Closed Form of Complex FunctionDate: 03/24/2003 at 12:05:03 From: Richard H. Warren Subject: Complex Variables I would like a closed form (not a power series) for f(z) such that f is analytic and f(z) = 0 when z = (k*pi)^3, z = ((k*pi)^3)e^i*2*pi/3, and z = ((k*pi)^3)e^i*4*pi/3 where k is a positive integer. A geometric description of the zeros is that they occur on three rays and that there are 3 zeros on circles of radius (k*pi)^3. If z is not described above, then f(z) is not 0, except possibly f(0) may be 0. Note that k is any positive integer in the description of the problem. This means that each of the 3 rays has an infinite number of zeros. Date: 03/24/2003 at 12:12:31 From: Doctor Douglas Subject: Re: Complex Variables Hi, Richard, Thanks for submitting your question to the Math Forum. f(z) = sin(z^(1/3)) actually is very close to our needs. Since the argument of a negative real number is pi, when this is plugged into f(z) the argument will become pi/3, and so these do *not* form zeroes of f (because sin(w) only has roots on the real-w axis). We can choose the argument of z such that -pi < Arg(z) <= pi. This will force the cube root to an even smaller range of angles around zero, i.e., -pi/3 < Arg(z^(1/3)) <= pi/3, and so f(z) only has zeroes on the positive real axis (and the origin) at values z = (k*pi)^3 for k a nonnegative integer. That takes care of the roots on the real axis. Now we can also multiply f(z) by a function that has similar roots on the ray at Arg(z) = 2*pi/3. We know that for numbers on this ray, Arg(z^(1/3)) will be 2*pi/9, and we can rotate these onto the real axis by a multiplication with exp(-i*2*pi/9): g(z) = sin[exp(-2*pi*i/9)*z^(1/3)]. And you can develop a similar expression for h(z) having zeroes on the ray at Arg(z) = -2*pi/3. Then the final function F(z) = f(z)*g(z)*h(z) will have the requisite zeroes at |z|=(k*pi)^3, Arg(z)={0,+2*pi/3, -2*pi/3}, and k a nonnegative integer. It seems clear to me that except possibly at the origin, where F(0)=0, F is analytic. You can remove this special point from the domain of F, if you wish. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ Date: 03/25/2003 at 14:59:20 From: Richard H. Warren Subject: Thank you (Complex Variables) Thank you very much for the answer. I understand it. |
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