# Inverse Z-transform mystic simplification

I have the following expression:

$$X(z) = \frac{16}{15}\frac{1}{1-\frac14z^{-1}} - \frac{16}{15}\frac{1}{1-4z^{-1}}$$

According to my understanding this should become:

$$x(n) = \frac{16}{15}\left(\frac14\right)^n u(n) - \frac{16}{15} 4^n u(n)$$

But according to my source it is:

$$x(n) = \frac{16}{15}\left(\frac14\right)^n u(n) + \frac{16}{15} 4^n u(-n-1)$$

Are these expressions equal?

If they are, how can $-u(n) = u(-n-1)$ ?

If not, why?

I used $a^nu(n)\Longleftrightarrow \frac{1}{1-az^{-1}}$.

This is probably homework, so I'll just give you a few hints to get you on the right track. First of all, given an expression $X(z)$, the corresponding sequence is usually not unique. Your solution is correct if the region of convergence (ROC) of $X(z)$ is $|z|>4$, which gives you an exponentially increasing sequence. If this were the impulse response of a discrete-time system, then the system would be causal but unstable.
The other solution assumes that the ROC is $\frac14<|z|<4$, which includes the unit circle. This corresponds to a two-sided decaying sequence. As an impulse response it would correspond to a non-causal but stable system.
So, to answer your question, both solutions are correct but they are not identical, because they are inverse transforms of $X(z)$ assuming different regions of convergence.