Partial decomposition of $\frac{z^2}{(2z^2+3)((s^2+1)z^2+1)}$

Question

Recently I have come arcross the following fraction

$$\dfrac{z^2}{(2z^2+3)((s^2+1)z^2+1)}$$

Hence I have encountered this fraction within a task of integration I want to do a partial decomposition. First of all I rewrote it as following

$$\dfrac{z^2}{(2z^2+3)((s^2+1)z^2+1)}=\dfrac{Az+B}{2z^2+3}+\dfrac{Cz+D}{(s^2+1)z^2+1} $$

which yields to a system of equations for $A,B,C$ and $D$

$$\begin{align} 0&=A(1+s^2)+3C\\ 0&=B(1+s^2)+3D\\ 0&=s^2A+2C\\ 1&=s^2B+2D \end{align}$$

Solving this system gives $A=C=0$ and $B=\dfrac{3}{s^2-2}$ and $D=-\dfrac{s^2-1}{s^2-2}$. By plugging in these values we arrive at

$$\dfrac{3}{(s^2-2)(2z^2+3)}-\dfrac{s^2-1}{(s^2-2)((s^2+1)z^2+1)}=\dfrac{s^2(z^2-3)+5z^2+6}{(s^2-2)(2z^2+3)((s^2+1)z^2+1)}$$

which is in fact not the wanted fraction. From the original task(solution to Problem 2 by ysharifi) I am going through I know, that the right partial decomposition should look like

$$\dfrac1{3s^2+1}\left(\frac3{2z^2+3}-\frac1{(s^2+1)z^2+1}\right)$$

but honestly speaking I have no clue how to get to this. Could someone please go through the whole process with me. I am quite confused right now.

Thanks in advance.

Answer 1

Even though we could solve this using the method of residues, I'll follow your method \begin{equation} \frac{z^2}{(2z^2+3)((s^2+1)z^2+1)} = \frac{Az + B}{2z^2 + 3} + \frac{Cz + D}{(s^2+1)z^2 + 1} \tag{1} \end{equation} which is \begin{equation} \frac{z^2}{(2z^2+3)((s^2+1)z^2+1)} = \frac{(Az + B)((s^2+1)z^2 + 1) + (Cz + D)(2z^2 + 3)}{(2z^2 + 3)((s^2+1)z^2+1)} \end{equation} Expanding \begin{equation} \frac{ B+3D + (A+3C)z + (B+2D+Bs^2)z^2 + (A+2C+As^2)z^3}{(2z^2 + 3)((s^2+1)z^2+1)} \end{equation} By identification of coefficients, we have \begin{align} B+3D &= 0\\ A+3C &= 0 \\ B+2D+Bs^2 &= 1 \\ A+2C+As^2 &= 0 \end{align} From the first two, we get $A = -3C$ and $B= - 3D$, so replace in the last two, \begin{align} -D -3s^2D &= 1\\ -C -3s^2C &= 0 \end{align} which means that $C =0$ (also $A=0$) and \begin{equation} D = -\frac{1}{1+3s^2} \end{equation} So replacing in $(1)$, we get

\begin{equation} \frac{z^2}{(2z^2+3)((s^2+1)z^2+1)} = \frac{1}{1+3s^2} \Bigg( \frac{ 3}{2z^2 + 3} - \frac{ 1}{(s^2+1)z^2 + 1} \Bigg) \end{equation}

Answer 2

Based on your demand, here's another method using residues

Let \begin{equation} F(z)= \frac{z^2}{(2z^2+3)((s^2+1)z^2+1)} = \frac{\frac{z^2}{2(s^2+1)}}{(z^2 + \frac{3}{2})(z^2 + \frac{1}{s^2+1})} \end{equation} \begin{equation} F(z) = \frac{\frac{z^2}{2(s^2+1)}}{\Pi_{i=0}^3(z-z_i)} = \frac{r_0}{z-z_0} + \frac{r_1}{z-z_1} + \frac{r_2}{z-z_2} + \frac{r_3}{z-z_3} \end{equation} where \begin{equation} r_k = \lim_{z \rightarrow z_k} (z-z_k)F(z) \end{equation} The poles $z_k$ are the roots of the denominator, i.e. $ z^2 = -\frac{3}{2} $ that gives the first two poles $ z_{0,1} = \pm \frac{\sqrt{3}}{\sqrt{2}}i $ and similarly, we compute the second two poles as $ z_{3,4} = \pm \sqrt{\frac{1}{s^2+1}}i $ Now, let's compute $r_i$'s, as \begin{align} r_0 = \lim_{z \rightarrow z_0} (z-z_0) F(z) &= \frac{\frac{z_0^2}{2(s^2+1)}}{(z_0-z_1)(z_0-z_2)(z_0-z_3)} \\ r_1 = \lim_{z \rightarrow z_1} (z-z_1) F(z) &= \frac{\frac{z_1^2}{2(s^2+1)}}{(z_1-z_0)(z_1-z_2)(z_1-z_3)} \\ r_2 = \lim_{z \rightarrow z_2} (z-z_2) F(z) &= \frac{\frac{z_2^2}{2(s^2+1)}}{(z_2-z_0)(z_2-z_1)(z_2-z_3)} \\ r_3 = \lim_{z \rightarrow z_3} (z-z_3) F(z) &= \frac{\frac{z_3^2}{2(s^2+1)}}{(z_3-z_0)(z_3-z_1)(z_3-z_2)} \end{align} After you replace and compute your $r_i$'s, you get $F(z)$ in the most decomposed form. Then, you could combine $(z-z_0)(z-z_1) = 2z^2 + 3$ and $(z-z_1)(z-z_2) = (s^2+1)z^2 + 1$