When calculating the error of a Taylor series, the formula is as follows: $$R_n(x)=\frac{f^{(n+1)}(z)(x-c)^{(n+1)}}{(n+1)!}$$ $z$ is the maximum value of the expression on the interval between $x$ and $c$.
Is $z$ the maximum value on the error expression or the original expression? I have been assuming the error expression because we want to know the maximum error. Is this correct?
No, $z$ (that you'd better denote $\xi$ to avoid any ambiguity) is just the value r in the Taylor-Lagrange form of the remainder: the theorem asserts there exists some $\xi$ between $a$ and $x$ such that the error is $$ R_n(x)=\frac{f^{(n+1)}(\xi)(x-c)^{(n+1)}}{(n+1)!}, $$ so that, if we have a bound $M$ for $|f^{(n+1)}(\xi)|$ on some interval centred at $c$, we can deduce that $$ |R_n(x)|\le \frac{M|x-c|^{(n+1)}}{(n+1)!}. $$