Suppose we have an open-loop transfer function $$G(s) = \frac{1}{s(s+a)(s+b)}$$ If we plot the root locus for the closed-loop system we will get roughly something like this : Now the question is when I add a new zero to the system which is at $-a$ then the book says that we should plot the root-locus without cancelling the pole-zero ( $-a$ in this case ) . I have a practical doubt , in real systems suppose we add a new zero somehow obviously it will not be exactly at $-a$ but at some $-a+\epsilon$ where $\epsilon$ is very small , even in that case the root-locus will become something like this : Because now the aysmptotes will change cause $$n-m = 2$$ . Therefore the new asymptotes are at $$\frac{\pi}{2}$$ rad and $$\frac{3\pi}{2}$$ rad . These two plots become completely different , if I go strictly by the book then my system (even after the addition of new zero ) becomes unstable for some value of gain $K$ but if I consider the situation practically then my system is always stable . Please help me out , which one is correct .