Given a $\mathcal{Z}$ transformed function $E(z)=\frac{1}{z+4}$.
I know there are several ways to get the inverse $\mathcal{Z}$ transform of this function :
Using partial fraction
$$E(z)=\frac{1}{z+4}$$
$$\frac{E(z)}{z}=\frac{1}{z(z+4)}$$
apply partial fraction here,
$$\frac{E(z)}{z}= \tfrac14 \left(\frac{1}{z} - \frac{1}{z+4} \right)$$
so $E(z)$ is,
$$E(z)= \tfrac14 \left(1 - \frac{z}{z+4} \right)$$
as you know, it is easy to use inverse Z-transform here.
$$ e[n] = \tfrac14 ( \delta[n] - (-4)^n ) $$
inversion formula method
$E(z)$ has simple pole at $z=-4$, the residue is evaluated as
$$\begin{split} \text{residue}_{z=-4} &= ((z+4)\times E(z)\times z^{k-1})_{z=-4}\\ &= (1\times z^{k-1})_{z=-4}\\ &= (-4)^{k-1}\\ &= -(1/4)\times (-4)^k \end{split}$$
The problem is that inverse $\mathcal{Z}$ transform of the same function $E(z)$has 2 different answers. As you see, there is no delta function when I used inversion formula method.
At first, I thought it was because a simple pole, $z=-4$, is outside of the unit circle, but same things happen when I used a simple pole which is inside of the unit circle.
Why does this difference happen? What is the real answer?