It is easier to leave the roots in symbolic form.
You have already established that $$\frac{z^5 - 1}{z^4 + 1} = z - \frac{z + 1}{z^4 + 1}.$$ Then let the roots of $z^4 + 1 = 0$ be $$r_k = e^{(2k-1) \pi i/4}, \quad k \in \{1, 2, 3, 4\}.$$ We then seek coefficients $A_k \in \mathbb C$ satisfying
$$\frac{z + 1}{z^4 + 1} = \sum_{k=1}^4 \frac{A_k}{z - r_k}.$$ To this end we employ the "cover-up" method arising from the fact that when the RHS is expressed over the common denominator $z^4 + 1$, the numerator is the sum $$\sum_{k=1}^4 A_k \prod_{j \ne k} (z - r_j) = A_1 (z - r_2)(z - r_3)(z - r_4) + A_2 (z - r_1)(z-r_3)(z-r_4) + \cdots,$$ where each product of linear factors excludes the factor with $r_k$ as a root. Hence by choosing $z = r_k$ in the above expression for each $k$, we obtain the successive identities
$$\begin{align}
r_1 + 1 &= A_1(r_1 - r_2)(r_1 - r_3)(r_1 - r_4), \\
r_2 + 1 &= A_2(r_2 - r_1)(r_2 - r_3)(r_2 - r_4), \\
r_3 + 1 &= A_3(r_3 - r_1)(r_3 - r_2)(r_3 - r_4), \\
r_4 + 1 &= A_4(r_4 - r_1)(r_4 - r_2)(r_4 - r_3), \\
\end{align}$$
or more generally,
$$r_k + 1 = A_k \prod_{j \ne k} (r_k - r_j),$$ hence
$$A_k = \frac{r_k + 1}{\prod_{j \ne k} (r_k - r_j)} = \frac{e^{(2k-1)\pi i/4} + 1}{\prod_{j \ne k} (e^{(2k-1) \pi i/4} - e^{(2j-1) \pi i/4})}.$$
All that remains is to evaluate this expression: for instance,
$$A_1 = \frac{e^{\pi i/4} + 1}{(e^{\pi i/4} - e^{3\pi i/4})(e^{\pi i/4} - e^{-\pi i/4})(e^{\pi i/4} - e^{-3\pi i/4})} = \frac{1}{2(1 - \sqrt{2} + i)}.$$ The rest I leave as an exercise.
