``` wu :: forums (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi) general >> complex analysis >> About essential singularity (Message started by: immanuel78 on Nov 2nd, 2006, 11:55pm) ``` 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|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. 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 thath(z) - w = p(z) ebzfor some polynomial p, and constant b.  But |h(x)-w| is bounded for x real, while|p(x)ebx| ~ |x|d eRe(b)xwhere d=deg(p), so we must have Re(b)=0 and d=0.  But this givesh(z) - w = Cebz,which is plainly impossible (since h'(z)=0, or since h is even). 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|