phukraut

here's a taste of what i need help on:

{z : 0 < |z| < infinity} is conformally equivalent to {z : r < |z| < R} if and only if r=0 and R=infinity.

i need to prove this, but i don't know how to get started. can anyone give me some pointers?

thx.

rsl12

this is the complex plane you're talking about?

not that that helps--i would have no idea how to prove it...seems self evident...

or is this for a topology class?

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Giltwist

Thats not complex analysis, that is real analysis. I'd need the definition of "conformally equivalent"

kel

Well the sets are identical... one is a synonym of the other through substitution...
The formal way to prove this is if you take the opposite of union (I forgot the formal name for this operation) of the two sets then it should the empty set. This means they are equivalent (I don't know what conformally equivalent is).
As I will show, if you mess with R, or r then you won't get the empty set.

The basic deal is that if R is not equal to infinite, but to some finite number, then the set will always be of a finite size in the positive direction.... So it could never be equal to the inifinite positive set and when you take ~union (the set of all elements that exists on only one set or the other, but not in both) of the two you will not have an empty set. If it is -infinite then you will have an empty set. Whether r is infinite or not makes no difference, I am not sure but in this case if r were infinite or -infinite we would get the empty set, which ~union with the original, is non empty.

If little r were set to anything other then zero then then again the set's ~union would have elements in it, either elements <0, or elements >0.

That basically takes care of the two cases I see here... r != 0, or R != infinity


Don't know if this proof holds water in the crazy set universe.

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phukraut

conformally equivalent means there is a conformal mapping between the sets in the complex plane. conformal means there is a biholomorphic map.. a holomorphic map f(z) is a function with a complex derivative such that df/dzbar=0. thanks for the help.

i attempted to prove this by contradiction. i assumed there existed a map, then by a compositions of maps from this beginning (a composition of transforms) showed that the end result would have to be something that is NOT conformally equivalent to the first set. this would imply that the only conformal mapping of that set which is of the form in the second set, would have to be the identity map.

Giltwist

Ah with that definition of conformally equivalent, that is complex analysis. You must be in a grad program for math, because I am a junior in an undergrad math program and just learning real analysis. I think you are pretty much out of luck finding help here.

phukraut

the set theoretic approach was interesting and i would have never even considered an argument along those lines, so it was not a waste to post.

this was a difficult set of questions that i was working on, so i thought it couldn't hurt to try and see if anyone else out there was doing it .

thanks again for the help.

phukraut

at first i thought you meant intersection, but now i know you must mean the minus, ie: for two sets A and B, A\B={x in A: x not in B}. therefore, if the sets are equivalent, then A\B=null.

well, at least for standard axioms (ZF?), it works.. to recap your steps:

if we name the two sets A and B respectively, then, if R is less than infinity, then we must have |A\B|>0 because some elements (an infinite number) will not be removed at the positive tail.

if r>0, then once again, |A\B|>0 since some elements (possibly infinite again if we are talking about say hte real numbers) will not be removed about the point 0 (zero).

therefore, A\B=null only if A=B. if A=B, then r=0 and R=infinity.

that's a clever way to prove seemingly obvious facts about sets, i'll use it some day, thx..

i wonder if the same would hold for complex numbers.. since there is no actual lessthan/greaterthan relation on complex numbers, only on their real moduli, could we still use this argument (putting aside "conformally" for a moment and just talking about actual set equivalence)?

rsl12

can you use a topologic approach?

a donut with one hole cannot be conformally equivalent to a plane; therefore, r=0.

Not sure how you could prove that R=infinity though...

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rsl12

actually, i just noticed that the signs are less than signs, not less than or equal to. Is this right?

If so, can't you map onto a region

{z':0<|z'|<1} using z'=1/(R+1)*e^it

where R=|z| and t = the angle for z?

i'm stoopid though--i haven't used complex analysis in years...

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