

Title: About essential singularity Post by immanuel78 on Nov 2^{nd}, 2006, 11:55pm Let f(z)=z*cos(1/z) and g(z)=z*sin(1/z) Then f and g have an essential singularity at z=0. This is not difficult to me. Then detremine A=f({z:0<z<d}) and B=g({z:0<z<d}) for arbitrarily small values of d. By Picard's theorem, A and B are C(complex number) except for at most one point. I also want to know this one point if the point exists. How can I get it? 

Title: Re: About essential singularity Post by Icarus on Nov 3^{rd}, 2006, 3:45pm Generally, you find the missing points by examining the functions and figuring out what they would miss. For instance, it is fairly obvious that in every neighborhood of 0, e^{1/z} is not zero. For your functions, I do not believe there is a missed point, so the answer to your first question is A = B = C. 

Title: Re: About essential singularity Post by Eigenray on Nov 3^{rd}, 2006, 6:21pm Intuitively, no single point is special enough to be the only value missed by either f or g. Since f is odd, this is easy to make rigorous: If f missed some point w, then since f(z)=f(z), it would have to miss w as well. So the only point f could miss is 0, but it hits that at 1/[(n+1/2)pi]. Since g is even, it's a little trickier. But since g(zbar)=g(z)bar, if g missed a single point it would have to be real. For convenience work with h(z)=g(1/z)=sin(z)/z. We need to show that any real number is hit by h(z) for some arbitrarily large z. h(x+iy) = (sin x cosh y + i cos x sinh y)/(x+iy), so Im h(z) = (x cos x sinh y  y sin x cosh y)/z^{2}, and we see h(z) is real whenever tan(x)/x = tanh(y)/y. As y goes from 0 to +/ infinity, tanh(y)/y goes from 1 to 0. As x goes from n[pi] to (n+1/2)[pi], tan(x)/x goes from 0 to +infinity. Thus for any given y, there is a unique x, n[pi]<x<(n+1/2)[pi], with h(x+iy) real. The set of such (x,y) is a single curve C_{n}, with a vertical asymptote at x=n[pi], and symmetric about the xaxis, crossing it at the point x_{n} (which satisfies tan(x_{n})=x_{n}). Now, along C_{n}, x = y tan(x)/tanh(y), so Re h(z) = (x sin x cosh y + y cos x sinh y)/(x^{2}+y^{2}) = [y sin^{2}x/cos x cosh^{2}y/sinh y + ycos x sinh y]/(x^{2}+y^{2}) = y [ sin^{2}x cosh^{2}y + cos^{2}x sinh^{2}y ]/[z^{2}cos x sinh y ]. As y goes to 0, x goes to x_{n}, and h(z) goes to h(x_{n}) = sin(x_{n})/x_{n}. As y goes to +/ infinity, x goes to n[pi], and it's easy to see that h(z) goes to (1)^{n}*infinity. Thus h(C_{n}) is the interval [sin(x_{n})/x_{n}, +infinity), if n is even, or (infinity, sin(x_{n})/x_{n}], if n is odd. Since sin(x_{n})/x_{n}>0 as n>infinity, any nonzero real number is in h(C_{n}) for arbitrarily large n, and all points of C_{n} have norm at least n[pi]. Hence any nonzero real number is of the form h(z) for arbitrarily large z, or equivalently, g(z) for arbitrarily small z. And of course 0=g(1/(n pi)) as well. So A = B = C. 

Title: Re: About essential singularity Post by Eigenray on Nov 3^{rd}, 2006, 6:58pm Actually, there's a much easier solution. Suppose that for some w, and some R>0, there is no z with h(z)=w and z>R. Then the zeros of h(z)w are contained in { z: z < R }, hence are finite in number. Since log h(z) = O(z), it follows from the strong version of the Weierstrass factorization theorem that h(z)  w = p(z) e^{bz} for some polynomial p, and constant b. But h(x)w is bounded for x real, while p(x)e^{bx} ~ x^{d} e^{Re(b)x} where d=deg(p), so we must have Re(b)=0 and d=0. But this gives h(z)  w = Ce^{bz}, which is plainly impossible (since h'(z)=0, or since h is even). 

Title: Re: About essential singularity Post by immanuel78 on Nov 10^{th}, 2006, 12:37am Let f(z)=z^(2k1) * sin(1/z) for k in Z(integer). Let A={z:0<z<d} for arbitrarily small values of d. Then f(A) seems to be C(complex number) How can I know it? ps) I don't know Weierstrass factorization theorem so that I want to know the solution without this theorem. 

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