Maximum peak current swing (Class A amplifier)
Okay. I finally took a moment to download and read a few pages of your book. I wasn't sure before, but I really think you did a good job in writing up the question now that I've had a chance to skim it. You selected out the right parts.

The two equations you are discussing in the Output Power section are prefaced by the comment that the "maximum unclipped ac signal occurs when the Q-point is centered." This is important and you selected out the right figure, Figure 7.2(b), which diagrams this exact situation. 

So let me take your diagram and add something to it:



I believe the authors have defined \$I_{c}=I_{C}-I_{C_Q}\$ and \$V_{c}=V_{CE}-V_{{CE}_Q}\$. And they are then saying that \$I_{c(max)}\$ is the peak magnitude above (or below) the center-line of \$I_{C_Q}\$ and that \$V_{c(max)}\$ is similarly the peak magnitude above (or below) the center-line of \$V_{{CE}_Q}\$.



As \$V_{CE}=V_{{CE}_Q}+R_c\cdot\left(I_{C_Q}-I_{C}\right)\$ and therefore \$V_c=R_c\cdot\left(I_{C_Q}-I_{C}\right)\$, \$V_c\$ has a maximum value when \$I_C=0\:\text{A}\$ and thus it must be that \$V_{c(max)}=R_c\cdot\left(I_{C_Q}-0\:\text{A}\right)=R_c\cdot I_{C_Q}\$



As \$I_C=I_{C_Q}+\frac{V_{{CE}_Q}-V_{C}}{R_c}\$ and therefore \$I_c=\frac{V_{{CE}_Q}-V_{C}}{R_c}\$, \$I_c\$ has a maximum value when \$V_C=0\:\text{V}\$ and thus it must be that \$I_{c(max)}=\frac{V_{{CE}_Q}-0\:\text{V}}{R_c}=\frac{V_{{CE}_Q}}{R_c}\$



I believe that's what the book is telling you.

They probably should have said \$V_{ce}=V_{CE}-V_{{CE}_Q}\$ instead of using \$V_c\$ and also should have said \$V_{ce(max)}\$ instead of \$V_{c(max)}\$.

That was what confused me until I skimmed the four pages I needed to see. I thought they were talking about the collector voltage in the first equation.