Sinopsis
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1914 Excerpt: ...note that the circles J?, F in 3), 4) may be replaced by any circle $ in the ring R. For the integrand of 3) is analytic in the ring E--$, and that of 4) is analytic in $--F. 4. Let us now prove the important theorem: Iff(z) can be developed in a two-way power series /() = inO-«)", (5--9 this series must be the series of Laurent. For the function defined by the series 5) satisfies the conditions of Laurent's theorem in 2. Thus / admits the development (Z-«)" (6 where the coefficients ln are the coefficients of Laurent given in 3), 4). Subtracting 5) and 6) we get 0=Ji»0-«)n, bn = an-ln. (7 Let us multiply 7) by (z--a)Hm+I) and integrate around a circle S lying within the ring of convergence of 7). Then by 101, 11) 0 = 2 7rtim... bm = 0, To = 0, ± 1, ± 2... Thus, «« = '» and the coefficients in 5) are the coefficients of Laurent. 118. Zeros and Poles. 1. Let /(z) be a one-valued and analytic function within a circle $ about z--a. Then Taylor's development is valid within $ and we have For 2 = a, this gives /(a) = a0. If a0 = 0, / vanishes at z = a. We say z = a is a root or a zero of /(z). Suppose flO = al = = a--l = ' flm == 0. Then 1) becomes /()-(«-a)»(am + am+,(z-a) + «m+2(z-a)2 + ) = (z-a)".;(z). Here #(z) is analytic within $ and does not vanish at z = a. We say z = a is a root or zero of order m. Since g does not vanish at a it cannot = 0 in some circle about this point. We have thus the theorem: "Let f(z) be one-valued and analytic about the point z = a, and vanish at this point, but not identically. Then there exists a positive integer m such that,,, Nx,n /(z) = (z-a)-£(z) (2 where g(z) is analytic about z = a and does not vanish in some domain about a. 2. Suppose n...
Reseña del editor
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1914 Excerpt: ...note that the circles J?, F in 3), 4) may be replaced by any circle $ in the ring R. For the integrand of 3) is analytic in the ring E--$, and that of 4) is analytic in $--F. 4. Let us now prove the important theorem: Iff(z) can be developed in a two-way power series /() = inO-«)", (5--9 this series must be the series of Laurent. For the function defined by the series 5) satisfies the conditions of Laurent's theorem in 2. Thus / admits the development (Z-«)" (6 where the coefficients ln are the coefficients of Laurent given in 3), 4). Subtracting 5) and 6) we get 0=Ji»0-«)n, bn = an-ln. (7 Let us multiply 7) by (z--a)Hm+I) and integrate around a circle S lying within the ring of convergence of 7). Then by 101, 11) 0 = 2 7rtim... bm = 0, To = 0, ± 1, ± 2... Thus, «« = '» and the coefficients in 5) are the coefficients of Laurent. 118. Zeros and Poles. 1. Let /(z) be a one-valued and analytic function within a circle $ about z--a. Then Taylor's development is valid within $ and we have For 2 = a, this gives /(a) = a0. If a0 = 0, / vanishes at z = a. We say z = a is a root or a zero of /(z). Suppose flO = al = = a--l = ' flm == 0. Then 1) becomes /()-(«-a)»(am + am+,(z-a) + «m+2(z-a)2 + ) = (z-a)".;(z). Here #(z) is analytic within $ and does not vanish at z = a. We say z = a is a root or zero of order m. Since g does not vanish at a it cannot = 0 in some circle about this point. We have thus the theorem: "Let f(z) be one-valued and analytic about the point z = a, and vanish at this point, but not identically. Then there exists a positive integer m such that,,, Nx,n /(z) = (z-a)-£(z) (2 where g(z) is analytic about z = a and does not vanish in some domain about a. 2. Suppose n...
"Sobre este título" puede pertenecer a otra edición de este libro.
