Chapter9

# Chapter9 - Chapter 9 Isolated Singularities and the Residue...

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Unformatted text preview: Chapter 9 Isolated Singularities and the Residue Theorem 1 /r 2 has a nasty singularity at r = 0 , but it did not bother Newton—the moon is far enough. Edward Witten 9.1 Classification of Singularities What is the difference between the functions sin z z , 1 z 4 , and exp ( 1 z ) ? All of them are not defined at 0, but the singularities are of a very different nature. For complex functions there are three types of singularities, which are classified as follows. Definition 9.1. If f is holomorphic in the punctured disk { z ∈ C : 0 < | z- z | < R } for some R > 0 but not at z = z then z is an isolated singularity of f . The singularity z is called (a) removable if there is a function g holomorphic in { z ∈ C : | z- z | < R } such that f = g in { z ∈ C : 0 < | z- z | < R } , (b) a pole if lim z → z | f ( z ) | = ∞ , (c) essential if z is neither removable nor a pole. Example 9.2. The function sin z z has a removable singularity at 0, as for z 6 = 0 sin z z = 1 z X k ≥ (- 1) k (2 k + 1)! z 2 k +1 = X k ≥ (- 1) k (2 k + 1)! z 2 k . and the power series on the right-hand side represents an entire function (you may meditate on the fact why it has to be entire). Example 9.3. The function 1 z 4 has a pole at 0, as lim z → 1 z 4 = ∞ . 93 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM 94 Example 9.4. The function exp ( 1 z ) does not have a removable singularity (consider, for example, lim x → + exp ( 1 x ) = ∞ ). On the other hand, exp ( 1 z ) approaches 0 as z approaches 0 from the negative real axis. Hence lim z → exp ( 1 z ) 6 = ∞ , that is, exp ( 1 z ) has an essential singularity at 0. To get a feel for the different types of singularities, we start with the following results. Proposition 9.5. Suppose z is a isolated singularity of f . Then (a) z is removable if and only if lim z → z ( z- z ) f ( z ) = 0; (b) if z is a pole then lim z → z ( z- z ) n +1 f ( z ) = 0 for some positive integer n . Remark. The smallest possible n in (b) is the order of the pole. We will see in the proof that “near the pole z ” we can write f ( z ) as h ( z ) ( z- z ) n for some function h which is holomorphic (and not zero) at z . This is very similar to the game we played with zeros in Chapter 8: f has a zero of order (or multiplicity) m at z if we can write f ( z ) = ( z- z ) m h ( z ), where h is holomorphic and not zero at z . We will make use of the notions of zeros and poles of certain orders quite extensively in this chapter. Proof. (a) Suppose z is removable, and g is holomorphic on D R ( z ), the open disk with radius R centered at z such that f = g for z 6 = z . Then we can make use of the fact that g is continuous at z : lim z → z ( z- z ) f ( z ) = lim z → z ( z- z ) g ( z ) = g ( z ) lim z → z ( z- z ) = 0 ....
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## This note was uploaded on 06/18/2011 for the course MATH 375 taught by Professor Marchesi during the Spring '10 term at Binghamton.

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Chapter9 - Chapter 9 Isolated Singularities and the Residue...

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