While carrying out the long division method, it always converts the given expression in the simplest form. That means as far as possible, it should be in form of Z divided by some polynomial. |Z| > 1 it is causal sequence. For the causal sequence the denominator polynomial should have maximum power of Z transform on its left. Thus the given expression is in the total proper form.

Find inverse Z transform of X(z) is = Z/Z-1 |Z|>1

X(Z) = 1 + Z

^{-1 }+^{ }Z^{-2 }+ Z^{-3}+..........
∞

X(Z) = ∑ x(n) Z

^{-n}
n=0

X(Z) is = x(0) + x(1) Z

^{-1 }+ x(2) Z^{-2 }+ x(3) Z^{-3}^{}

^{}

^{X(n) = {}x(0) , x(1), x(2), x(3) }

x(n) = { 1, 1, 1, 1 }

x(n) = u(n)

Z

^{-1}{Z^{-1}/Z-1} = u(n)
This is a standard Z transform pair.