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Show that if $z_0$ is a zero of $p_{0}(z)$ of order $d$, then for all $z$ sufficiently near $z_0$, there are positive constants $c_1$ and $c_2$ such that $c_1|z-z_0|^d\le |p_{n}(z)|\le c_2|z-z_0|^d$

In the reasoning for the solution of the above exercise, I'm told that $p(z)=(z-z_0)^{d}q(z)$ and $q(z)$ is a polynomial with no zeros for all $z$ sufficiently near $z_0$.

So since $q(z)$ is a polynomial in the $z$-plane, it follows that $q(z)$ is continuous in the $z$-plane. But why can $q(z)$ not have any zeros for all $z$ sufficiently near $z_0$? If it were to have zeros sufficiently near $z_0$ and since $q(z)$ is continuous, that means that $q(z_1)$, where $z_1$ is a zero near $z_0$, equals $0$. I think that we would eventually arrive at the conclusion that $z_0$ is a zero of $q(z)$, which is a contradiction. But, I don't really know how to show this step-by-step and using tools from analysis.

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Hints: Basically you can show in general that if $f(z)$ is continuous at $z_0$ and $f(z_0) = a$ where $a\ne 0$ (i.e. $f(z_0)\ne 0$), then there is a neighbourhood around $z_0$ where $f(z)$ is never $0$. Try using $\varepsilon$-$\delta$ definition of continuity, for example, along with triangle inequalities. Use the fact that $$\color{blue}{|f(z)| \ge |a| - |f(z) - a|}$$ (because $\color{blue}{|u-v| \ge \left| |u| - |v|\right| \ge |v| - |u|}$) and there is some neighbourhood of $z_0$ such that $|f(z)-a| < |a|/2$ for all $z$ in this neighbourhood.

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Since $q(z)$ is non-zero polynomial, it can only have finitely many zeros (If it would of had more zeros than it's degree, then we would get a homogenous linear system of equations with more equations than variables, which would imply that every coefficient of our polynomial is zero, so that $q$ would be zero). Just set $\varepsilon = \min\{|z_0-z|: q(z) = 0\}$. From our assumption $\varepsilon > 0$. Then $q$ doesn't have any zeros for $z$ such that $|z-z_0|<\varepsilon $.

Another approach would be to work with continuity of $q$, but since it's a polynomial, our argument is simpler.

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