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   Author  Topic: analytic and one-one  (Read 5857 times)
immanuel78
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analytic and one-one  
« on: Oct 9th, 2006, 12:06am »
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Let G be a region.
Let f : G -> C be analytic and one to one.
Then f'(z) (f prime) is not zero for all z in G.
 
This problem seems to be easy to me at first but the proof...??  
« Last Edit: Oct 9th, 2006, 12:08am by immanuel78 » IP Logged
towr
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Re: analytic and one-one  
« Reply #1 on: Oct 9th, 2006, 12:39am »
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If f'(z) is 0 for some z, then there will be points a =\= b such that f(a)=f(b).
I'm not entirely sure how to prove it though.
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Michael Dagg
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Re: analytic and one-one  
« Reply #2 on: Oct 9th, 2006, 8:49pm »
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Think more carefully about this problem. Several of the
other problems that you have posted here are harder,  
all of which someone has wrote solutions to.
 
Besides the idea that   f'   be   <>   0  throughout   G,   what else  
can you say about   f   or   f'  when    f: G --> C   is analytic
and   1-to-1 ?
 
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Re: analytic and one-one  
« Reply #3 on: Oct 19th, 2006, 8:42pm »
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I guess I'm too rusty to remember what Michael is hinting at, but here are my thoughts:
 
The theorem is easily equivalent to the statement that if f is analytic at a point a, then f is one-to-one in some neighborhood of a only if f'(a) != 0, which in turn follows from the special case a = f(a) = 0.
 
If f'(0) = 0, then f(z) = zng(z) for some n > 1 and analytic g with g(0) = b != 0. Because of this, there is some neigborhood of 0 in which g is nearly constant, but zn takes on every value but 0 multiple times. A continuity argument should be able to demonstrate that some values of the product occur more than once.
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Re: analytic and one-one  
« Reply #4 on: Nov 3rd, 2006, 12:38pm »
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on Oct 9th, 2006, 12:39am, towr wrote:
If f'(z) is 0 for some z, then there will be points a =\= b such that f(a)=f(b).
I'm not entirely sure how to prove it though.

Not necesarilly. If we take f(x) = x^3, f'(x) = (x^2)/3, and f'(0) = 0. But, there are no values a and b with a f(a)=f(b), but a!=b.
Also, if I am right, I believe this one is also analytic...
 
And does the C imply a region, or being constant? (correction, wikipedia tells me it is a function that is continuous). Also, I am not quite sure how to interpret the statement.
 
Let G be a region.
Let f : G -> C be analytic and one to one.
Then f'(z) (f prime) is not zero for all z in G.
 
1) For all z in G f'(z) != 0 ( [not zero] for all )
2) Not for all z in G f'(z) = 0 (not [zero for all])
 
Statement 1:
\forall {z \in G} f'(z) \neq 0
x^3 is analytic, and one-to-one. But, f'(0) = 0. False.
 
Statement 2:
f(z) is one-to-one. Thus, if a \ne b f(a) \ne f(b). Because f(z) is continuous, we can state (using the mean value theorem) that there exists a c such that f'(c) = (f(a)-f(b))/(a-b). Because f(a) \ne f(b), f'(c) \ne 0. And thus it is not [zero for all] z.
 
Question: Can we use the mean-value-theorem?
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Re: analytic and one-one  
« Reply #5 on: Nov 3rd, 2006, 3:32pm »
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Sjoerd, C is the set of complex numbers, and f is a function from G, an open subset of C (ie, for any point x of G, G also contains all points within some distance r(x) of x) into C.
 
In the complex numbers x3 is not one-to-one, but is in fact "3-to-1" for every point except 0. That is, for every z =/= 0, there are three values w such that w3 = z.
 
Complex functions of complex variables are much better behaved than their real cousins. If a complex function is differentiable at all on a region, then it not only has derivatives of all orders, but it is also analytic there. Compare this to the real numbers, where a function can have any desired number of derivatives before it breaks down, and even if the function has derivatives of all orders, it still need not be analytic.
 
Another example of how complex functions are better behaved is the result involved here: a differentiable complex function is one-to-one in a neighborhood of a point if and only if its derivative is non-zero.
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Re: analytic and one-one  
« Reply #6 on: Nov 3rd, 2006, 9:41pm »
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Suppose f has a zero of order k at 0, i.e., f(z) = zkg(z), where g is holomorphic and g(0) != 0.  Then there's some small disk D={z : |z|<R} around 0 in which g has no zeroes, so f has exactly k zeroes in D, counting multiplicity.
 
Let r = inf { |f(z)| : |z|=R } > 0.  Then for any w with |w|<r, it follows from Rouche's theorem that f(z)-w has exactly k zeroes within C.  That is, f takes on every value w, with |w|<r, exactly k times on D.
 
It follows f is locally injective at 0 iff k=1.  Hence f is locally injective iff f'(z) !=0 for all z.
 
One can also phrase the above argument in terms of the argument principle, using the fact that winding numbers are locally constant.
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