# holomorphic function convergent sequence identically zeroWatch

#1
Hi

Theorem attached and proof.

I am stuck on:

1) Where we get comes from so is the first non-zero fourier coeffient. So I think this term is , from the radius of the open set, but I don't know how to take care of the rest of the higher tems through , is this some theorem or?

2) The conclusion thus has only one zero at I think i'm being stupid but what is this being made from? We know and , but I dont understand.

Thanks
0
2 years ago
#2
(Original post by xfootiecrazeesarax)
Hi

Theorem attached and proof.
No it isn't...
0
#3
(Original post by DFranklin)
No it isn't...
0
2 years ago
#4
For (1), it's literally just continuity. g is cts and |g(z_0)| > 0. By cty of g,, we can find R s.t.. |z-z0| < R => |g(z) - g(z_0)| < |g(z_0)|/2. Shrink r so that |z-z0| <R is always satisfied and you're done.

For (2), well, if we can write f as (z-z0)^m g(z), and g is never zero near z0, how can f be 0 unless z = z0?
0
#5
(Original post by DFranklin)
and g is never zero near z0,
Thanks

I don't understand how implies this though

This choice seems a bit random to be, I dont understand it's significance.

Thanks
0
2 years ago
#6
(Original post by xfootiecrazeesarax)
Thanks

I don't understand how implies this though

This choice seems a bit random to be, I dont understand it's significance.

Thanks
g is never zero near z0 because it's non-zero at z0, and because it's cts, we can find a neighbourhood of z0 where g(z) is always close to g(z0). Then since g(z0) isn't 0, g(z) isn't either.

This is a basic continuity argument.

Can I ask what your background is, because it seems you're attempting a postgraduate course in pure mathematics with gaps in your knowledge that should have been covered in the first year of an undergraduate degree.
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#7
(Original post by DFranklin)
g is never zero near z0 because it's non-zero at z0, and because it's cts, we can find a neighbourhood of z0 where g(z) is always close to g(z0). Then since g(z0) isn't 0, g(z) isn't either.

This is a basic continuity argument.

Can I ask what your background is, because it seems you're attempting a postgraduate course in pure mathematics with gaps in your knowledge that should have been covered in the first year of an undergraduate degree.
ta

doing masters in math
the course listed background material as complex analysis only
but yes it seems to have links to pure/algebra which were not listed as prerequisites and I did very little pure during my undergrad.

*in fact if done in first year I've most likely forgotten some
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2 years ago
#8
(Original post by xfootiecrazeesarax)
ta

doing masters in math
the course listed background material as complex analysis only
Thing is, I would assume real analysis as a prerequisite for complex analysis (and I'd be very surprised if your university doesn't).

Certainly this would have been regarded as a straightforward argument in my complex analysis course.
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#9
(Original post by DFranklin)
Thing is, I would assume real analysis as a prerequisite for complex analysis (and I'd be very surprised if your university doesn't).

Certainly this would have been regarded as a straightforward argument in my complex analysis course.
fair point, I'm pretty sure we didn't use much analysis in the complex analysis course, if so I've forgot it.
0
2 years ago
#10
(Original post by xfootiecrazeesarax)
fair point, I'm pretty sure we didn't use much analysis in the complex analysis course, if so I've forgot it.
To be honest, if you can't follow this, I'm not really sure what to say to you at this point. The problem is that your course is expecting you to be able to use certain tools from complex/real analysis. I mean, we could go back and forth for an hour and I could persuade you why this is true, but at this level, you really are expected to be able to look at something like this and say "yeah, gonna be true by continuity". And my feeling (from having done a post grad course involving a lot of Complex Variable stuff) is that this is going to be an on going issue for you.

At the least, I would read back over your complex variable notes, or if you really didn't do anything on the "analysis" side for complex variable, see if there's a text book your lecturer can recommend.
0
2 years ago
#11
(Original post by DFranklin)
For (1), it's literally just continuity. g is cts and |g(z_0)| > 0. By cty of g,, we can find R s.t.. |z-z0| < R => |g(z) - g(z_0)| < |g(z_0)|/2. Shrink r so that |z-z0| <R is always satisfied and you're done.
Maybe I'm confused, but don't you want something like this here:

Else you don't get the lower bound on g(z)?
0
2 years ago
#12
(Original post by xfootiecrazeesarax)
fair point, I'm pretty sure we didn't use much analysis in the complex analysis course, if so I've forgot it.
Can I suggest that you buy a copy of Schaum's Complex Variables text:

https://www.amazon.co.uk/Schaums-Out.../dp/0071615695

It's a fairly fast-paced but decent overview of complex analysis, and is presented in the standard Schaum "solved problems" format. It may be a bit light on the hard-core analysis side (e.g. I don't think it mentions the identity theorem, which I think you're using/proving here), but that may be what you need at the moment.
0
2 years ago
#13
(Original post by atsruser)
Maybe I'm confused, but don't you want something like this here:

Else you don't get the lower bound on g(z)?
I don't think there's a problem with what I wrote;. You'll need to use |a-b| < C implies |b|>|a|-C but that should be obvious at this level.

[I'm aware I may seem a bit harsh, but having done a post graduate course that was fairly similar to this, I'm aware how it's assumed you can "fill in the details" on things like this (and indeed much more complicated than this)].
0
2 years ago
#14
(Original post by DFranklin)
I don't think there's a problem with what I wrote;. You'll need to use |a-b| < C implies |b|>|a|-C but that should be obvious at this level.
No, there isn't a problem - it's just that when you said "And you're done", it looked like you'd omitted half the argument, and claimed that it was the finished item. In fact I see that you were only talking about the continuity bit - fair enough, though maybe confusing to the OP.
0
2 years ago
#15
(Original post by atsruser)
No, there isn't a problem - it's just that when you said "And you're done", it looked like you'd omitted half the argument, and claimed that it was the finished item. In fact I see that you were only talking about the continuity bit - fair enough, though maybe confusing to the OP.
Maybe a bit terse, but this kind of argument is *very* common in complex analysis.
0
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