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        general >> complex analysis >> analytic and one-one
        
(Message started by: immanuel78 on Oct 9^th, 2006, 12:06am)

Title: analytic and one-one
Post by immanuel78 on Oct 9^th, 2006, 12:06am

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...??

Title: Re: analytic and one-one
Post by towr on Oct 9^th, 2006, 12:39am

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.

Title: Re: analytic and one-one
Post by Michael_Dagg on Oct 9^th, 2006, 8:49pm

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 ?

Title: Re: analytic and one-one
Post by Icarus on Oct 19^th, 2006, 8:42pm

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.

Title: Re: analytic and one-one
Post by Sjoerd Job Postmus on Nov 3^rd, 2006, 12:38pm

on 10/09/06 at 00:39:17, 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:
[hide]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.[/hide]

Question: Can we use the mean-value-theorem?

Title: Re: analytic and one-one
Post by Icarus on Nov 3^rd, 2006, 3:32pm

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.

Title: Re: analytic and one-one
Post by Eigenray on Nov 3^rd, 2006, 9:41pm

Suppose f has a zero of order k at 0, i.e., f(z) = z^kg(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.