Proving the polynomial $P_t(z) = \sum_{j=0}^N a_j(t) z^j$ has exactly one zero if $P_0$ has a simple zero with $a_j(t)$ and $P_t(z)$ continuous in $t$

Question

Let $P_t(z)$ be a polynomial in $z$ of degree $\leq N$ for each value of $t \in [0,1]$. Suppose that $P_t(z)$ depends continuously on $t$ in the sense that $$P_t(z) = \sum_{j=0}^N a_j(t) z^j$$ and each $a_j(t)$ continuous for $t \in [0,1]$.

1. Suppose $z_0$ is a simple zero of $P_0$. Show that there exist $r>0$ and $t_0>0$ s.t. $\forall t<t_0$ the polynomial $P_t$ has exactly one zero in $D_r(z_0)$.
2. What can you say if $z_0$ is a zero of order $k \geq 2$ of $P_0$?

I tried to solve question 1 but can't get the rigorous proof, this is my idea: I want to use Rouche's theorem, thus need to find $f$ and $g$ such that $|f(z)|>|g(z)|$ for all $z \in C$ , so that $f$ and $f+g$ have the same number of zeros inside the circle $0$.

Let $f = P_0$ and $g = P_0-P_t$, then let $I = \{j = 0,1,...,N | a_j \neq 0\}$, then let $\epsilon = \min_{j \in I} {{a_j(0)}}>0$. Since $a_j(t)$ is continuous for every t, we know that there exists a $t_0$ such that for all $t \in [0,t_0]$ $|a_j(0)-a_j(t)|<\frac{\epsilon}{N}$. Then we get the following (with $g(z) = P_t-P_0$), \begin{align} |g(z)| &= |\sum_{j=0}^N(a_j(0)-a_j(t))z^j|\\ &\leq \sum_{j=0}^N|a_j(0)-a_j(t)|\,|z^j|\\ &\leq \sum_{j=0}^N\frac{\epsilon}{N}\,|z^j|\\ &= \epsilon \sum_{j=0}^N|z^j| \end{align}

But I can't seem to get that it is smaller than $P_0(z) = \sum_{j=0}^N a_j(0) z^j$, so maybe I need to choose my $\epsilon$ differently, I also noticed that I don't use $r>0$ and the disk $D_r(z_0)$ anywhere, as well as the fact that the zero is simple which makes me think I'm overlooking something.

Can somebody help me with these 2 questions?

Answer 1

For problem 1) and 2) Rouche's theorem does apply here, but you need to choose $\epsilon$ differently.

More precisely, by the discreteness of zero, for any sufficiently small disk $D_{r}(z_0)$ the polynomial $P_0(z)$ have the unique zero $z_0$ on $D_r(z_0)$. In particular, $P_0(z)$ does not vanish on the boundary $\partial D_r(z_0)$, so by compactness $\lvert P_0(z) \rvert$ is lower bound on the boundary. Let $\epsilon > 0$ be smaller than this lower bound.

Then by continuity of coefficients, for sufficiently small $t_0$ the function $\lvert P_{t_0}(z) - P_0(z) \rvert$ will have value less than $\epsilon$ on the boundary $\partial D_r(z_0)$. So Rouche's theorem applies, and $P_{t_0}(z)$ has the same number of zeros as $P_0(z)$ inside $D_r(z_0)$.

Thus we conclude that for sufficiently small $t_0$, $P_{t_0}(z)$ has the same number of zeros inside $D_r(z_0)$ as the multiplicity of $P_0$'s zero at $z_0$.