Zeros of $f + t g$ goes to the zeros of $g$ as $t \to \infty$?

Question

Suppose $f, g \in \mathbb C[x]$ with $\deg(f) > \deg(g)$. Without loss of generality, let $\deg(f) = n$ and $\deg(g) = n-1$. I read a statement without proof: If $h(x) = f(x) + t g(x)$ with $t \in [0, \infty)$, the zeros of $h$ will start from the zeros of $f(x)$ ($t=0$) and converge to the zeros of $g$ ($t \to \infty$) and of course, one of them will tend to infinity. Intuitively, I understand this should be true since if $t$ is sufficiently large, the zeros should be the ones of $g(x)$. I am wondering what a rigorous way to argue this point?

Answer 1

Let $a$ be a zero of $g$ and take $r>0$ such that $g(z)\ne0$ if $0<|z-a|\le r$. Then, using Rouché-Frobenius, it is possible to prove that for $t$ large enough, $g$ and $f+t\,g$ have the same number of zeros on $\{|z-a|<r\}$. Letting $r\to0$, this shows that $f-t\,g$ has $n-1$ zeroes as close to the zeroes of $g$ as desired.