# Why does $q(z)$ not have any zeros for all $z$ sufficiently near $z_0$?

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.

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.
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.