What conditions would guarantee $\Psi: \mathbb{R}^n \rightarrow \mathbb{R}^2$ to be non-zero in a small neighbourhood? Suppose I have $\Psi: \mathbb{R}^n \rightarrow \mathbb{R}^2$ where 
$\Psi(\mathbf{0}) = \mathbf{0}$. I would like to show that there exists a small open set $U$ around $\mathbf{0}$ such that it is non-zero for all points in $U \backslash \{ \mathbf{0} \}$. 
I am wondering what kind of conditions on $\Psi$ would ensure this is satisfied? Any comments are appreciated. Thank you.  

Here is a relevant fact: 

Theorem Suppose that $\psi\colon \mathbb{R}^n \to \mathbb{R}$ is $C^3$, satisfies $\psi(\mathbf{0}) = 0$, and that the matrix of second partial derivatives $\left[\frac{\partial^2 \psi}{\partial x_i\,\partial x_j}(\mathbf{0})\right]$ is positive-definite.  Then $\psi$ has a local minimum at $\mathbf{0}$, meaning there is an open neighborhood $U$ of $\mathbf{0}$ such that $\psi(\mathbf{x}) > 0$ for all $\mathbf{x} \in U$, $\mathbf{x} \neq \mathbf{0}$.  Similarly, if the matrix of second partial derivatives is negative definite, then $\psi$ has a local maximum at $\mathbf{0}$.

The proof requires Taylor's theorem and some linear algebra.  
Now suppose $\mathbf{\Psi}\colon \mathbb{R}^n \to \mathbb{R}^2$ is of class $C^3$, and $\mathbf{\Psi}(\mathbf{0}) = \mathbf{0}$.  Writing $\mathbf{\Psi} = (\psi_1,\psi_2)$, suppose further that the matrix of second partial derivatives $\left[\frac{\partial^2 \psi_k}{\partial x_i\,\partial x_j}(\mathbf{0})\right]$ is positive-definite or negative-definite for each $k$.  Then applying the theorem to each component gives neighborhoods $U_1$ and $U_2$ of $\mathbf{0}$ for which $\psi_k(\mathbf{x}) \neq 0$ for all $\mathbf{x} \in U_k \setminus\{\mathbf{0}\}$.  Let $U = U_1 \cap U_2$.
