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Topic: Elementary complex analysis
Replies: 19   Last Post: Mar 9, 2013 11:35 AM

 Messages: [ Previous | Next ]
 quasi Posts: 12,067 Registered: 7/15/05
Re: Elementary complex analysis
Posted: Mar 8, 2013 3:55 AM

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 all
the 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 have
no 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

Date Subject Author
3/5/13 Paul
3/5/13 bacle
3/5/13 W^3
3/5/13 Scott Berg
3/5/13 J. Antonio Perez M.
3/6/13 Scott Berg
3/6/13 Frederick Williams
3/6/13 Robin Chapman
3/6/13 David C. Ullrich
3/6/13 Paul
3/7/13 AP
3/7/13 David C. Ullrich
3/7/13 William Hughes
3/8/13 quasi
3/8/13 William Hughes
3/8/13 quasi
3/8/13 AP
3/8/13 David C. Ullrich
3/8/13 W^3
3/9/13 David C. Ullrich