Closed Form of Complex Function

Date: 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.