# How do you write the partial fraction decomposition of the rational expression z/(z^4-1) ?

Dec 13, 2015

#### Explanation:

Dec 13, 2015

$\frac{z}{{z}^{4} - 1} = \frac{1}{4 \left(z + 1\right)} + \frac{1}{4 \left(z - 1\right)} - \frac{z}{2 \left({z}^{2} + 1\right)}$

#### Explanation:

The expression equals z/((z+1)(z-1)(z^2+1).

$\frac{z}{\left(z + 1\right) \left(z - 1\right) \left({z}^{2} + 1\right)} = \frac{A}{z + 1} + \frac{B}{z - 1} + \frac{C z + D}{{z}^{2} + 1}$

$z = A \left(z - 1\right) \left({z}^{2} + 1\right) + B \left(z + 1\right) \left({z}^{2} + 1\right) + \left(C z + D\right) \left({z}^{2} - 1\right)$

IF $z = 1$:

$1 = 4 B$
$B = \frac{1}{4}$

IF $z = - 1$:

$- 1 = - 4 A$
$A = \frac{1}{4}$

Plug in two values for $z$:

IF $z = 2$:

$- 1 = 2 C + D$

IF $z = 3$:

$- \frac{3}{2} = 3 C + D$

Solve the system to find that $C = - \frac{1}{2}$ and $D = 0$.

Thus:

$\frac{z}{{z}^{4} - 1} = \frac{1}{4 \left(z + 1\right)} + \frac{1}{4 \left(z - 1\right)} - \frac{z}{2 \left({z}^{2} + 1\right)}$