Difficulty proving a statement about polynomials using the argument principle. Hello I am having some difficulty with the following problem, I believe it is meant to be a rather simple exercise. 
Suppose that $P_{t}(z)$ is a continuous polynomial in $z$ for any fixed $0 \le t \le 1$ and that it is continuous in $t$, in the sense that the degree of $P_t(z)$ is bounded independent of $t$ and the coefficients are continuous functions of $t$.
Let $Z=\{(z,t):P_{t}(z)=0\}$ (note that it is closed in $\mathbb{C} \times [0,1]$ ) if $P_{t_{0}}(z_{0})=0$ and its partial derivative with respect to $z$ is not zero at $z_{0}$ then use the argument principle to show that there exists an $\epsilon \gt 0$ such that for t sufficiently close to $t_{0}$ there is a unique $z$ in the open disk with centre $z_{0}$ and radius $\epsilon$ with $P_{t}(z)=0$ , moreover what can be said about $P_{t_{0}}$ if it vanishes to order k at $z_{0}$.
So I am not sure how to proceed.
Here is what I have thought:
Well the polynomial are certainly holomorphic, so I have a result that says that $(1/2\pi)$ times the contour around the boundary integral of $\frac{p'}{p}$ will be equal to the sum of the multiplicities of the zeros minus the sum of the multiplicities of the poles. But I dont understand how I can use this to show what is being asked. Ie, I get what the general theorem states, but I am not sure how it relates to this question
 A: $P_{t_0}(z)$ has a (simple or multiple) zero at $z_0$. Thus there is an $\epsilon$ so that the disk of radius $\epsilon$ has no zero of $P_{t_0}(z)$ in its interior or its boundary (other than $z_0$). Simply because zeros are isolated. 
Now the claim is that there is a $\delta$ such that for $|t-t_0|<\delta$ the polynomial $P_t(z)$ has no zero on the boundary of the circle. This is just compactness, one could say that if there were $t_n\to t_0$ with zeros $z_n$ on the boundary then some subsequence of those zeros would converge to a zero of $P_{t_0}$.
Therefore for $|t-t_0|<\delta$ the function
$$n(t)=\frac{1}{2\pi i}\int_C\frac{P^{\prime}_t(z)}{P_t(z)} dz$$ is well defined and is a continuous integer valued function of $t$.
Thus this function is constant and the polynomials $P_t(z)$ all have the same number of zeros inside the circle.
Note that in the case of a zero of multiplicity greater than $1$, this zero may split into many zeros, but the sum of their multiplicities will be constant.
