I got two method to take inversion of $X(z)$
1. Inversion Integral Formula
$$ X(z) = x(0) + x(T)z^{-1} + x(2T)z^{-2} + ... + x(kT)z^{-k} $$
where we try to isolate $x(kT)$ as a residue of $x(kT)z^{-1}$
$$ X(z)z^{k-1} = x(0)z^{k-1} + x(T)z^{k-2} + x(2T)z^{k-3} + ... + x(kT)z^{-1} $$
$$ x(kT) = \frac{1}{j2\pi}\oint_{C}X(z)z^{k-1}dz $$
Thus $$ x(kT) = \sum{Res(X(z)z^{k-1})} $$
2. Partial Fraction Method
I don't mention about this method let me go through my quesetion.
I got a $X(z)$ like this:
$$ X(z) = \frac{z(1-e^{-aT})}{(z-1)(z-e^{-aT})} $$
Using Inversion Integral Formula:
$$ X(z)z^{k-1} = \frac{z^{k}(1-e^{-aT})}{(z-1)(z-e^{-aT})} $$
$$ Res(X(z)z^{k-1}, 1) = 1 $$ $$ Res(X(z)z^{k-1}, e^{-aT}) = -e^{-akT} $$
Then,
$$ x(kT) = u(kT) - e^{-akT}u(kT) $$
Using Partial Fraction Method:
$$ X(z) = \frac{z(1-e^{-aT})}{(z-1)(z-e^{-aT})} = \frac{Res(X(z), 1)}{z-1} + \frac{Res(X(z), e^{-aT})}{z-e^{-aT}} $$
$$ X(z) = \frac{z^{-1}}{(1-z^{-1})} + \frac{-e^{-aT}z^{-1}}{1-e^{-aT}z^{-1}} $$
Then,
$$ x(kT) = u((k-1)T) - e^{-aT}(e^{-aT})^{(k-1)}u((k-1)T) $$
$$ x(kT) = u((k-1)T) - (e^{-aT})^{k}u(kT) $$
I expect to get same result but I got different although they are similar. My question is why?