞ bn n=0 such that ∞ m n (z − z0) f(z)= bn(z − z0) ,z ∈ D (z0,r). n=0 → − m Let z z0 to obtain b0 = limz→z0 (z z0) f(z) =0.Thusf(z)can be written as −m −m+1 b0(z − z0) + b1(z − z0) + ···, showing that f has a pole of order m at z0. The remaining equivalence in (b)is also left as an exercise (Problem 1). (c)If f(z)does not have a limit in Cˆ as z → z0, then by (a)and (b), f must have an essential singularity at z0. Conversely, if f has an essential singularity at z0, then (a)and Cˆ ♣ (b)again imply that lim z→z0 f(z)cannot exist in . The behavior of a function near an essential singularity is much more pathological even than (4.1.5c)suggests, as the next theorem shows. 4.1.6 Casorati-Weierstrass Theorem Let f have an isolated essential singularity at z0. Then for any complex number w, f(z)comes arbitrarily close to w in every deleted neighborhood of z0. That is, for any δ>0,f(D (z0,δ)) is a dense subset of C. Proof. Suppose that for some δ>0,f(D (z0,δ)) is not dense in C. Then for some w ∈ C, there exists >0 such that D(w, )does not meet f(D (z0,δ)). For z ∈ D (z0,δ), put 4.1. SINGULARITIES 5 g(z)=1/(f(z)−w)).
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