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From: José Carlos Santos on 21 Jul 2010 19:38
On 21-07-2010 23:53, David Bernier 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.
>>>
>>> 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.
>
> 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?

I suggest that you see the statement of the Casorati-Weierstrass this
way: for *any* neighborhood V of _a_, g(V\{a}) is dense. The word "any"
is important.

Best regards,

Jose Carlos Santos
From: David Bernier on 22 Jul 2010 01:49
Jos� Carlos Santos wrote:
> On 21-07-2010 23:53, David Bernier 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.
>>>>
>>>> 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.
>>
>> 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?
>
> I suggest that you see the statement of the Casorati-Weierstrass this
> way: for *any* neighborhood V of _a_, g(V\{a}) is dense. The word "any"
> is important.
[...]

If g has an essential singularity at 0, shouldn't it be:
for any neighborhood V of 0, g(V\{0}) is dense in C ?

Regards,

David Bernier

From: David Bernier on 22 Jul 2010 02:21
David Bernier wrote:
> Jos� Carlos Santos wrote:
>> On 21-07-2010 23:53, David Bernier 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.
>>>>>
>>>>> 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.
>>>
>>> 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 the inverse function of g is discontinuous at g(a) ...

So g can't have an analytic inverse.

'a' can be any number in the range of g.

Then, one might try to show that the inverse of f itself
is discontinuous.


>>> So at this point I'm stuck. Might Rouche's Theorem be of
>>> any use for this sub-problem?
>>
>> I suggest that you see the statement of the Casorati-Weierstrass this
>> way: for *any* neighborhood V of _a_, g(V\{a}) is dense. The word "any"
>> is important.
> [...]
>
> If g has an essential singularity at 0, shouldn't it be:
> for any neighborhood V of 0, g(V\{0}) is dense in C ?
>
> Regards,
>
> David Bernier
>

From: José Carlos Santos on 22 Jul 2010 02:22
On 22-07-2010 6:49, David Bernier wrote:

>>> 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.
>>>
>>> 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?
>>
>> I suggest that you see the statement of the Casorati-Weierstrass this
>> way: for *any* neighborhood V of _a_, g(V\{a}) is dense. The word "any"
>> is important.
> [...]
>
> If g has an essential singularity at 0, shouldn't it be:
> for any neighborhood V of 0, g(V\{0}) is dense in C ?

Right. I forgot that the singularity was at 0.

Besides: keep in mind the open mapping theorem.

Best regards,

Jose Carlos Santos
From: David C. Ullrich on 22 Jul 2010 11:18
On Wed, 21 Jul 2010 18:53:41 -0400, David Bernier
<david250(a)videotron.ca> wrote:

>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.

You want the C-W theorem plus the Open Mapping Theorem.
The image of, say, the disk D(1, 1/2) is open. The image of
the deleted disk D'(0, 1/2) = D(0, 1/2) \ {0} is dense.
Hence the function is not 1-1.

>[ 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

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