hw11_soln - Complex Analysis Fall 2007 Homework 11...

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Complex Analysis Fall 2007 Homework 11: Solutions 3.3.16 If f has a zero of multiplicity k at z 0 then we can write f ( z ) = ( z - z 0 ) k φ ( z ), where φ is analytic and φ ( z 0 ) 6 = 0. Differentiating this expression yields f 0 ( z ) = k ( z - z 0 ) k - 1 φ ( z ) + ( z - z 0 ) k φ 0 ( z ). Therefore f 0 ( z ) f ( z ) = k ( z - z 0 ) k - 1 φ ( z ) + ( z - z 0 ) k φ 0 ( z ) ( z - z 0 ) k φ ( z ) = k z - z 0 + φ 0 ( z ) φ ( z ) . Since φ ( z 0 ) 6 = 0, the function φ 0 ( z ) ( z ) is analytic at z 0 and therefore has a convergent Taylor series in a neighborhood of z 0 . Thus f 0 ( z ) f ( z ) = k z - z 0 + φ 0 ( z ) φ ( z ) = k z - z 0 + X n =0 a n ( z - z 0 ) n . for all z in a deleted neighborhood of z 0 . Uniqueness of such expressions implies that this is the Laurent series for f 0 /f in a deleted neighborhood of z 0 and therefore f 0 /f has a simple pole at z 0 with residue k . 3.3.20(a) The closure of a set A is the intersection of all the closed sets that contain A . It is not hard to show that an element belongs to the closure of A if and only if every neighborhood of that point intersects A . Therefore, what we need to prove the following: given any w C and any ± > 0, there exists z U so that f ( z ) D ( w ; ± ). So, let w C and ± > 0. According to the Casorati-Weierstrass Theorem, there is a sequence z 1 , z 2 , z 3 , ··· ∈ C so that z n z 0 and f ( z n ) w . Since z 0 U and U is open, there is a δ > 0 so that D ( z 0 ; δ ) U . Choose N 1 Z + so that | z n - z 0 | < δ for n N 1 and choose N 2 Z + so that | f ( z n ) - w | < ± . Let N = max { N 1 , N 2 } . Then z N D ( z 0 ; δ ) U and f ( z N ) D ( w ; ± ). That is, z N satisfies the required conditions. Since w C and ± > 0 were arbitrary, we conclude that the necessary condition holds for all w C and ± > 0. Thus, the closure of f ( U ) is C . 3.R.2 The function 1 / cos z fails to be analytic precisely where cos z = 0. The latter occurs if and only if z = + π/ 2 for some n Z . Since the derivative of cos z is - sin z and sin( + π/ 2) = ( - 1) n 6 = 0, we see that cos z has simple zeros at the points z = + π/ 2, n Z . Consequently, 1 / cos z has simple poles at these points. 3.R.6(a)
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hw11_soln - Complex Analysis Fall 2007 Homework 11...

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