■solution 1
$Z[x(nT+T)]=z^{T}{\displaystyle\sum_{n=0}^{+\infty} x(nT+T)z^{-(nT+T)}+x(0)-x(0)}$
$=z^{T}{X(z)-x(0)}=\underline{z\cdot X(z)-z\cdot x(0)}$
■solution 2
$X(z)=\displaystyle\lim_{n \to \infty} [x(0)+x(T)z^{-1}+x(2T)z^{-2}+\cdot\cdot\cdot x(nT)z^{-n}]$-(i)
$Z[x(nT+T)]=\displaystyle\lim_{n \to \infty}[x(T)+x(2T)z^{-1}+ \cdot\cdot\cdot x(nT)z^{-(n-1)}+x(nT+T)z^{-n}]$-(ii)
(ii)に$z^{-1}$を乗算．
$z^{-1}\cdot Z[x(nT+T)]=\displaystyle\lim_{n \to \infty}[x(T)z^{-1}+x(2T)z^{-2}+ \cdot\cdot\cdot x(nT)z^{-n}+x(nT+T)z^{-(n+1)}]$-(iii)
(iii)-(i)を実行．
$z^{-1}\cdot Z[x(nT+T)]-X(z)=\displaystyle\lim_{n \to \infty}[-x(0)+x(nT+T)z^{-(n+1)}]=-x(0)$
(∵$\displaystyle\lim_{n \to \infty}z^{-(n+1)}=0$)
$z^{-1}\cdot Z[x(nT+T)]-X(z)=-x(0)$より,
$\underline{Z[x(nT+T)]=z\cdot X(z)-z\cdot x(0)}$