$11\mathrm{.}15.$ Given the $z-$transform

$E(z)=\{e(t)\}=\frac{4z}{(z-1)(z-0\mathrm{.}4)(z-0\mathrm{.}8)}$

$($a$)$ By inspection we can see that $e(0)=e(1)=0$ and $e(2)=4.$ Show that this statement is true.