```Date: Mar 8, 2013 3:55 AM
Author: quasi
Subject: Re: Elementary complex analysis

William Hughes wrote:>David C. Ullrich wrote:>>Paul wrote:>> >>> >I suspect there's a theorem about entire complex function>> >f which have the property that the absolute value of f(z) >> >tends to infinity as the absolute value of z tends to >> >infinity. What does this theorem say?  I don't know of any >> >such functions besides polynomials of degree >= 1.  Is it >> >the case that the set of  functions which have this >> >property is just the set of polynomials of degree >= 1.>>>> Yes.>>>> Non-elementary proof: Look up the Piicard theorems. This is >> immediate even from the "Little" Picard theorem.>>>> Elementary proof: Let g = 1/f. Since f has only finitely many>> zeroes, g is entire except for finitely many poles. Let R be >> a rational function with the same poles as g, and with the >> same principal part at each pole. Then g - R is an entire >> function that tends to 0 at infinity, so g = R.>>Ok, I see why g-R is entire but not why it tends to 0>at infinity.  What am I missing?I think the following variation of David Ullrich's argument repairs the flaw.Let R be the rational function consisting of the sum of allthe principal parts of g at its poles.Then R approaches to 0 at infinity, hence since g also approaches 0 at infinity, so does g - R.As in David's argument, g - R is entire, hence, since g - R appoaches 0 at infinity, it follows that g = R.Write R = p/q as a quotient of polynomials where p,q haveno common zeros.Then f = 1/g = 1/R = q/p.Since f is entire, p has no zeros, hence p is a nonzero constant.Therefore f is a polynomial, as was to be shown.quasi
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