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I recently came across the following argument regarding the uniqueness of the zeros of a complex polynomial.

Please note that the proof that a complex polynomial of degree $m$ has $m$ zeros has been established at this point. The following is also not the complete proof of the uniqueness of the roots. I just want to focus on a particular passage from the proof that I need help clarifying.

Consider the following equation of two factorizations of the same complex polynomial of degree $m$, where $z, \lambda_i, \tau_i\in\mathbb{C}$, for $i=1\dots m$:

$$\left(z - \lambda_1\right)\left(z - \lambda_2\right)\dots\left(z - \lambda_m\right) = \left(z - \tau_1\right)\left(z - \tau_2\right)\dots\left(z - \tau_m\right)$$

Therefore, all $\lambda_i$ and $\tau_i$ are the zeros of the polynomial. Moreover, substituting $z=\lambda_i$, the resulting equation implies that $\lambda_i = \tau_j$ for some $j \in \left\{1\dots m\right\}$. To make it simple, let's relabel $\tau_j$ so that $\lambda_i = \tau_i$.

Now, consider $i = 1$. Dividing both sides by $z - \lambda_1$, we get

$$\left(z - \lambda_2\right)\dots\left(z - \lambda_m\right) = \left(z - \tau_2\right)\dots\left(z - \tau_m\right)$$

for all $z\in\mathbb{C}$ except possibly $z = \lambda_1$.

So far, with the exception of the word "possibly", it has been straight-forward and obvious to me. However, what comes next in the argument puzzles me:

"Actually, the equation above [after dividing by $z - \lambda_1$] holds for all $z\in\mathbb{C}$ because, otherwise, by subtracting the right side from the left side, we would get a non-zero polynomial that has infinitely many zeros."

That passage is a part of the proof of theorem 4.14 in "Linear Algebra Done Right", third edition (S. Axler).

Why would the alternative case imply a non-zero polynomial with infinitely many roots? Could someone please kindly show me?

Thank you.

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If $f(z)=g(z)$ for every $z$ with the possible exception of $\lambda_1$ then $f(z)-g (z)=0$ whenever $z \in \mathbb C \setminus \{\lambda_1\}$. The set $\mathbb C \setminus \{\lambda_1\}$ is an infinite set so $f(z)-g(z)=0$ has infinitely many solutions. This cannot happen if $f$ and $g$ are polynomials.

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  • $\begingroup$ I would rather use continuity of polynomials to conclude that the equation must hold for $z=\lambda_1$ also. $\endgroup$ – Kavi Rama Murthy Sep 11 at 9:37
  • $\begingroup$ Ah, yes. It's that simple. Thank you! My knowledge in Mathematics is very shallow. I don't know enough about continuity to prove it. $\endgroup$ – Niko Gambt Sep 11 at 9:54

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