Holomorphic?
in [Math]
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From: I.N. Galidakis on 27 Jul 2010 09:55 Axel Vogt wrote: > 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 :-) There's no math without head banging :-) > 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). I don't know, because I cannot see what idea you are using behind your argument. Perhaps if you explain the main idea in one-two words, I will be able to see it. > PS: your email bounces, wanted to reply directly ... If you mean the address: morpheus(a)olympus.mons, that's a fake address I use with all my posts, because of spam problems. -- I.
From: Axel Vogt on 28 Jul 2010 14:46
I.N. Galidakis wrote: > Axel Vogt wrote: >> 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. .... >> 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). > > I don't know, because I cannot see what idea you are using behind your > argument. Perhaps if you explain the main idea in one-two words, I will > be able to see it. .... I was to sloppy it seems: The fct is holomorphic iff the (real) Jacobi-Matrix in the usual coordinates is complex linear for the chosen point (x,y). And in that case it is the complex derivative (one has to take care what that should mean). Now a holomorphic fct has only isolated critical points. My fault was: those are the critical points of the real fct, which would be det(J) = 0. But one needs J = 0. |