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CHAPTER 7 Classification of Singularities Dr. Pulak Sahoo

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CHAPTER 7 Classification of Singularities Dr. Pulak Sahoo CHAPTER 7 Classification of Singularities BY Dr. Pulak Sahoo Assistant Professor Department of Mathematics University Of Kalyani West Bengal, India E-mail : sahoopulak1@gmail:com 1 Module-2: Zeros of an Analytic Function Definition 1. If z0 is a regular point of an analytic function f and if f(z0) = 0, then z0 is called a zero of f. The point z0 is called a zero of f of order m if in some neighbourhood of z0, f can be expanded in a Taylor series of the form 1 X n f(z) = an(z − z0) ; where am 6= 0: n=m Theorem 1. A point z = z0 is a zero of f of order m if and only if in some neighbourhood m of z0, f can be expressed in the form f(z) = (z − z0) φ(z), where φ(z) is analytic at z0 and φ(z0) 6= 0. Proof. First we assume that z0 is a zero of f of order m. Then in some neighbourhood of z0 we can expand f as 1 X n f(z) = an(z − z0) ; where am 6= 0: n=m Then 1 m X n−m f(z) = (z − z0) an(z − z0) n=m 1 m X p = (z − z0) bp(z − z0) ; where n − m = p and bp = ap+m p=0 m = (z − z0) φ(z); 1 X p where φ(z) = bp(z − z0) is analytic at z0 and φ(z0) = b0 = am 6= 0: p=0 Next we assume that in some neighbourhood of z0, f can be expressed as f(z) = m (z − z0) φ(z), where φ(z) is analytic at z0 and φ(z0) 6= 0. Then we can expand φ(z) in 2 Taylor series about z0 to obtain 1 X n φ(z) = an(z − z0) ; where a0 = φ(z0) 6= 0: n=0 Therefore, in some neighbourhood of z0, we have 1 1 m X n X n+m f(z) = (z − z0) an(z − z0) = an(z − z0) n=0 n=0 1 X p = bp(z − z0) ; where m + n = p and bp = ap−m: p=m Since bm = a0 6= 0; z0 is a zero of f of order m. This proves the theorem. Theorem 2. The zeros of a nonconstant analytic function are isolated points. Proof. Let z0 be a zero of f of order m. Then in some neighbourhood of z0, f can be m expressed in the form f(z) = (z − z0) φ(z), where φ(z) is analytic at z0 and φ(z0) 6= 0. 1 Let " = 2 j φ(z0) j> 0. Since φ(z) is continuous at z0, there exists δ > 0 such that 1 j φ(z) − φ(z ) j < " = j φ(z ) j whenever j z − z j < δ: 0 2 0 0 Therefore, 1 j φ(z ) j − j φ(z) j ≤| φ(z) − φ(z ) j < j φ(z ) j whenever j z − z j < δ: 0 0 2 0 0 1 From this we obtain 2 j φ(z0) j <j φ(z) j whenever j z − z0 j < δ: This means that φ(z) 6= 0 in j z −z0 j < δ: Hence f cannot vanish in 0 <j z −z0 j < δ: Thus z0 is an isolated zero of f. z0 being arbitrary, we thus conclude that the zeros of an analytic function are isolated points. This proves the theorem. Theorem 3. The number of zeros of an analytic function (not identically zero) interior to a simple closed contour are finite. Proof. If possible, we may assume that f(z) has infinite number of zeros inside a region en- closed by a closed contour. The region being bounded and closed, by Bolzano-Weierstrass theorem any sequence of zeros fa1; a2;::: g in the region has a limit point in the region. Let a be the limit point of the sequence, i.e. there exists a subsequence fank g such that ank ! a as k ! 1: This implies f(a) = lim f(an ) = 0: k!1 k 3 Therefore, a is a zero of the function, and in every neighbourhood of a, however small it may be, there are infinitely many zeros. This means that a is not isolated which contradicts the fact that the zeros of an analytic function are isolated points. Hence, the number of zeros of an analytic function interior to a simple closed contour are finite. This proves the theorem. Note 1. If f(z) and g(z) are analytic at z0 and have zeros of order m and n, respectively, then the product function h(z) = f(z)g(z) has a zero of order m + n at z0. Example 1. Find order of zero of the function f(z) = z4 sin z at z = 0. Solution. At z = 0, the function z4 has a zero of order 4 and the function sin z has zero of order 1. Therefore, the function f(z) = z4 sin z has a zero of order 4 + 1 = 5 at z = 0: Theorem 4. (Weierstrass and Casorati Theorem) If f(z) has an isolated essential singularity at z = z0, then f(z) comes arbitrarily close to every complex value in each deleted neighbourhood of z0 . Proof. Since z = z0 is an isolated essential singularity of f(z), it is analytic throughout a deleted neighbourhood of z0. Suppose, for some complex number c, that j f(z) − c j ≥ " > 0 1 for all z in a punctured disk 0 <j z − z0 j< δ: Let g(z) = f(z)−c : Then 1 1 j g(z) j =j j ≤ for 0 <j z − z j< δ: f(z) − c " 0 Hence by Riemann's theorem, g(z) has a removable singularity at z0 and we may write 1 g(z) = = a + a (z − z ) + a (z − z )2 + :::: f(z) − c 0 1 0 2 0 Observe that limz!z0 1=(f(z) − c) = a0: Now we consider two cases. Case (i). a0 6= 0: Then limz!z0 f(z) = c + 1=a0; and f(z) has a removable singularity at z0; a contradiction. Case (ii). a0 = 0. Assume that ak is the first nonzero coefficient. Then 1 2 k = ak + ak+1(z − z0) + ak+2(z − z0) + :::: (f(z) − c)(z − z0) 4 Therefore, k 1 lim (z − z0) (f(z) − c) = 6= 0; z!z0 ak and f(z) has a pole of order k at z0; a contradiction. This proves the theorem. Corollary 1. Suppose that f(z) has an isolated essential singularity at z0. Given any complex number c, there exists a sequence fzng such that zn ! z0 and f(zn) ! c: Proof. We choose a sequence fδng for which δn > 0 for each n and lim δn = 0: Therefore n!1 by Weierstrass-Casorati theorem, we can find a sequence of points fzng such that j f(zn) − c j < 1=n for 0 <j zn − z0 j< δn: Thus, f(zn) ! c as zn ! z0. This proves the corollary. Theorem 5. A point z0 is a pole of order m of a function f if and only if it is a zero of order m of the function 1=f. Proof. First we assume that z0 is a pole of f of order m. Then in some neighbourhood of z0, we can write g(z) f(z) = m (z − z0) 1 m 1 where g is analytic at z0 and g(z0) 6= 0: Therefore, f(z) = (z−z0) h(z) where h(z) = g(z) is analytic at z0 and h(z0) 6= 0: This shows that z0 is a zero of 1=f of order m. Next we assume that z0 is a zero of 1=f of order m. Then in some neighbourhood of z0, we can write 1 = (z − z )mg(z) f(z) 0 h(z) 1 where g is analytic at z0 and g(z0) 6= 0: Then f(z) = m where h(z) = is (z−z0) g(z) analytic at z0 and h(z0) 6= 0: This shows that z0 is a pole of f of order m. Theorem 6. If a function f has an essential singularity at z0, then the function 1=f has also an essential singularity at z0. Proof. If possible, we assume that z0 is a regular point of 1=f and it is not a zero of 1=f. Then z0 is a regular point of f contradicting the nature of z0. 5 Let z0 be a regular point of 1=f and it is a zero of 1=f of order m. Then z0 is a pole of f of order m which contradicts the nature of z0. Let z0 be a pole of 1=f of order m. Then z0 is a zero of f of order m which contradicts the nature of z0. Therefore the only possibility that remains for z0 is to be essential singularity of 1=f: Singularities at Point at Infinity 1 The behavior of a function f(z) at z = 1 is considered by making the substitution z = w and examining the nature of f(1=w) at w = 0. It is to be noted that f(z) is regular or has a pole or has an essential singularity at z = 1 if f(1=w) has the same property at w = 0: For instance, the function f(z) = z2 has a pole of order 2 at z = 1. Again the function f(z) = ez has an essential singularity at z = 1, since f(1=w) = e1=w has an essential singularity at w = 0: Theorem 7.
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