"Unit-Circle" Equivalent of Matlab's roots()?

Is there a function in Matlab that will take as input a complex
polynomial and return the points on the unit circle at which the
polynomial goes to zero?
-- 
%  Randy Yates                  % "...the answer lies within your soul
%% Fuquay-Varina, NC            %       'cause no one knows which side
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%%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*, ELO
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"Randy Yates" <yates@ieee.org> wrote in message
news:llfb6w5k.fsf@ieee.org...
> Is there a function in Matlab that will take as input a complex
> polynomial and return the points on the unit circle at which the
> polynomial goes to zero?

Randy,

I don't think so.  Not all polynomials have zeros on the unit circle - in
fact, not many unless the polynomical represents a filter design with a stop
band.

You can use something like "roots", convert them to polar coordinates and
then search for magnitudes at or near 1.0.

Fred

"Randy Yates" <yates@ieee.org> wrote in message
news:llfb6w5k.fsf@ieee.org...

> Is there a function in Matlab that will take as input a complex
> polynomial and return the points on the unit circle at which the
> polynomial goes to zero?

Have you tried the 90&#4294967295; rotation on the Riemann sphere
w = (-i*z+1)/(z-i)
?  You could make this substitution for P(w), multiply
by (z-i)^length(P), isolate the real and imaginary parts
of the resulting polynomial, and find roots of both parts.
A common real root z could then be converted to the
corresponding w on the unit circle.

-- 
write(*,*) transfer((/17.392111325966148d0,6.5794487871554595D-85, &
6.0134700243160014d-154/),(/'x'/)); end

>>>>> "Randy" == Randy Yates <yates@ieee.org> writes:

    Randy> Is there a function in Matlab that will take as input a complex
    Randy> polynomial and return the points on the unit circle at which the
    Randy> polynomial goes to zero?

No, I don't.  In addition to what others have said, you could do the
following, if P(z) is your polynomial.

Let z = exp(j*w), since you want roots on the circle.  If there are
roots on the circle, P(exp(j*w)) = 0.  Expand P(exp(j*w)) out in terms
of sin(w*n) and cos(w*n).  Since sin(w*n) and cos(w*n) can be written
as sums of powers of sin(w) and/or cos(w), you will have a polynomials
in powers of sin(w) and cos(w).  Solve one of these, and find possible
values of w.  Use these values in the other verify that it is a
solution.

Straight-forward, I think, but not builtin.  There might be some
gotcha's, though---I didn't work it all out in detail.

Ray