# Infinitely Many Zeros

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.

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.
• I would rather use continuity of polynomials to conclude that the equation must hold for $z=\lambda_1$ also. – Kavi Rama Murthy Sep 11 at 9:37