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Title: About essential singularity Post by immanuel78 on Nov 2nd, 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? |
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Title: Re: About essential singularity Post by Icarus on Nov 3rd, 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, e1/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. |
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Title: Re: About essential singularity Post by Eigenray on Nov 3rd, 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(z-bar)=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 Cn, with a vertical asymptote at x=n[pi], and symmetric about the x-axis, crossing it at the point xn (which satisfies tan(xn)=xn). Now, along Cn, x = y tan(x)/tanh(y), so Re h(z) = (x sin x cosh y + y cos x sinh y)/(x2+y2) = [y sin2x/cos x cosh2y/sinh y + ycos x sinh y]/(x2+y2) = y [ sin2x cosh2y + cos2x sinh2y ]/[|z|2cos x sinh y ]. As y goes to 0, x goes to xn, and h(z) goes to h(xn) = sin(xn)/xn. 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(Cn) is the interval [sin(xn)/xn, +infinity), if n is even, or (-infinity, sin(xn)/xn], if n is odd. Since sin(xn)/xn->0 as n->infinity, any non-zero real number is in h(Cn) for arbitrarily large n, and all points of Cn have norm at least n[pi]. Hence any non-zero 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. |
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Title: Re: About essential singularity Post by Eigenray on Nov 3rd, 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) ebz for some polynomial p, and constant b. But |h(x)-w| is bounded for x real, while |p(x)ebx| ~ |x|d eRe(b)x where d=deg(p), so we must have Re(b)=0 and d=0. But this gives h(z) - w = Cebz, which is plainly impossible (since h'(z)=0, or since h is even). |
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Title: Re: About essential singularity Post by immanuel78 on Nov 10th, 2006, 12:37am Let f(z)=z^(2k-1) * 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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