Holomorphic?
in [Math]
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From: I.N. Galidakis on 19 Jul 2010 01:11 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.
From: David C. Ullrich on 19 Jul 2010 06:45 On Mon, 19 Jul 2010 08:11:36 +0300, "I.N. Galidakis" <morpheus(a)olympus.mons> 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? Yes. You're using the "polar form of the C-R equations". That's the form they take in standard polar coordinates. But your r and phi are _not_ standard polar coordinates! 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". >Thanks,
From: I.N. Galidakis on 19 Jul 2010 12:57 David C. Ullrich wrote: > On Mon, 19 Jul 2010 08:11:36 +0300, "I.N. Galidakis" > <morpheus(a)olympus.mons> 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? > > Yes. You're using the "polar form of the C-R equations". That's > the form they take in standard polar coordinates. But your > r and phi are _not_ standard polar coordinates! > > 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". If I understand what you are saying, if I set: theta=m*phi R=r*sin(n/m*theta) then, under this transformation, x=R*cos(theta) y=R*sin(theta) I will find that the C-R equations (for R and theta) are satisfied? Right? Thanks, >> Thanks, -- I.
From: David Bernier on 19 Jul 2010 14:53 I.N. Galidakis wrote: > David C. Ullrich wrote: >> On Mon, 19 Jul 2010 08:11:36 +0300, "I.N. Galidakis" >> <morpheus(a)olympus.mons> 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? >> >> Yes. You're using the "polar form of the C-R equations". That's >> the form they take in standard polar coordinates. But your >> r and phi are _not_ standard polar coordinates! >> >> 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". > > If I understand what you are saying, if I set: > > theta=m*phi > R=r*sin(n/m*theta) > > then, under this transformation, > > x=R*cos(theta) > y=R*sin(theta) > > I will find that the C-R equations (for R and theta) are satisfied? Right? > > Thanks, > >>> Thanks, I looked at your question yesterday. One special case that is easier is m = 0; then x(r, phi) = r*sin(n*phi), y(r, phi) = 0 . So f(r,phi) = r*sin(n*phi) + 0*i . I'm not sure what phi is, though. David Bernier
From: David C. Ullrich on 20 Jul 2010 08:33
On Mon, 19 Jul 2010 19:57:08 +0300, "I.N. Galidakis" <morpheus(a)olympus.mons> wrote: >David C. Ullrich wrote: >> On Mon, 19 Jul 2010 08:11:36 +0300, "I.N. Galidakis" >> <morpheus(a)olympus.mons> 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? >> >> Yes. You're using the "polar form of the C-R equations". That's >> the form they take in standard polar coordinates. But your >> r and phi are _not_ standard polar coordinates! >> >> 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". > >If I understand what you are saying, if I set: > >theta=m*phi >R=r*sin(n/m*theta) > >then, under this transformation, > >x=R*cos(theta) >y=R*sin(theta) > >I will find that the C-R equations (for R and theta) are satisfied? Right? No. If R=r*sin(n/m*theta) then R and theta are not independent. >Thanks, > >>> Thanks, |