Chapter9

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

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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 Classifcation oF Singularities What is the diference between the Functions sin z z , 1 z 4 , and exp ° 1 z ± ? All oF them are not de±ned at 0, but the singularities are oF a very diferent nature. ²or complex Functions there are three types oF singularities, which are classi±ed as Follows. Defnition 9.1. IF f is holomorphic in the punctured disk { z C :0 < | z z 0 | <R } For some R> 0 but not at z = z 0 then z 0 is an isolated singularity oF f . The singularity z 0 is called (a) removable iF there is a Function g holomorphic in { z C : | z z 0 | <R } such that f = g in { z C :0 < | z z 0 | <R } , (b) a pole iF lim z z 0 | f ( z ) | = , (c) essential iF z 0 is neither removable nor a pole. Example 9.2. The Function sin z z has a removable singularity at 0, as For z ° =0 sin z z = 1 z ² k 0 ( 1) k (2 k + 1)! z 2 k +1 = ² k 0 ( 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 0 ³ ³ ³ ³ 1 z 4 ³ ³ ³ ³ = . 93

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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 0 + exp ° 1 x ± = ). On the other hand, exp ° 1 z ± approaches 0 as z approaches 0 from the negative real axis. Hence lim z 0 ² ² exp ° 1 z ±² ² ° = , that is, exp ° 1 z ± has an essential singularity at 0. To get a feel for the diFerent types of singularities, we start with the following results. Proposition 9.5. Suppose z 0 is a isolated singularity of f . Then (a) z 0 is removable if and only if lim z z 0 ( z z 0 ) f ( z ) = 0; (b) if z 0 is a pole then lim z z 0 ( z z 0 ) 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 0 ” we can write f ( z ) as h ( z ) ( z z 0 ) n for some function h which is holomorphic (and not zero) at z 0 . 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 0 if we can write f ( z )=( z z 0 ) m h ( z ), where h is holomorphic and not zero at z 0 . We will make use of the notions of zeros and poles of certain orders quite extensively in this chapter. Proof. (a) Suppose z 0 is removable, and g is holomorphic on D R ( z 0 ), the open disk with radius R centered at z 0 such that f = g for z ° = z 0 . Then we can make use of the fact that g is continuous at z 0 : lim z z 0 ( z z 0 ) f ( z )= l im z z 0 ( z z 0 ) g ( z )= g ( z 0 )l im z z 0 ( z z 0 )=0 . Conversely, suppose that lim z z 0 ( z z 0 ) f ( z ) = 0, and f is holomorphic on the punctured disk ˆ D R ( z 0 )= D R ( z 0 ) \{ z 0 } .Thende±ne g ( z )= ³ ( z z 0 ) 2 f ( z )i f z ° = z 0 , 0i f z = z 0
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## This note was uploaded on 02/27/2012 for the course MATH 417 taught by Professor Staff during the Fall '11 term at SUNY Albany.

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

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