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We consider an overdetermined set of equations, consisting of two equations for one complex variable $x$. I want to show that there are no roots for $x$ in the complex unit disc but without the origin, i.e. for $0 < |x| < 1$.

Consider the example system \begin{align} 0 &= F(x)^2 + A(x), \\ 0 &= F(x) - B(x), \end{align} where $A(x)$, $B(x)$ and $F(x)$ are all polynomials in $x$. The exact expressions for these three variables are unknown. However, we do know that $A(x)$ is at least of the order $x^2$ and $B(x)$ is at least of the order $x$. There is a possibility that $F(x) := 0$ and not a function of $x$ at all.

My proposed technique is taking the second equation and stating that $F(x) = B(x)$ and substituting this in the first equations to obtain \begin{align} 0 = B(x)^2 + A(x). \end{align}

If I can now show that the above equation has no roots in the complex unit disc, I am done. At least, that is what I thought. I can actually prove using Rouche's theorem that the above equation has a number of roots in the complex unit disc. However, none of these roots $x$ make the first two equations zero.

Clearly, there is an error in my reasoning. Could you help to point this out to me? My guess is that I am not allowed to substitute one equation to the other, but I do not know why.

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You'll need to make some further assumptions. What if $A,B$ and $F$ are all zero? Then every $x$ in the unit disc satisfies the equations. – goblin May 30 '13 at 8:40
Useful comment, thank you. I've edited the question with this information. – user60307 May 30 '13 at 10:25
You can't substitute one equation into the other because the equations do not hold for all $x$. They only hold for certain $x$, and this is the constraint you are losing when you substitute one into the other like that and "forget" about the second equation. – Antonio Vargas May 30 '13 at 16:53
Also, your constraints on $A(x)$ and $B(x)$ do not imply that there will be no roots in $|x| < 1$. Take, for example, $A(x) = (x^2-1)F(x)^2$ and $B(x) = (1-x)F(x)$. Then $x=0$ is a root of both equations. Or, if $F(x) \equiv 0$, just take $A(x) = x^2$ and $B(x) = x$. – Antonio Vargas May 30 '13 at 16:59
So, in general, you are never allowed to substitute one equation to the other in an overdetermined set of equations. Is my understanding correct? Would there be any literature or other references towards this statement? – user60307 May 31 '13 at 8:20

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