For a Pauli vector defined by $\vec{\sigma}=\sigma_1 \hat{x_1}+\sigma_2 \hat{x_2}+\sigma_3 \hat{x_3}$, wikipedia states that $\vec{a}\cdot\vec{\sigma}$ has two eigenvector given by
$$ \psi_+ = \frac{1}{\sqrt{2|\vec{a}|(a_3+|\vec{a}|)}} \begin{bmatrix} a_3 + |\vec{a}| \\ a_1 + ia_2 \end{bmatrix}; \qquad \psi_- = \frac{1}{\sqrt{2|\vec{a}|(a_3+|\vec{a}|)}} \begin{bmatrix} ia_2 - a_1 \\ a_3 + |\vec{a}| \end{bmatrix} ~ $$
However, these formula is invalid in the case when $\vec{a}=(0,0,-a)$ where $a$ is a positive number. I worked out other form of these eigenvectors such as
$$ \psi_+ = \frac{1}{\sqrt{2|\vec{a}|(|\vec{a}|-a_3 )}} \begin{bmatrix} ia_2-a_1 \\ a_3-|\vec{a}| \end{bmatrix}; \qquad \psi_- = \frac{1}{\sqrt{2|\vec{a}|(|\vec{a}|-a_3 )}} \begin{bmatrix} a_3 - |\vec{a}| \\ a_1+ia_2 \end{bmatrix} ~ $$
but these again are invalid in the case when $\vec{a}=(0,0,a)$.
My question is whether there exist a expression for each of these two eigenvectors that is well defined for arbitary vector $\vec{a}\in\mathbb{R}^3$?