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Unformatted text preview: Remarks: • The function f ( z ) = tan(1 /z ) has a nonisolated singularity at z = 0. We will not discuss this type of singularity here. • f has an essential singularity z ∈ C if and only if the principal part of the Laurent series of f at z contains infinitely many terms. • Be careful to consider the right Laurent series when analyzing singular- ities! It must be the one which is valid in the immediate neighborhood of z . Examples: 1. The function f ( z ) = sin z/z has a removable singularity at z = 0. It is removed by assigning f (0) = 1. This can be seen from the Laurent series of f at 0, which we obtain from (869): f ( z ) = ∞ summationdisplay n =0 ( − 1) n z 2 n (2 n + 1)! = 1 − z 2 3! + z 4 5! − + ··· , z ∈ C . (905) 2. The function f ( z ) = 1 z ( z − 2) 5 + 3 ( z − 2) 2 (906) has a simple pole at z = 0 and a pole of order 5 at z = 2. From the examples in Section 16.1 we know that f ( z ) = z − 5 sin z has a fourth- order pole at z = 0 and that f ( z ) = 1 / ( z 3 − z 4 ) has a third order pole at z = 0 3. The functions exp parenleftbigg 1 z parenrightbigg = summationdisplay n = −∞ z n ( − n )! , | z | > , (907) sin parenleftbigg 1 z parenrightbigg = summationdisplay n = −∞ ( − 1) − n z 2 n − 1 ( − 2 n + 1)! , | z | > , (908) have an (isolated) essential singularity at z = 0. The behavior of complex functions near singularities is characterized by the following theorems. 159 Theorem 29 If a holomorphic function f has a pole at z = z , then | f ( z ) | → ∞ as z → z in any manner. Example: The function f ( z ) = 1 /z 2 hs a second-order pole at z = 0 We have | f ( z ) | = | z | − 2 → ∞ as z → 0 in any manner. Theorem 30 (Picard) A holomorphic function f with an isolated singularity at a point z takes on every value, with at most one exceptional value, in B ε ( z ) , for any ε > . Example: Consider the function f ( z ) = exp(1 /z ), which has an essential singularity at z = 0. For a given complex number c = c exp( iα ) negationslash = 0, we want to find z ∈ C with f ( z ) = c . We set z = r exp( iϕ ) and obtain the equation exp(1 /z ) = exp parenleftBig cos ϕ r parenrightBig exp parenleftbigg − i sin ϕ r parenrightbigg = c exp( iα ) , (909) so that we need to solve cos ϕ = r log c , sin ϕ = − rα. (910) We have 1 = cos 2 ϕ + sin 2 ϕ = r 2 ( log 2 c + α 2 ) , so that r = 1 radicalbig log 2 c + α 2 . (911) Now we may replace α by α +2 πn , n ∈ Z , without changing the given number c . This allows us to make r > 0 arbitrarily small, and so a complex number z with f ( z ) = c can be found in B ε (0), for any ε > 0. Zeros of Holomorphic Functions Definition 14 A complex function f which is holomorphic in a domain U ⊆ C has a zero at z ∈ U if f ( z ) = 0 . A zero of f has order n ∈ N if f ( m ) ( z ) = 0 , m < n and f ( n ) ( z ) negationslash = 0 . If n = 1 , z is called a simple zero....
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This note was uploaded on 11/17/2011 for the course MATH 529 taught by Professor Staff during the Spring '08 term at UNC.
- Spring '08