Existence of a neighborhood in ${C}^4$ of $(1,1,1,1)$ such that if $a\in U$ then $p_a(z)$ has a root $r(a)$ close to -1. Here $p_a(z)=a_0+a_1z+a_2z^2+a_3z^3$ is a polynomial with coefficient vector $a=(a_0,a_1,a_2,a_3)$. Given that $p_{(1,1,1,1)}(z)$ has a simple root at $z=-1$, we want to show there is a neighborhood $U\subset\mathbb{C}^4$ of $(1,1,1,1)$ such that if $a\in U$ then $p_a(z)$ has a unique root $r(a)$ close to $-1$.
Moreover show that if the neighborhood $U$ is sufficiently small then $a\mapsto r(a)$ is a continuous function on $U$.
I would like to use Rouche's theorem, but am not sure how to deal with $U$ in $\mathbb{C}^4$.
 A: Let us denote by $v\in\mathbb{C}^4$ a $4$-tuple of coefficients and by $u$ the $4$-tuple $(1,1,1,1)$. We know that $p_u(z)$ ha a simple root at $z=-1$, in fact $p_u(z)=(z+1)(z^2+1)$ so it has $3$ simple roots, namely $-1, i, -i$; in particular, if we consider the compact set (a disc) $K=\{z\in\mathbb{C}\ :\ |z+1|\leq 1\}$, the polynomial $p_u(z)$ has only one root (counted with multiplicity) in $K$.
Now, take $v\in\mathbb{C}^4$, such that $\|u-v\|< \epsilon$ (with respect to the usual hermitian structure on $\mathbb{C}^4$); then $q(z)=p_v(z)-p_u(z)=b_0+b_1z+b_2z^2+b_3z^3$ with $|b_j|<\epsilon$. So, given $z\in bK$, $|q(z)|<\epsilon(1+|z|+|z|^2+|z|^3)<15\epsilon$ (as, at most, if $z\in bK$, $|z|=2$) and, again if $z\in bK$,
$$|p_u(z)|=|(z+1)(z^2+1)|=|z^2+1|=|z+i||z-i|>1/\sqrt{2}\;.$$
So, if $15\epsilon<1/\sqrt{2}$, i.e. if $\epsilon<\sqrt{2}/30$ ($\sim 0.047\ldots$), for every $z\in bK$ we have
$$|p_u(z)-p_v(z)|<|p_u(z)|$$
therefore, by Rouché's theorem, $p_u$ and $p_v$ have the same number of zeroes in $K$, that is, only $1$.

I should say that the estimate given above is incredibly rough. You can do much better, for example, by writing everything with respect to the root you already know and trying and estimating the ratio $(p_v-p_u)/p_u$.
Consider a polynomial $p_v(z)$, with $v=(a_0,a_1,a_2,a_3)$, and suppose that $\|v-u\|<\epsilon$ is not "too large", whatever that means, then the Euclidean division algorithm gives
$$p_v(z)=a_3p_u(z)+(a_2-1)z^2+(a_1-1)z+(a_0-1)$$
so
$$\frac{|p_v(z)-p_u(z)|}{|p_u(z)|}\leq |a_3-1|+\frac{|(a_2-1)z^2+(a_1-1)z+(a_0-1)|}{|z+1||z^2+1|}\;.$$
If $|z+1|=1$, i.e. if $z\in bK$, then we reduce ourselves to
$$\left|\frac{(a_2-1)z^2+(a_1-1)z+(a_0-1)}{z^2+1}\right|=\left|(a_2-1)+\frac{(a_1-1)z+(a_0-1)-(a_2-1)}{z^2+1}\right|$$
which is bounded by
$$|a_2-1|+4\sqrt{2}\|u-v\|$$
so
$$\frac{|p_v(z)-p_u(z)|}{|p_u(z)|}\leq 2(1+2\sqrt{2})\epsilon\;.$$
We can apply Rouché's theorem if $\epsilon<\frac{2\sqrt{2}-1}{14}$ ($\sim 0.13\ldots$)

So, there exists an open set $U$ of the form
$$U=\{v\in\mathbb{C}^4\ :\ \|v-u\|\leq \epsilon_0\}$$
such that $v\in U\Rightarrow p_v(z)$ has a simple root in $K$, let us call it $r(v)$.
As for the continuity, consider the compact set $K_\delta=\{z\in\mathbb{C}\ :\ |z+1|<\delta\}$. We can perform again the same computations (the rough ones or the refined ones), taking into account that, if $z\in bK_\delta$, then
$$|z^2+1|>\mathrm{dist}(i,bK)*\mathrm{dist}(-i,bK)=\left(\sqrt{2}-\delta\right)^2\;.$$
So,for example, the rough estimate becomes
$$|q(z)|<\epsilon(1+(1+\delta)+(1+\delta)^2+(1+\delta)^3)=\epsilon((1+\delta)^4-1)/\delta$$
$$|p_u(z)|>\delta\left(\sqrt{2}-\delta\right)^2$$
Therefore, $p_u$ and $p_v$ have the same number of roots in $K_\delta$ whenever
$$\epsilon<\delta^2\frac{(\sqrt{2}-\delta)^2}{(1+\delta)^4-1}=g(\delta)$$
where $g:[0,1]\to\mathbb{R}_{\geq 0}$ is a continuous function with $g(0)=0$.
So, for every $\delta\in(0,1]$ there exists $\epsilon>0$ such that, if $\|u-v\|<\epsilon$, then $|r(u)-r(v)|<\delta$.
This shows that $r$ is continuous in $u$. To show that $r$ is continuous in $U$, we repeat this reasoning for two $4$-tuples $v, v'\in\mathbb{C}^4$ which, however, is quite painful!

The "right" way to do it
You have a continuous map $v\mapsto p_v(z)$ from $\mathbb{C}^4$ to the polynomials of degree $\leq 3$, with respect to the compact-open topology on the polynomials, i.e. for every $v\in\mathbb{C}^4$, given $F\Subset \mathbb{C}$ compact and $\eta>0$, we can find $\epsilon=\epsilon(v,F,\eta)>0$ such that, if $\|v-v'\|<\epsilon$, then 
$$\max_{z\in F} |p_v(z)-p_{v'}(z)|<\eta\;.$$
Therefore, given $F$ such that $p_v(z)\neq 0$ if $z\in F$, we can consider
$$m=\min_{z\in F} |p_v(z)|$$
and ask that $\eta<m$. Then we obtain $\epsilon$ such that, if $\|v-v'\|<\epsilon$, then $|p_v(z)-p_{v'}(z)|<|p_v(z)|$ for all $z\in F$.
We apply this with $F=bK=b\{z\in\mathbb{C}\ :\ |z+1|<1\}$ and we obtain the first part and an open set $U$ around $(1,1,1,1)$ in $\mathbb{C}^4$ such that $p_v(z)$ has a simple root in $K$.
We apply this with $F=bK_{\delta}(v)=b\{z\in\mathbb{C}\ :\ |z+r(v)|<\delta\}$ and we obtain the second part, i.e. the continuity of $r$ in a possibly smaller open set $U$.
