Question: 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.

Answer: 
Here is a relevant fact: 

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> **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 $\mathbf{0}$.
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> 
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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 $U = U\_1 \cap U\_2$.
Let $h(x)=\|\Psi(x)\|^2\_2$. Then $h(x)=0$ iff $\Psi(x)=0$ and $h:\mathbb{R}^n \to \mathbb{R}$.

If $\Psi \in C^2$, $\nabla h(0)=0$, and $\text{Hess}\_0 h$ (the Hessian of $h$ at $x=0$) is positive definite. Then $0$ is a local minimum. Hence there is no other point such that $h(x) = 0 $ (i.e. $\Psi(x)=0$) for $x$ in a neighborhood of $0$.