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Topic: About essential singularity (Read 6143 times) |
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immanuel78
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About essential singularity
« on: Nov 2nd, 2006, 11:55pm » |
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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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| « Last Edit: Nov 3rd, 2006, 12:01am by immanuel78 » |
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Icarus
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Re: About essential singularity
« Reply #1 on: Nov 3rd, 2006, 3:45pm » |
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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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Eigenray
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Re: About essential singularity
« Reply #2 on: Nov 3rd, 2006, 6:21pm » |
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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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| « Last Edit: Nov 3rd, 2006, 6:26pm by Eigenray » |
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Eigenray
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Re: About essential singularity
« Reply #3 on: Nov 3rd, 2006, 6:58pm » |
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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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immanuel78
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Re: About essential singularity
« Reply #4 on: Nov 10th, 2006, 12:37am » |
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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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