Holomorphic? [Math]

Prev: JSH: So what happened?
Next: The weight of a singularity

From: I.N. Galidakis on 23 Jul 2010 05:22

David C. Ullrich wrote:
> On Fri, 23 Jul 2010 04:57:23 +0300, "I.N. Galidakis"
> <morpheus(a)olympus.mons> wrote:
>
>> David C. Ullrich wrote:
>> [snip for brevity]
>>
>>> What I meant was exactly what I said, no more and no less.
>>>
>>> If x = r cos(phi) and y = r sin(phi) then you should find that
>>> x + iy _does_ satisfy the "polar form of the C-R equations".
>>
>> Yes, that's what you said, except that in this case I don't see how it
>> applies, since x and y are _not_ given by these equations. x and y are given
>> by the equations I give in my first post (also below). The curve in
>> "standard" Polar coordinates is descibed by:
>>
>> r=cos(k*phi), with k=m/n\in Q
>>
>> and I cannot see what your "if x = r cos(phi)..." has to do with the above.
>
> You were applying the C-R equations "in polar form" in a situation
> where they're simply not applicable, since your coordinates are
> not polar coordinates. I was trying to clarify in which situation
> those equations _would_ imply the function was holomorphic.

Ok.

>>>> One question if I may: Can you enlighten us on how did you see that this
>>>> curve was holomorphic to begin, with without doing the calculations?
>>>
>>> x + iy is what's typiically known as "z", and dz/dz exists.
>>
>> I assume you mean x+yi, with:
>>
>> x = r*cos(n*phi)*sin(m*phi)
>
> When I say "if x = r cos(phi)"if x = r cos(phi)" why
> would you assume that I meant x = r*cos(n*phi)*sin(m*phi)?
> Especially after I clarified that what I meant was what
> I wrote, not what you assume I meant?
>
> As of about the second line of your last post I said
> I didn't follow what you meant since you hadn't
> specified the relation between something and
> something else. Instead of explaining you
> "snipped for brevity".

Well, I assumed that since you "got lost" on a simple argument which transformed
my coordinates to the "standard" polar coordinates you probably meant, there'd
be no reason for me to repeat it.

> Good luck.
--
I.

From: Axel Vogt on 23 Jul 2010 15:56

I.N. Galidakis wrote:
> I am checking the holomorphicity for the complex function
> f(r,phi)=x(r,phi)+y(r,phi)*i on the disk D={z:|z|<r+eps}. x and y are
> given as:
>
> x(r,phi)=r*sin(n*phi)*cos(m*phi)
> y(r,phi)=r*sin(n*phi)*sin(m*phi),
>
> m,n\in N.
>
> Checking the Polar form of the Cauchy-Riemann equations for these, I find:
>
> dx/dr=sin(n*phi)*cos(m*phi)
>
> while,
>
> 1/r*dy/dphi=cos(n*phi)*n*sin(m*phi)+sin(n*phi)*cos(m*phi)*m
>
> and these don't look equal. Does this mean that f(r,phi) is not
> holomorphic on the disk D?
>
> Am I missing something?
>
> Thanks,
> --
> I.

I tried your task as well, guessing it is non-holomorph (and failed):

The C-R condition means, that the Jacobi matrix is complex linear
But the problem is, how to recover the different coordinate system
(or else: how to write I*z as input using polar coordinates) and
it needs *care* to mix real and complex linearity working with
coordinates.

Same for (complex) difference quotients (writing additions in the
polar coordinates is a mess).

The last thing might be to try Morera's theorem (integral over
closed curves = 0 ---> analytic), but have no idea what to take
to show it is false, since for a circle it is true ...

From: Axel Vogt on 23 Jul 2010 16:18

From: I.N. Galidakis on 25 Jul 2010 08:48

Axel Vogt wrote:
> I.N. Galidakis wrote:
>> I am checking the holomorphicity for the complex function
>> f(r,phi)=x(r,phi)+y(r,phi)*i on the disk D={z:|z|<r+eps}. x and y are
>> given as:
>>
>> x(r,phi)=r*sin(n*phi)*cos(m*phi)
>> y(r,phi)=r*sin(n*phi)*sin(m*phi),
>>
>> m,n\in N.
>>
>> Checking the Polar form of the Cauchy-Riemann equations for these, I find:
>>
>> dx/dr=sin(n*phi)*cos(m*phi)
>>
>> while,
>>
>> 1/r*dy/dphi=cos(n*phi)*n*sin(m*phi)+sin(n*phi)*cos(m*phi)*m
>>
>> and these don't look equal. Does this mean that f(r,phi) is not
>> holomorphic on the disk D?
>>
>> Am I missing something?
>>
>> Thanks,
>> --
>> I.
>
> I tried your task as well, guessing it is non-holomorph (and failed):
>
> The C-R condition means, that the Jacobi matrix is complex linear
> But the problem is, how to recover the different coordinate system
> (or else: how to write I*z as input using polar coordinates) and
> it needs *care* to mix real and complex linearity working with
> coordinates.
>
> Same for (complex) difference quotients (writing additions in the
> polar coordinates is a mess).
>
> The last thing might be to try Morera's theorem (integral over
> closed curves = 0 ---> analytic), but have no idea what to take
> to show it is false, since for a circle it is true ...

Axel,

Here's another way to see what David is saying. Have Maple ready, because it
will be handy, as the calculations are erudite.

The initial function is given as x+y*i, with:

x(r,phi)=r*sin(n*phi)*cos(m*phi)
y(r,phi)=t*sin(n*phi)*sin(m*phi)

To change to a standard polar coordinate system: (r, theta), set:

r*exp(theta*i)=x+y*i, from which,

theta = -i*ln(sin(n*phi)*cos(m*phi)+sin(n*phi)*sin(m*phi)*i)

(In Maple:
> x:=(r,theta,phi)->r*sin(theta)*cos(phi);
> y:=(r,theta,phi)->r*sin(theta)*sin(phi);
> z:=(r,theta)->r*cos(theta);

> eq:=r*exp(theta*I)=x(r,n*phi,m*phi)+y(r,n*phi,m*phi)*I;
> sol:=solve(eq,{r,theta});

> theta:=rhs(sol[2,1]);

)

On the new polar coordinate system x+y*i = u'+v'*i, with:

u'=r*cos(theta)
v'=r*sin(theta)

and now,

du'/dr:

> simplify(diff(r*cos(theta),r));

and dv'/dtheta = dv'/dphi*dphi/dtheta:

> simplify(1/r*diff(r*sin(theta),phi)/diff(theta,phi));

and they are the same.

(To David: Thanks for being (intentionally) terse, as this forced me to bang my
head until I got it, instead of being spoon-fed with the answer).
--
I.

From: Axel Vogt on 26 Jul 2010 16:31

I.N. Galidakis wrote:
> Axel Vogt wrote:
>> I.N. Galidakis wrote:
>>> I am checking the holomorphicity for the complex function
>>> f(r,phi)=x(r,phi)+y(r,phi)*i on the disk D={z:|z|<r+eps}. x and y are
>>> given as:
>>>
>>> x(r,phi)=r*sin(n*phi)*cos(m*phi)
>>> y(r,phi)=r*sin(n*phi)*sin(m*phi),
>>>
>>> m,n\in N.
>>>
>>> Checking the Polar form of the Cauchy-Riemann equations for these, I
>>> find:
>>>
>>> dx/dr=sin(n*phi)*cos(m*phi)
>>>
>>> while,
>>>
>>> 1/r*dy/dphi=cos(n*phi)*n*sin(m*phi)+sin(n*phi)*cos(m*phi)*m
>>>
>>> and these don't look equal. Does this mean that f(r,phi) is not
>>> holomorphic on the disk D?
>>>
>>> Am I missing something?
>>>
>>> Thanks,
>>> --
>>> I.
>>
>> I tried your task as well, guessing it is non-holomorph (and failed):
>>
>> The C-R condition means, that the Jacobi matrix is complex linear
>> But the problem is, how to recover the different coordinate system
>> (or else: how to write I*z as input using polar coordinates) and
>> it needs *care* to mix real and complex linearity working with
>> coordinates.
>>
>> Same for (complex) difference quotients (writing additions in the
>> polar coordinates is a mess).
>>
>> The last thing might be to try Morera's theorem (integral over
>> closed curves = 0 ---> analytic), but have no idea what to take
>> to show it is false, since for a circle it is true ...
>
> Axel,
>
> Here's another way to see what David is saying. Have Maple ready,
> because it will be handy, as the calculations are erudite.
>
> The initial function is given as x+y*i, with:
>
> x(r,phi)=r*sin(n*phi)*cos(m*phi)
> y(r,phi)=t*sin(n*phi)*sin(m*phi)
>
> To change to a standard polar coordinate system: (r, theta), set:
>
> r*exp(theta*i)=x+y*i, from which,
>
> theta = -i*ln(sin(n*phi)*cos(m*phi)+sin(n*phi)*sin(m*phi)*i)
>
> (In Maple:
>> x:=(r,theta,phi)->r*sin(theta)*cos(phi);
>> y:=(r,theta,phi)->r*sin(theta)*sin(phi);
>> z:=(r,theta)->r*cos(theta);
>
>> eq:=r*exp(theta*I)=x(r,n*phi,m*phi)+y(r,n*phi,m*phi)*I;
>> sol:=solve(eq,{r,theta});
>
>> theta:=rhs(sol[2,1]);
>
> )
>
> On the new polar coordinate system x+y*i = u'+v'*i, with:
>
> u'=r*cos(theta)
> v'=r*sin(theta)
>
> and now,
>
> du'/dr:
>
>> simplify(diff(r*cos(theta),r));
>
> and dv'/dtheta = dv'/dphi*dphi/dtheta:
>
>> simplify(1/r*diff(r*sin(theta),phi)/diff(theta,phi));
>
> and they are the same.
>
> (To David: Thanks for being (intentionally) terse, as this forced me to
> bang my head until I got it, instead of being spoon-fed with the answer).

Hm ... I do not want to bang my head :-)

But what about my (almost parallel) reply:

J = (real) JacobiMatrix of x() + y()*I w.r.t. (r,phi).

Then det(J) = -1/2*(-1+cos(2*n*phi))*r*m, which is 0
for phi = 2*anyInteger/n and all r,m.

Thus the critical points are not isolated and thus the
function can not be holomorphic choosing such phi.

where I computed det(J) using Maple.

I think the argument is correct (but does not give the
exact points, where the fct is (non)-holomorphic).

PS: your email bounces, wanted to reply directly ...

First | Prev | Next | Last
Pages: 1 2 3 4
Prev: JSH: So what happened?
Next: The weight of a singularity