Prev: Quasi-orthogonalizing 1, x, x^2, ,, for large x
Next: Geometric problem
From: dushya on 21 Jul 2010 10:17
Is following statement true --
If f:C-->C is analytic bijective function with analytic inverse then
f is of the form f(z) = az + b (C is set of complex numbers and a, b
are constants and a is nonzero).

If it is true then can anyone suggest some proof.
From: José Carlos Santos on 21 Jul 2010 10:23
On 21-07-2010 15:17, dushya wrote:

> Is following statement true --
> If f:C-->C is analytic bijective function with analytic inverse then
> f is of the form f(z) = az + b (C is set of complex numbers and a, b
> are constants and a is nonzero).

Yes.

> If it is true then can anyone suggest some proof.

Consider the function g:C\{0} ---> C defined by g(z) = f(1/z). It is
injective and analytic. Deduce from this that the singularity of _g_
at 0 is either a pole or a removable singularity. Use this to prove
that _f_ is a polynomial function and then prove that it must be a
polynomial function of the type that you mentioned.

Best regards,

Jose Carlos Santos
From: dushya on 21 Jul 2010 12:18
On Jul 21, 7:23 am, José Carlos Santos <jcsan...(a)fc.up.pt> wrote:
> On 21-07-2010 15:17, dushya wrote:
>
> > Is following statement true --
> >   If  f:C-->C is analytic bijective function with analytic inverse then
> > f is of the form f(z) = az + b (C is set of complex numbers and a, b
> > are constants and a is nonzero).
>
> Yes.
>
> > If it is true then can anyone suggest some proof.
>
> Consider the function g:C\{0} ---> C defined by g(z) = f(1/z). It is
> injective and analytic. Deduce from this that the singularity of _g_
> at 0 is either a pole or a removable singularity. Use this to prove
> that _f_ is a polynomial function and then prove that it must be a
> polynomial function of the type that you mentioned.
>
> Best regards,
>
> Jose Carlos Santos

thank you Jose :-). I tried but could not prove that singularity of g
at zero is either pole or a removable singularity :-)
From: Robert Israel on 21 Jul 2010 16:37

> On Jul 21, 7:23=A0am, Jos=E9 Carlos Santos <jcsan...(a)fc.up.pt> wrote:
> > On 21-07-2010 15:17, dushya wrote:
> >
> > > Is following statement true --
> > > =A0 If =A0f:C-->C is analytic bijective function with analytic inverse
> > > ==
> then
> > > f is of the form f(z) =3D az + b (C is set of complex numbers and a, b
> > > are constants and a is nonzero).
> >
> > Yes.
> >
> > > If it is true then can anyone suggest some proof.
> >
> > Consider the function g:C\{0} ---> C defined by g(z) =3D f(1/z). It is
> > injective and analytic. Deduce from this that the singularity of _g_
> > at 0 is either a pole or a removable singularity. Use this to prove
> > that _f_ is a polynomial function and then prove that it must be a
> > polynomial function of the type that you mentioned.
> >
> > Best regards,
> >
> > Jose Carlos Santos
>
> thank you Jose :-). I tried but could not prove that singularity of g
> at zero is either pole or a removable singularity :-)

Well, the only other possibility is an essential singularity.
Strange things happen near an essential singularity. It's easy to
rule this out using Picard's "Great" Theorem, but if you're not willing
to use that you could use the Casorati-Weierstrass theorem to show
that the inverse function can't be continuous.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: David Bernier on 21 Jul 2010 18:53
Robert Israel wrote:
>> On Jul 21, 7:23=A0am, Jos=E9 Carlos Santos<jcsan...(a)fc.up.pt> wrote:
>>> On 21-07-2010 15:17, dushya wrote:
>>>
>>>> Is following statement true --
>>>> =A0 If =A0f:C-->C is analytic bijective function with analytic inverse
>>>> ==
>> then
>>>> f is of the form f(z) =3D az + b (C is set of complex numbers and a, b
>>>> are constants and a is nonzero).
>>>
>>> Yes.
>>>
>>>> If it is true then can anyone suggest some proof.
>>>
>>> Consider the function g:C\{0} ---> C defined by g(z) =3D f(1/z). It is
>>> injective and analytic. Deduce from this that the singularity of _g_
>>> at 0 is either a pole or a removable singularity. Use this to prove
>>> that _f_ is a polynomial function and then prove that it must be a
>>> polynomial function of the type that you mentioned.
>>>
>>> Best regards,
>>>
>>> Jose Carlos Santos
>>
>> thank you Jose :-). I tried but could not prove that singularity of g
>> at zero is either pole or a removable singularity :-)
>
> Well, the only other possibility is an essential singularity.
> Strange things happen near an essential singularity. It's easy to
> rule this out using Picard's "Great" Theorem, but if you're not willing
> to use that you could use the Casorati-Weierstrass theorem to show
> that the inverse function can't be continuous.

I'm doing this as an exercise. I see how to rule out an essential
singularity at zero using Picard's "Great" Theorem: as Jose Carlos
Santos pointed out, his g:C\{0} ---> C is both injective and analytic.

Now I'm trying to rule out the same using only the Casorati-Weierstrass theorem.

[ small spoiler space in case others want to try using Casorati-Weierstrass only]
-
-
-
-
-
-
-
-
.....






Attempt using Casorati-Weierstrass:


Suppose g has an essential singularity at zero. By the
Casorati-Weierstrass theorem, if 'a' is in the range of g,
there exists a sequence of non-zero complex numbers

{z_j}_{j = 1... oo} with |z_{j+1}|< |z_j|, plus the two
following conditions:

(a) |z_j| < 1/j and
(b) | g(z_j) - a| < 1/j .

Then for f:

let w_j = 1/(z_j).

(a) |w_j| > j
(b) | f( w_j ) - a| < 1/j .

We want to conclude that f isn't injective [contradiction].

So at this point I'm stuck. Might Rouche's Theorem be of
any use for this sub-problem?

David Bernier
 |  Next  |  Last
Pages: 1 2
Prev: Quasi-orthogonalizing 1, x, x^2, ,, for large x
Next: Geometric problem