What does $\operatorname{Tr}>0$ imply about complex eigenvalues of a Jacobi matrix?

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I am currently taking a mathematical biology course and am working through the notes. We are covering stability analysis using the Jacobian matrix. One of the conditions for checking the stability is as follows: $$\text{Tr}(J)^2<4\det(J)\implies \lambda=\mu+i\omega.$$ It is also written that $\operatorname{Tr}(J)<0\implies \text{Stable Spiral}$ and $\operatorname{Tr}(J)>0\implies \text{Unstable Spiral}$.

Why does the trace have this effect on the stability of the spiral? I understand that $\operatorname{Tr}(J)^2<4\det(J)$ gives complex eigenvalues from a negative determinant when solving for them. If the real part $\mu<0$ does this generally imply stability? Is the only effect of the complex eigenvalues creating the spiral?