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I hope this is not too long. Thanks in advance!

Edit: I edited it for a great deal, most of the information was unnecessary.

Let us define a function $h(z) = f(z) + g(z)$. We know that $f(z)$ has $K$ roots for $z$ within the complex unit circle and $g(z)$ also has $K$ roots for $z$ in the complex unit circle.

Rouche's theorem states that if $|f(z)| > |g(z)|, ~ |z| = 1$, $h(z)$ has the same number of roots in the complex unit disc that $f(z)$ has. In my case, this inequality does not always hold, it depends on other coefficients.

My question is, does it matter that $|f(z)|$ can be greater, equal, or smaller than $|g(z)|$? My gut feeling tells me that this should not change the number of roots $h(z)$ has, namely $K$.

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I assume all your functions are analytic functions - otherwise anything in the world can happen. – Greg Martin May 13 '13 at 17:14
Yes, all functions are analytic functions. – user60307 May 13 '13 at 17:23
up vote 0 down vote accepted

It absolutely matters. Anything can happen without that inequality.

For example, you might have $g=-f$, in which case $h$ is identically zero! Or, you could have $f(z) = 2z$ and $g(z) = 1-2z$, so that both $f$ and $g$ have one root inside the unit circle but $h(z) = 1$ has no roots.

On general linear algebra grounds, if you have any polynomial $h(z)$ of degree $2K$ whatsoever, then you should be able to write $h(z) = f(z) + g(z)$ where $f$ and $g$ are polynomials of degree $2K$, each of which has $K$ zeros at any points you want to specify in advance. (I admit that this construction doesn't rule out additional zeros of $f$ and $g$, but it might be illuminating nonetheless.)

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