For proper node or improper node -> node or spiral point how can you shift to get a spiral point? I thought spiral points were for complex eigenvalues. And you can't have only one eigenvalue in the complex case, so how do complex eigenvalues fall under this case?

Consider system $x'=f(x,y)$, $y'=g(x,y)$. If $f,g\in C^2$ (twice continuously differentiable, this requirement could be weakened significantly, then rather simple arguments show that node remains a node and its type (PN or IN) is preserved. But if $f,g\in C^1$ (once continuously differentiable) this is no longer true.

*Example*\begin{equation}\begin{aligned}

&x'=-x-yf( r),\\

&y'=-y+xf( r)

\end{aligned}\label{K}\end{equation}

with $f(r ) = 1/\ln(r )$ where $\theta, r$ are polar coordinates. Linearized system at $0$ has PN but (\ref{K}) has a spiral point.

Indeed, in polar coordinates (\ref{K}) becomes

\begin{equation}\begin{aligned}

&r'=-r,\\

&\theta'=f(r)

\end{aligned}\label{L}\end{equation}

and then $r=e^{c-t}$, $\theta'=1/(c-t)$ and $\theta =-\ln (t-c) +c_2$ so it makes an infinite number of rotations around $0$ as $t\to +\infty$ (but this rotation slows down).