With this opened-loop transfer function,
$$ \begin{align*} L(s) & = \dfrac{K(s+2)}{s(s+1)} \end{align*}$$
I would like to plot the root locus of this system. Based on the textbook, I got the following:
$$ \begin{align*}& 1 + L(s) = s^2 + (K+1)s + 2K = 0 \\\& n = 2, m = 1, p_1 = 0, p_2 = -1, z_1 = -2 \end{align*}$$
$$ \begin{align*}& θ = \dfrac{Nπ}{n - m} = \dfrac{Nπ}{2-1} = Nπ, \quad N = \pm1, \pm3,.. \end{align*}$$
Therefore, I learned that the locus is indeed a circle. Besides, the system with only two poles and only one zero has a center at its zero. In this case, the center is,
$$ O(-2, 0)$$
However, I also learned from a textbook, it said that the center of gravity can be calculated via:
$$ \sigma_c = \dfrac{1}{n-m}(\sum_{i=1}^np_i - \sum_{j=1}^mz_j) = \dfrac{1}{2-1}(0+(-1)-(-2)) = 1$$
This contradicts what I know, and I believe I must have mixed something different together, probably the branches. What is wrong with my idea?