Same z transformed function, but different answers of inverse z transform? 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

$$\begin{split}
\text{residue}_{z=-4} &amp;= ((z+4)\times E(z)\times z^{k-1})_{z=-4}\\
&amp;= (1\times z^{k-1})_{z=-4}\\ &amp;= (-4)^{k-1}\\ &amp;= -(1/4)\times (-4)^k \end{split}$$



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?
Here's a much simpler method. We have

$$E (z) := \frac{1}{z+4} = z^{-1} \left(\frac{1}{1 - (-4) \, z^{-1}}\right)$$

which we assume to be the $\mathcal Z$ transform of a causal discrete-time signal. Consulting a table of $\mathcal Z$ transforms, we conclude that

$$\mathcal Z^{-1} \left\{ \frac{1}{1 - (-4) \, z^{-1}} \right\} = (-4)^n \, u (n)$$

where $u$ is the discrete-time step function. Hence, the inverse $\mathcal Z$ transform of $E$ is simply the discrete-time signal above after a unit delay, i.e.,

$$e (n) = (-4)^{n-1} \, u (n-1)$$

which is zero at $n=0$. Hence, your 1st candidate answer is (almost) correct. Your 2nd candidate answer is only correct if $n \geq 1$. When working with functions, do not forget what their domains are.