First the case of Kn:
As every vertex in the graph is adjacent no two vertices can share a color. Therefore the chromatic number is at least n. The graph is trivially colored with n colors, each vertex a different color. Therefore χ(Kn)=n.
Second the case of Kn,m:
As the there are adjacent vertices in the graph the chromatic number cannot be 1. We can color the bipartite graph with two colors by coloring each of the two sections with one color. Therefore χ(Km,n)=2.
Third the case of Cm:
Begin by labeling the vertices around the graph v1 through vm such that vk and v(k+1) are adjacent for all k between 1 and m and v1 and vm are adjacent.
The chromatic number is trivially not 1 as v1 and v2 are adjacent.
If m is even we can color Cm such that vk is blue if k is odd and red if k is even. Therefore χ(Cm)=2 if m is even.
If m is odd we first show we cannot color Cm with two colors. Assume we only have two colors, then if vk is blue then v(k+1) must be red and vice versa. Letting v1 be blue we iterate this argument and find that vm must be blue, a contradiction since v1 and vm are adjacent. Thus χ(Cm) is at least 3 if m is odd.
Finally we find that it is exactly 3 by coloring as described earlier except vm is colored green.
Poincare
