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eugene
science forum Guru

Joined: 24 Nov 2005
Posts: 331

Posted: Mon Jul 10, 2006 9:35 am    Post subject: Dense subset of C

It is well known fact that if a,b are real and a/b isn't in Q, then the
set { ma + n b | m,n \in Z} is
dense in R(usually it is called Kronecker's theorem - it's weaker
version). I have the following question: if v and w are complex
non-zero numbers such that v/w is non-real. Is it true that
{ m v + n w | m,n \in Z} is dense in C ?

Thanks
William Elliot
science forum Guru

Joined: 24 Mar 2005
Posts: 1906

Posted: Mon Jul 10, 2006 9:51 am    Post subject: Re: Dense subset of C

On Mon, 10 Jul 2006, eugene wrote:

 Quote: It is well known fact that if a,b are real and a/b isn't in Q, then the set { ma + n b | m,n \in Z} is dense in R(usually it is called Kronecker's theorem - it's weaker version). I have the following question: if v and w are complex non-zero numbers such that v/w is non-real. Is it true that { m v + n w | m,n \in Z} is dense in C ? How much thought have you given this?

Did you look for any counter example?
eugene
science forum Guru

Joined: 24 Nov 2005
Posts: 331

Posted: Mon Jul 10, 2006 10:01 am    Post subject: Re: Dense subset of C

William Elliot wrote:
 Quote: On Mon, 10 Jul 2006, eugene wrote: It is well known fact that if a,b are real and a/b isn't in Q, then the set { ma + n b | m,n \in Z} is dense in R(usually it is called Kronecker's theorem - it's weaker version). I have the following question: if v and w are complex non-zero numbers such that v/w is non-real. Is it true that { m v + n w | m,n \in Z} is dense in C ? How much thought have you given this? Did you look for any counter example?

Yes, you are right? Sorry, i really didn't thought much on this.

We may put v = 3 * i and w = 3 for example and this set doesn't even
have accumulation points.

In fact, this question was arised from the problem:

f: C -> C -analytic and v, w in C such that v/w is not real such that
f(z + v) = f(z) = f(z + w) for all z in C. Prove that f is contsant.

There should probably be another apprach for this.

Thanks and sorry
Lasse

Joined: 29 Apr 2005
Posts: 71

Posted: Mon Jul 10, 2006 10:04 am    Post subject: Re: Dense subset of C

eugene wrote:
 Quote: It is well known fact that if a,b are real and a/b isn't in Q, then the set { ma + n b | m,n \in Z} is dense in R(usually it is called Kronecker's theorem - it's weaker version). I have the following question: if v and w are complex non-zero numbers such that v/w is non-real. Is it true that { m v + n w | m,n \in Z} is dense in C ?

HINT: Think of two such complex numbers, and try to work out what the
given set will look like in the complex number plane.
Lasse

Joined: 29 Apr 2005
Posts: 71

Posted: Mon Jul 10, 2006 10:08 am    Post subject: Re: Dense subset of C

 Quote: In fact, this question was arised from the problem: f: C -> C -analytic and v, w in C such that v/w is not real such that f(z + v) = f(z) = f(z + w) for all z in C. Prove that f is contsant. There should probably be another apprach for this.

Hint: any entire function which is bounded is also constant.
Lee Rudolph
science forum Guru

Joined: 28 Apr 2005
Posts: 566

Posted: Mon Jul 10, 2006 2:33 pm    Post subject: Re: Dense subset of C

"eugene" <jane1806@rambler.ru> writes:

 Quote: It is well known fact that if a,b are real and a/b isn't in Q, then the set { ma + n b | m,n \in Z} is dense in R(usually it is called Kronecker's theorem - it's weaker version). I have the following question: if v and w are complex non-zero numbers such that v/w is non-real. Is it true that { m v + n w | m,n \in Z} is dense in C ?

Maybe try, y'know, a few *examples* of v and w? Say, w = 1, and v...
well, gosh, what's some non-real complex number? i wonder.

Lee Rudolph
eugene
science forum Guru

Joined: 24 Nov 2005
Posts: 331

Posted: Mon Jul 10, 2006 2:37 pm    Post subject: Re: Dense subset of C

Lee Rudolph wrote:
 Quote: "eugene" writes: It is well known fact that if a,b are real and a/b isn't in Q, then the set { ma + n b | m,n \in Z} is dense in R(usually it is called Kronecker's theorem - it's weaker version). I have the following question: if v and w are complex non-zero numbers such that v/w is non-real. Is it true that { m v + n w | m,n \in Z} is dense in C ? Did you actually *think* about this question for, say, *one second*? Maybe try, y'know, a few *examples* of v and w? Say, w = 1, and v... well, gosh, what's some non-real complex number? i wonder. Lee Rudolph

I gave example some posts above.
Lee Rudolph
science forum Guru

Joined: 28 Apr 2005
Posts: 566

Posted: Mon Jul 10, 2006 2:48 pm    Post subject: Re: Dense subset of C

"eugene" <jane1806@rambler.ru> writes:

 Quote: Lee Rudolph wrote: "eugene" writes: It is well known fact that if a,b are real and a/b isn't in Q, then the set { ma + n b | m,n \in Z} is dense in R(usually it is called Kronecker's theorem - it's weaker version). I have the following question: if v and w are complex non-zero numbers such that v/w is non-real. Is it true that { m v + n w | m,n \in Z} is dense in C ? Did you actually *think* about this question for, say, *one second*? Maybe try, y'know, a few *examples* of v and w? Say, w = 1, and v... well, gosh, what's some non-real complex number? i wonder. Lee Rudolph I gave example some posts above.

So you did; apparently the post I replied to arrived here before
the various replies to it.

Anyway, to expand on Lasse's second hint: first, with v and w as
above, L = { m v + n w | m,n \in Z} is NEVER dense in C. It is, in
fact, a discrete subgroup of the abelian topological group C (isomorphic
to Z/2Z), and the quotient group C/L is therefore itself both a topological
group (just because L is normal in C) and a 2-manifold (because L is
discrete). Consequently C/L is, topologically, a torus. If (as you
asked in another of the late-arriving posts) f is a holomorphic
function on C such that f(z+v) = f(z) = f(z+w) for all z in C,
then f descends to a well-defined function g on C/L, which is clearly
continuous since f is. (In fact, there is a natural way to put the
structure of a complex curve on C/L so that g is holomorphic, but we
don't need this.) Because C/L, being homeomorphic to a torus, is compact,
|g| has a local maximum on C/L. Therefore |f| has local maxima on C.
Therefore f is constant.

Lee Rudolph
science forum Guru

Joined: 24 Mar 2005
Posts: 790

Posted: Tue Jul 11, 2006 2:45 am    Post subject: Re: Dense subset of C

In article <e8tpbv$mgn$1@panix2.panix.com>,
lrudolph@panix.com (Lee Rudolph) wrote:

 Quote: "eugene" writes: Lee Rudolph wrote: "eugene" writes: It is well known fact that if a,b are real and a/b isn't in Q, then the set { ma + n b | m,n \in Z} is dense in R(usually it is called Kronecker's theorem - it's weaker version). I have the following question: if v and w are complex non-zero numbers such that v/w is non-real. Is it true that { m v + n w | m,n \in Z} is dense in C ? Did you actually *think* about this question for, say, *one second*? Maybe try, y'know, a few *examples* of v and w? Say, w = 1, and v... well, gosh, what's some non-real complex number? i wonder. Lee Rudolph I gave example some posts above. So you did; apparently the post I replied to arrived here before the various replies to it. Anyway, to expand on Lasse's second hint: first, with v and w as above, L = { m v + n w | m,n \in Z} is NEVER dense in C. It is, in fact, a discrete subgroup of the abelian topological group C (isomorphic to Z/2Z), and the quotient group C/L is therefore itself both a topological group (just because L is normal in C) and a 2-manifold (because L is discrete). Consequently C/L is, topologically, a torus. If (as you asked in another of the late-arriving posts) f is a holomorphic function on C such that f(z+v) = f(z) = f(z+w) for all z in C, then f descends to a well-defined function g on C/L, which is clearly continuous since f is. (In fact, there is a natural way to put the structure of a complex curve on C/L so that g is holomorphic, but we don't need this.) Because C/L, being homeomorphic to a torus, is compact, |g| has a local maximum on C/L. Therefore |f| has local maxima on C. Therefore f is constant.

For those who wish to stay in C, notice C = U_{m,n in Z} (mv +
nw) + K, where K = {av + bw : a, b in [0,1]}. So f(C) = f(K), and
we're done by Liouville.
Lee Rudolph
science forum Guru

Joined: 28 Apr 2005
Posts: 566

Posted: Tue Jul 11, 2006 10:39 am    Post subject: Re: Dense subset of C

 Quote: In article , lrudolph@panix.com (Lee Rudolph) wrote: .... Because C/L, being homeomorphic to a torus, is compact, |g| has a local maximum on C/L. Therefore |f| has local maxima on C. Therefore f is constant. For those who wish to stay in C, notice C = U_{m,n in Z} (mv + nw) + K, where K = {av + bw : a, b in [0,1]}. So f(C) = f(K), and we're done by Liouville.

And for those who wish both to stay in C and to restrict their tools to
the maximum modulus principle (instead of appealing to Liouville), notice
that |f| has a maximum somewhere on the compact set K (a parallelogram
and its interior) and therefore has a local maximum somewhere in the
interior of the compact set U_{m,n in Z, |m|\le 1, |n|\le 1} (mv+nw)+K.

Lee Rudolph

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