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By Raghavan Narasimhan

The unique variation of this publication has been out of print for a few years. The look­ ance of the current moment variation owes a lot to the initiative of Yves Nievergelt at jap Washington college, and the aid of Ann Kostant, arithmetic Editor at Birkhauser. because the ebook was once first released, a number of humans have remarked at the absence of workouts and expressed the opinion that the ebook may were extra necessary had workouts been integrated. In 1997, Yves Nievergelt knowledgeable me that, for a decade, he had frequently taught a direction at japanese Washington in response to the booklet, and that he had systematically compiled workouts for his direction. He kindly positioned his paintings at my disposal. hence, the current variation appears to be like in elements. the 1st is basically only a reprint of the unique version. i've got corrected the misprints of which i've got turn into conscious (including these mentioned to me by way of others), and feature made a small variety of different minor alterations.

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Example text

Z-a Theorem 4. Let Q be open in 0 (independent oj J) depending only on a and Q, and a sequence {cn}n",O oj complex numbers such that L: n~O cn(z-a)n converges to J(z) Jor zED(a, r). Elementary theory of holomorphic functions 18 Remark. The proof which follows does not give the best possible value of r. We shall find this best value later (§3, Theorem 10 and its corollary). Proof. Let aEQ.

Necessity 01 the condition. Let a EE, V=D(a, p) a disc around a with V ( l E = {a} on which there exist h, g E Yf (V), h ¥= such that hi = g on V- {a}. From the Taylor expansions of g and h, we can write g(z)= (z-a)k¢(z), h(z)=(z-a)fl/J(z), where ¢, I/JEYf(V) and ¢(a)#O, l/J(a)#O. If UC Vis a disc D(a,r) with small enough r, then ¢(z)#O, l/J(z)#O on U. I/J(z) for ZE U-{a}. It then follows that if k~t, I is bounded on D(a,r')- {a} for r'< r; if k< t, I/(z)l-oo as z-a, z#a. ° Definition 4. Let/be a meromorphic function on an open set Q (defined by a discrete set E c Q and IE Yf (Q - E»).

X, y do not both lie in the same complementary interval of I - Eo. In this case, there is a point ~ E E with x< ~ < y, since otherwise, the open interval (x, y) would be disjoint from Eu {a} u {b}, so that the closed interval [x, yJ would be contained in a complementary interval of Eo. If xEEo , we have IIjJ(x)-IjJ(~)I=I¢(x)-¢(~)I~M(~-x) since ~EE. If x ¢ Eo, let J be the complementary interval of Eo containing x, and let x' be the right-hand endpoint of J. Then IIjJ(x) -1jJ(~)1 ~ IIjJ(x) -1jJ(x')1 + IIjJ(x') -1jJ(~)I; now IIjJ(x) -1jJ(x')1 ~ M(x' -x) by Case 1, and IIjJ(x')-IjJ(OI ~ E E, x' E Eo.

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