Re: Number of roots of an algebraic function

"John L. Barber" <jlbarber@xxxxxxxx> writes:

A quick question for the math wallahs here from a poor, ignorant
physicist. A pair of questions, really, with the second being the
more important.

Suppose we have an algebraic function f(z), where all of the parameters
(by which I mean numbers) in the function are real.

Since you're a poor, ignorant physicist you will be pardoned if you
don't know that there's a technical meaning for "algebraic function".
Specifically, a (genuine, i.e., not multivalued) function f of a
complex variable z in a domain D in the complex numbers C is
*algebraic* if there is a complex POLYNOMIAL function F of two
variables z, w such that F(z,f(z)) = 0 for all z in D. More
classically, the "multivalued function" which associates to
each z in a domain D in C the set-with-multiplicities of complex
numbers w such that F(z,w) = 0 is a "complete algebraic function"
(of which the original f is a "branch", at least, if it's
continuous--which probably should have been required of f
earlier). Can I assume that your condition "all of the parameters
....in the function are real" is equivalent to "there exists such
an F which is a function with REAL coefficients"?

In that case, yes, complex roots of the complete algebraic
function come in conjugate pairs.

2) What results are there regarding the number of roots of such
a function f(z)?

We should take the "simplest" F that works for your given f.
Then write F(z,w) as a polynomial in z with coefficients that
are polynomials in w; e.g., F(z,w) = (w^2-1)z^3 + wz + 1.
A root of the complete algebraic function is a value z=z_0 such
that F(z_0,0)=0; in this example F(z,0)=-z^3+1, which has 3
roots, and in general the number of roots (counting multiplicities,
and making technical corrections when the "leading coefficient"
is divisible by w) is exactly the highest power of z that appears
in F(z,w).

Specifically, for the problem I have in mind, I have a function
f(z) that is (I'm not quite sure how to describe this) like a
polynomial in z, except with a few multiplicative factors of
sqrt(1 + a z) thrown in here and there, where "a" is a positive constant.

Okay. Suppose we have the expression f(z)=z^2+ sqrt(1+az)z + 1.
Call that w. Then (w-z^2-1)^2=(1+az)z^2, so we can take F(z,w)=
(w-z^2-1)^2-(1+az)z^2. The highest power of z that appears in F(z,w)
is the fourth power, and the coefficient of z^4 is 1, so there are
exactly four roots.

Lee Rudolph
.

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