Find the exact values of A, B, C, and D in the following partial fraction decomposition. z z4 − 1

We are tasked with finding the exact values of $A$, $B$, $C$, and $D$ in the partial fraction decomposition of the following rational expression:

$\frac{z}{z^4 - 1} = \frac{A}{z+1} + \frac{B}{z-1} + \frac{Cz + D}{z^2 + 1}$

Step 1: Factor the denominator

The denominator $z^4 - 1$ can be factored as:

$z^4 - 1 = (z^2 - 1)(z^2 + 1) = (z-1)(z+1)(z^2+1)$

So, the partial fraction decomposition is:

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

Step 2: Multiply both sides by the denominator

To eliminate the denominators, multiply both sides by $(z+1)(z-1)(z^2+1)$:

$z = A(z-1)(z^2+1) + B(z+1)(z^2+1) + (Cz+D)(z^2-1)$

Step 3: Expand both sides

Expand the right-hand side of the equation:

1. First term: [ A(z-1)(z^2+1) = A[z(z^2+1) - (z^2+1)] = A[z^3 + z - z^2 - 1] ] $= A(z^3 - z^2 + z - 1)$

2. Second term: [ B(z+1)(z^2+1) = B[z(z^2+1) + (z^2+1)] = B[z^3 + z + z^2 + 1] ] $= B(z^3 + z^2 + z + 1)$

3. Third term: $(Cz+D)(z^2-1) = Cz(z^2-1) + D(z^2-1) = Cz(z^2 - 1) + D(z^2 - 1)$ $= Cz^3 - Cz + Dz^2 - D$

Step 4: Combine like terms

Now, collect all powers of $z$ from the right-hand side:

$z = (A+B+C)z^3 + (B - A + D)z^2 + (A + B - C)z + (-A + B - D)$

Step 5: Set up the system of equations

Since the left-hand side is simply $z$, this means the coefficients of each power of $z$ on the right-hand side must match the corresponding powers on the left-hand side. Equating coefficients:

• Coefficient of $z^3$: $A + B + C = 0$
• Coefficient of $z^2$: $-A + B + D = 0$
• Coefficient of $z^1$: $A + B - C = 1$
• Constant term: $-A + B - D = 0$

Step 6: Solve the system of equations

We have the following system:

1. $A + B + C = 0$
2. $-A + B + D = 0$
3. $A + B - C = 1$
4. $-A + B - D = 0$

Solve these equations step by step:

• From (1): $C = -A - B$

• From (2) and (4): $-A + B + D = 0$ and $-A + B - D = 0$, add these two equations to get: $-2A + 2B = 0 \implies A = B$

• Substitute $A = B$ into (1): $A + A + C = 0$, so $2A + C = 0$ or $C = -2A$.

• Substitute $A = B$ and $C = -2A$ into (3): $A + A - (-2A) = 1 \implies 4A = 1 \implies A = \frac{1}{4}$

Thus, $A = \frac{1}{4}$, $B = \frac{1}{4}$, and $C = -\frac{1}{2}$.

• Now substitute $A = \frac{1}{4}$ and $B = \frac{1}{4}$ into (2): $-\frac{1}{4} + \frac{1}{4} + D = 0 \implies D = 0$

Step 7: Final values

Thus, the exact values are:

$A = \frac{1}{4}, \quad B = \frac{1}{4}, \quad C = -\frac{1}{2}, \quad D = 0$

Would you like to explore further details or ask additional questions? Here are five related questions:

1. How can we verify the result of a partial fraction decomposition?
2. What is the significance of each term in the partial fraction decomposition?
3. Can we use this method for any polynomial denominator?
4. How do we handle repeated roots in partial fraction decomposition?
5. How would this process change for a more complex numerator?

Tip: Always check if the denominator can be factored into simpler terms to facilitate the decomposition process.