Magnitude of the open loop transfer function can be written as $K\lvert\frac{N(s)}{D(s)}\rvert$ (this is an assumption) and for magnitude condition to be satisfied, that expression must equal 1. Ending points are when K approaches infinity and for MC to be satisfied, $\lvert\frac{N(s)}{D(s)}\rvert$ must approach 0.
That's what I've read here: https://lpsa.swarthmore.edu/Root_Locus/DeriveRootLocusRules.html##section10
and then they claim: "So the zeros of the loop gain (which occur at N(s)=0, and perhaps as s→∞) are the ending points for the loci (when K→∞)."
But that seems problematic to me.
1) In order for magnitude to approach 1, if K is approaching infinity, the absolute value would have to approach 0 with the same "speed"(I don't know the right term, but if K behaves like x, and the absolute value like $x^2$, the limit is not going to be equal to 1). How do they know that K and the absolute value approach their limits with the same "speed"? If I'm not making any sense, think about this: if A = x and B = $\frac{1}{x^5}$, as x -> $\infty$, A -> $\infty$ and B -> 0, but A*B does not -> 1
2) What if N(s) and D(s) are both approaching 0? For example,
$\frac{N(s)}{D(s)} = \frac{(s+1)}{(s+1)(s+3)}$.
As s -> -1, $\lvert\frac{N(s)}{D(s)}\rvert$ -> $\frac{1}{2}$ which is not equal to 0, so the MC isn't satisfied. Matlab still plots a branch there, starting and ending at the same point, s = -1, but why is that point a part of the root locus? It doesn't satisfy the magnitude criterion.
This is because root locus plot gives the roots of the closed loop system with respect to $K$. With the unit feedback the closed loop system becomes $$ G_c(s) = \frac{N_c(s)}{D_c(s)} = \frac{K N(s)}{D(s) + K N(s)} $$ Root locus plot is essentially the plot of the zeros of the polynomial $D_c(s)$, i.e. values of $s \in \mathbb{C}$ where $D_c(s)=0$. Now, it should be obvious that root locus plot starts from open loop roots $\{s \in \mathbb{C} | D(s)=0\}$, and "ends up" at open loop zeros $\{s \in \mathbb{C} | N(s)=0\}$ as $K$ ranges from $0$ to $\infty$.