For $\lambda_1=-1+4i:$ $$V_{\lambda_1}=\ker (A-\lambda_1 I)\equiv\left \{ \begin{matrix} -4ix_1+4x_2=0\\ -4x_1-4ix_2=0\end{matrix}\right.\qquad (*)$$ As $\lambda_1$ is simple, $\dim V_{\lambda_1}=1$ and $v=(i,-1)\neq (0,0)$ is a solution of $(*).$ As a consequence, $\{v\}$ is a basis of $V_{\lambda_1}.$

On the other hand, a basis of $\lambda_2=-1-4i$ is $\bar{v}=(-i,-1)$ according to a well known property.