Electrical – Stability & error from nyquist POLAR plot

You can ignore it because it is small, and the error comes because the pole has not been that precisely determined.

With the pole value you give (0.719 + 0.215 j), the solution from the characteristic equation for k is 0.0626793 - 0.0000849631 j. As the precision in the pole value is increased the imaginary part will become even closer to zero.

[I should also add that you are not plotting the polar (Nyquist) plot but rather the root locus plot.]

If there exists ω=φ such that |KG(jφ)| = 1 and ∠(KG(jφ)) = -180°, then we know that s=jφ must be a pole.

This is incorrect, and is leading to a misunderstanding.

The poles of the system are the roots of the characteristic equation of the Open Loop transfer function $$KG(s)=0$$

the roots are of the form $$s=\sigma+j\omega$$

These are the poles and zeros that are analysed in the Bode Plot.

Once the loop is closed the poles move location to be the roots of the characteristic equation $$1+KG(s)=0$$ You can view loop compensation as a method of moving the open loop poles into more suitable (stable) locations in the closed loop. This can even by performed directly using pole placement.

However, Bode stability analysis is based on the Nyquist Stability Criterion. So the condition for oscillation in a negative feedback system is unity gain and 180 degree phase shift :$$KG(s) = -1$$ and therefore the Bode plot illustrates the "stability condition" by rearranging $$KG(s)+1=0$$

This happens to be the same equation as the characteristic equation of the Closed Loop. But it is a misunderstanding to see this as relating to Bode stability plots, which are an Open Loop analysis.

The Nyquist Stability Criterion also tells us that in general (but there are exceptions), a closed loop system is stable if the unity gain crossing of the magnitude plot occurs at a lower frequency than the -180 degree crossing of the phase plot.