1. Must those A, B and C be constants? Or some of them can be first order polynomial?

$$\frac{1}{(z-1)^3(z+1)^3}=\frac{A_1}{(z-1)^3}+\frac{B_1}{(z-1)^2}+\frac{C_1}{(z-1)}+\frac{A_2}{(z+1)^3}+\frac{B_2}{(z+1)^2}+\frac{C_2}{(z+1)}.$$ $$\tag{1}$$

Since the denominator of the left hand side of $(1)$ has real roots only, all $A_i,B_i$, $i=1,2$, on the right hand side are real constants.

If the denominator has complex roots as e.g. in $(2z+1)/((z-1)(z^2+z+1))$, then the expansion into partial fractions would be of the form

$$\frac{2z+1}{(z-1)(z^2+z+1)}=\frac{A}{z-1}+\frac{Bz+C}{z^2+z+1},\tag{A}$$

where $a,b,C$ are real constants, that is, the factor $(z^2+z+1)$ of the denominator generates the partial fraction $\frac{Bz+C}{z^2+z+1}$, whose numerator is a first degree polynomial.

2. How to decompose it faster? (...) One of my friend suggests that keep differentiating the equation $1= A_1(z+1)^3 + ... + C_2(z-1)^3(z+1)^2$, is this a good idea?

Yes, it is because each differentiation decreases the degree of the current equation by one.

Another special method, addapted from the Residue method, consists of multiplying the given equation by a factor of the denominator and applying adequate limits, which we illustrate as follows, without reducing the speed of the computation very much though.

1. As an illustration, we can start by multiplying $(1)$ by the factor $(z-1)^3$ and let $z$ tend to the root $1$ of this factor: \begin{eqnarray*} \lim_{z\rightarrow 1}\frac{1}{(z+1)^{3}} &=&\lim_{z\rightarrow 1}A_{1}+\left( z-1\right) B_{1}+\left( z-1\right) ^{2}C_{1} \\ &&+\lim_{z\rightarrow 1}\frac{\left( z-1\right) ^{3}A_{2}}{\left( z+1\right) ^{3}}+\frac{\left( z-1\right) ^{3}B_{2}}{\left( z+1\right) ^{2}}+\frac{ \left( z-1\right) ^{3}C_{2}}{z+1}. \end{eqnarray*} On the left hand side $\lim_{z\rightarrow 1}\frac{1}{(z+1)^{3}}=\frac{1}{8}$ . Since the right hand side tends to $A_{1}$, we find $A_{1}=1/8$.
2. Similarly multiplying by $(z+1)^{3}$ and letting $z$ tend to the root $-1$, we get on the left hand side $\lim_{z\rightarrow -1}\frac{1}{(z-1)^{3}}=- \frac{1}{8}$ and $A_{2}$ on the right hand side. So $A_{2}=-\frac{1}{8}$. Multiplying both side by $z$ and letting $z$ tend to $\infty $, yields \begin{equation*} 0=C_{1}+C_{2}. \end{equation*}
3. To reduce the degree of the denominator we can move $\frac{A_{1}}{\left( z-1\right) ^{3}}+\frac{A_{2}}{\left( z+1\right) ^{3}}$ to the left hand side (described here) \begin{equation*} \frac{1}{\left( z-1\right) ^{3}(z+1)^{3}}-\frac{1}{8\left( z-1\right) ^{3}}+ \frac{1}{8\left( z+1\right) ^{3}}=\frac{8-(z+1)^{3}+\left( z-1\right) ^{3}}{ 8\left( z-1\right) ^{3}(z+1)^{3}}=\frac{-6\left( z-1\right) \left( z+1\right) }{8\left( z-1\right) ^{3}(z+1)^{3}}=-\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}}. \end{equation*} Consequently \begin{equation*} -\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}}=\frac{B_{1}}{\left( z-1\right) ^{2}}+\frac{C_{1}}{z-1}+\frac{B_{2}}{\left( z+1\right) ^{2}}- \frac{C_{1}}{z+1}.\tag{2} \end{equation*}
4. We can proceed likewise to above, multiplying by $\left( z+1\right) ^{2}$ and let $z\rightarrow -1$ \begin{eqnarray*} \lim_{z\rightarrow -1}-\frac{3}{4\left( z-1\right) ^{2}} &=&\lim_{z \rightarrow -1}\frac{B_{1}\left( z+1\right) ^{2}}{\left( z-1\right) ^{2}}+ \frac{C_{1}\left( z+1\right) ^{2}}{z-1}+B_{2}-\frac{C_{1}\left( z+1\right) ^{2}}{z+1} \\ -\frac{3}{16} &=&0+B_{2}. \end{eqnarray*}
5. Similarly, if we multiply by $\left( z-1\right) ^{2}$ and let $z\rightarrow 1 $, we find $B_{1}=-\frac{3}{16}=B_{2}$ \begin{equation*} -\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}}=-\frac{3}{16\left( z-1\right) ^{2}}+\frac{C_{1}}{z-1}-\frac{3}{16\left( z+1\right) ^{2}}-\frac{ C_{1}}{z+1}. \end{equation*}
6. Setting e.g. $z=0$ on both sides, and solving for $C_{1}$ we finally get $C_{1}= \frac{3}{16}=-C_{2}$. Thus \begin{align} \frac{1}{\left( z-1\right) ^{3}(z+1)^{3}}&=\frac{1}{8\left( z-1\right) ^{3}}- \frac{3}{16\left( z-1\right) ^{2}}+\frac{3}{16\left( z-1\right) }\\&-\frac{1}{ 8\left( z+1\right) ^{3}}-\frac{3}{16\left( z+1\right) ^{2}}-\frac{3}{ 16\left( z+1\right) }.\tag{3} \end{align}

1. Must those A, B and C be constants? Or some of them can be first order polynomial?

$$\frac{1}{(z-1)^3(z+1)^3}=\frac{A_1}{(z-1)^3}+\frac{B_1}{(z-1)^2}+\frac{C_1}{(z-1)}+\frac{A_2}{(z+1)^3}+\frac{B_2}{(z+1)^2}+\frac{C_2}{(z+1)}.$$ $$\tag{1}$$

Since the denominator of the left hand side of $(1)$ has real roots only, all $A_i,B_i$, $i=1,2$, on the right hand side are real constants.

If the denominator has complex roots as e.g. in $(2z+1)/((z-1)(z^2+z+1))$, then the expansion into partial fractions would be of the form

$$\frac{2z+1}{(z-1)(z^2+z+1)}=\frac{A}{z-1}+\frac{Bz+C}{z^2+z+1},\tag{A}$$

where $A,B,C$ are real constants, that is, the factor $(z^2+z+1)$ of the denominator generates the partial fraction $\frac{Bz+C}{z^2+z+1}$, whose numerator is a first degree polynomial.

2. How to decompose it faster? (...) One of my friend suggests that keep differentiating the equation $1= A_1(z+1)^3 + ... + C_2(z-1)^3(z+1)^2$, is this a good idea?

Yes, it is, because each differentiation decreases the degree of the current equation by one.

Another special method, addapted from the Residue method, consists of multiplying the given equation by a factor of the denominator and applying adequate limits, which we illustrate as follows, without reducing the speed of the computation very much though.

1. As an illustration, we can start by multiplying $(1)$ by the factor $(z-1)^3$ and let $z$ tend to the root $1$ of this factor: \begin{eqnarray*} \lim_{z\rightarrow 1}\frac{1}{(z+1)^{3}} &=&\lim_{z\rightarrow 1}A_{1}+\left( z-1\right) B_{1}+\left( z-1\right) ^{2}C_{1} \\ &&+\lim_{z\rightarrow 1}\frac{\left( z-1\right) ^{3}A_{2}}{\left( z+1\right) ^{3}}+\frac{\left( z-1\right) ^{3}B_{2}}{\left( z+1\right) ^{2}}+\frac{ \left( z-1\right) ^{3}C_{2}}{z+1}. \end{eqnarray*} On the left hand side $\lim_{z\rightarrow 1}\frac{1}{(z+1)^{3}}=\frac{1}{8}$ . Since the right hand side tends to $A_{1}$, we find $A_{1}=1/8$.
2. Similarly multiplying by $(z+1)^{3}$ and letting $z$ tend to the root $-1$, we get on the left hand side $\lim_{z\rightarrow -1}\frac{1}{(z-1)^{3}}=- \frac{1}{8}$ and $A_{2}$ on the right hand side. So $A_{2}=-\frac{1}{8}$. Multiplying both side by $z$ and letting $z$ tend to $\infty $, yields \begin{equation*} 0=C_{1}+C_{2}. \end{equation*}
3. To reduce the degree of the denominator we can move $\frac{A_{1}}{\left( z-1\right) ^{3}}+\frac{A_{2}}{\left( z+1\right) ^{3}}$ to the left hand side (described here) \begin{equation*} \frac{1}{\left( z-1\right) ^{3}(z+1)^{3}}-\frac{1}{8\left( z-1\right) ^{3}}+ \frac{1}{8\left( z+1\right) ^{3}}=\frac{8-(z+1)^{3}+\left( z-1\right) ^{3}}{ 8\left( z-1\right) ^{3}(z+1)^{3}}=\frac{-6\left( z-1\right) \left( z+1\right) }{8\left( z-1\right) ^{3}(z+1)^{3}}=-\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}}. \end{equation*} Consequently \begin{equation*} -\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}}=\frac{B_{1}}{\left( z-1\right) ^{2}}+\frac{C_{1}}{z-1}+\frac{B_{2}}{\left( z+1\right) ^{2}}- \frac{C_{1}}{z+1}.\tag{2} \end{equation*}
4. We can proceed likewise to above, multiplying by $\left( z+1\right) ^{2}$ and let $z\rightarrow -1$ \begin{eqnarray*} \lim_{z\rightarrow -1}-\frac{3}{4\left( z-1\right) ^{2}} &=&\lim_{z \rightarrow -1}\frac{B_{1}\left( z+1\right) ^{2}}{\left( z-1\right) ^{2}}+ \frac{C_{1}\left( z+1\right) ^{2}}{z-1}+B_{2}-\frac{C_{1}\left( z+1\right) ^{2}}{z+1} \\ -\frac{3}{16} &=&0+B_{2}. \end{eqnarray*}
5. Similarly, if we multiply by $\left( z-1\right) ^{2}$ and let $z\rightarrow 1 $, we find $B_{1}=-\frac{3}{16}=B_{2}$ \begin{equation*} -\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}}=-\frac{3}{16\left( z-1\right) ^{2}}+\frac{C_{1}}{z-1}-\frac{3}{16\left( z+1\right) ^{2}}-\frac{ C_{1}}{z+1}. \end{equation*}
6. Setting e.g. $z=0$ on both sides, and solving for $C_{1}$ we finally get $C_{1}= \frac{3}{16}=-C_{2}$. Thus \begin{align} \frac{1}{\left( z-1\right) ^{3}(z+1)^{3}}&=\frac{1}{8\left( z-1\right) ^{3}}- \frac{3}{16\left( z-1\right) ^{2}}+\frac{3}{16\left( z-1\right) }\\&-\frac{1}{ 8\left( z+1\right) ^{3}}-\frac{3}{16\left( z+1\right) ^{2}}-\frac{3}{ 16\left( z+1\right) }.\tag{3} \end{align}

1. Must those A, B and C be constants? Or some of them can be first order polynomial?

$$\frac{1}{(z-1)^3(z+1)^3}=\frac{A_1}{(z-1)^3}+\frac{B_1}{(z-1)^2}+\frac{C_1}{(z-1)}+\frac{A_2}{(z+1)^3}+\frac{B_2}{(z+1)^2}+\frac{C_2}{(z+1)}$$ $$\tag{1}$$

Since the denominator of the left hand side of $(1)$ has real roots only, all $A_i,B_i$, $i=1,2$, on the right hand side are real constants.

If the denominator has complex roots as e.g. in $z/((z-1)(z^2+z+1))$, then the expansion into partial fractions would be of the form

$$\frac{2z+1}{(z-1)(z^2+z+1)}=\frac{A}{z-1}+\frac{Bz+C}{z^2+z+1},\tag{A}$$

where $a,b,C$ are real constants, that is, the factor $(z^2+z+1)$ of the denominator generates the partial fraction $\frac{Bz+C}{z^2+z+1}$, whose numerator is a first degree polynomial.

2. How to decompose it faster? (...) One of my friend suggests that keep differentiating the equation $1= A_1(z+1)^3 + ... + C_2(z-1)^3(z+1)^2$, is this a good idea?

Yes, it is because each differentiation decreases the degree of the current equation by one.

Another special method, addapted from the Residue method, consists of multiplying the given equation by a factor of the denominator and applying adequate limits, which we illustrate as follows, without reducing the speed of the computation very much though.

1. As an illustration, we can start by multiplying $(1)$ by the factor $(z-1)^3$ and let $z$ tend to the root $1$ of this factor: \begin{eqnarray*} \lim_{z\rightarrow 1}\frac{1}{(z+1)^{3}} &=&\lim_{z\rightarrow 1}A_{1}+\left( z-1\right) B_{1}+\left( z-1\right) ^{2}C_{1} \\ &&+\lim_{z\rightarrow 1}\frac{\left( z-1\right) ^{3}A_{2}}{\left( z+1\right) ^{3}}+\frac{\left( z-1\right) ^{3}B_{2}}{\left( z+1\right) ^{2}}+\frac{ \left( z-1\right) ^{3}C_{2}}{z+1} \\ \frac{1}{8} &=&A_{1}+0 \end{eqnarray*} On the left hand side $\lim_{z\rightarrow 1}\frac{1}{(z+1)^{3}}=\frac{1}{8}$ . Since the right hand side tends to $A_{1}$, we find $A_{1}=1/8$.
2. Similarly multiplying by $(z+1)^{3}$ and letting $z$ tend to the root $-1$, we get on the left hand side $\lim_{z\rightarrow -1}\frac{1}{(z-1)^{3}}=- \frac{1}{8}$ and $A_{2}$ on the right hand side. So $A_{2}=-\frac{1}{8}$. Multiplying both side by $z$ and letting $z$ tend to $\infty $, yields \begin{equation*} 0=C_{1}+C_{2}. \end{equation*}
3. To reduce the degree of the denominator we can move $\frac{A_{1}}{\left( z-1\right) ^{3}}+\frac{A_{2}}{\left( z+1\right) ^{3}}$ to the left hand side (described here) \begin{equation*} \frac{1}{\left( z-1\right) ^{3}(z+1)^{3}}-\frac{1}{8\left( z-1\right) ^{3}}+ \frac{1}{8\left( z+1\right) ^{3}}=\frac{8-(z+1)^{3}+\left( z-1\right) ^{3}}{ 8\left( z-1\right) ^{3}(z+1)^{3}}=\frac{-6\left( z-1\right) \left( z+1\right) }{8\left( z-1\right) ^{3}(z+1)^{3}}=-\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}} \end{equation*} Consequently \begin{equation*} -\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}}=\frac{B_{1}}{\left( z-1\right) ^{2}}+\frac{C_{1}}{z-1}+\frac{B_{2}}{\left( z+1\right) ^{2}}- \frac{C_{1}}{z+1}\tag{2} \end{equation*}
4. We can proceed likewise to above, multiplying by $\left( z+1\right) ^{2}$ and let $z\rightarrow -1$ \begin{eqnarray*} \lim_{z\rightarrow -1}-\frac{3}{4\left( z-1\right) ^{2}} &=&\lim_{z \rightarrow -1}\frac{B_{1}\left( z+1\right) ^{2}}{\left( z-1\right) ^{2}}+ \frac{C_{1}\left( z+1\right) ^{2}}{z-1}+B_{2}-\frac{C_{1}\left( z+1\right) ^{2}}{z+1} \\ -\frac{3}{16} &=&0+B_{2} \end{eqnarray*}
5. Similarly, if we multiply by $\left( z-1\right) ^{2}$ and let $z\rightarrow 1 $, we find $B_{1}=-\frac{3}{16}=B_{2}$ \begin{equation*} -\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}}=-\frac{3}{16\left( z-1\right) ^{2}}+\frac{C_{1}}{z-1}-\frac{3}{16\left( z+1\right) ^{2}}-\frac{ C_{1}}{z+1} \end{equation*}
6. Setting e.g. $z=0$ on both sides, and solving for $C_{1}$ we finally get $C_{1}= \frac{3}{16}=-C_{2}$. Thus \begin{equation*} \frac{1}{\left( z-1\right) ^{3}(z+1)^{3}}=\frac{1}{8\left( z-1\right) ^{3}}- \frac{3}{16\left( z-1\right) ^{2}}+\frac{3}{16\left( z-1\right) }-\frac{1}{ 8\left( z+1\right) ^{3}}-\frac{3}{16\left( z+1\right) ^{2}}-\frac{3}{ 16\left( z+1\right) } \end{equation*} $$\tag{3}$$

1. Must those A, B and C be constants? Or some of them can be first order polynomial?

$$\frac{1}{(z-1)^3(z+1)^3}=\frac{A_1}{(z-1)^3}+\frac{B_1}{(z-1)^2}+\frac{C_1}{(z-1)}+\frac{A_2}{(z+1)^3}+\frac{B_2}{(z+1)^2}+\frac{C_2}{(z+1)}.$$ $$\tag{1}$$

Since the denominator of the left hand side of $(1)$ has real roots only, all $A_i,B_i$, $i=1,2$, on the right hand side are real constants.

If the denominator has complex roots as e.g. in $(2z+1)/((z-1)(z^2+z+1))$, then the expansion into partial fractions would be of the form

$$\frac{2z+1}{(z-1)(z^2+z+1)}=\frac{A}{z-1}+\frac{Bz+C}{z^2+z+1},\tag{A}$$

where $a,b,C$ are real constants, that is, the factor $(z^2+z+1)$ of the denominator generates the partial fraction $\frac{Bz+C}{z^2+z+1}$, whose numerator is a first degree polynomial.

2. How to decompose it faster? (...) One of my friend suggests that keep differentiating the equation $1= A_1(z+1)^3 + ... + C_2(z-1)^3(z+1)^2$, is this a good idea?

Yes, it is because each differentiation decreases the degree of the current equation by one.

Another special method, addapted from the Residue method, consists of multiplying the given equation by a factor of the denominator and applying adequate limits, which we illustrate as follows, without reducing the speed of the computation very much though.

1. As an illustration, we can start by multiplying $(1)$ by the factor $(z-1)^3$ and let $z$ tend to the root $1$ of this factor: \begin{eqnarray*} \lim_{z\rightarrow 1}\frac{1}{(z+1)^{3}} &=&\lim_{z\rightarrow 1}A_{1}+\left( z-1\right) B_{1}+\left( z-1\right) ^{2}C_{1} \\ &&+\lim_{z\rightarrow 1}\frac{\left( z-1\right) ^{3}A_{2}}{\left( z+1\right) ^{3}}+\frac{\left( z-1\right) ^{3}B_{2}}{\left( z+1\right) ^{2}}+\frac{ \left( z-1\right) ^{3}C_{2}}{z+1}. \end{eqnarray*} On the left hand side $\lim_{z\rightarrow 1}\frac{1}{(z+1)^{3}}=\frac{1}{8}$ . Since the right hand side tends to $A_{1}$, we find $A_{1}=1/8$.
2. Similarly multiplying by $(z+1)^{3}$ and letting $z$ tend to the root $-1$, we get on the left hand side $\lim_{z\rightarrow -1}\frac{1}{(z-1)^{3}}=- \frac{1}{8}$ and $A_{2}$ on the right hand side. So $A_{2}=-\frac{1}{8}$. Multiplying both side by $z$ and letting $z$ tend to $\infty $, yields \begin{equation*} 0=C_{1}+C_{2}. \end{equation*}
3. To reduce the degree of the denominator we can move $\frac{A_{1}}{\left( z-1\right) ^{3}}+\frac{A_{2}}{\left( z+1\right) ^{3}}$ to the left hand side (described here) \begin{equation*} \frac{1}{\left( z-1\right) ^{3}(z+1)^{3}}-\frac{1}{8\left( z-1\right) ^{3}}+ \frac{1}{8\left( z+1\right) ^{3}}=\frac{8-(z+1)^{3}+\left( z-1\right) ^{3}}{ 8\left( z-1\right) ^{3}(z+1)^{3}}=\frac{-6\left( z-1\right) \left( z+1\right) }{8\left( z-1\right) ^{3}(z+1)^{3}}=-\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}}. \end{equation*} Consequently \begin{equation*} -\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}}=\frac{B_{1}}{\left( z-1\right) ^{2}}+\frac{C_{1}}{z-1}+\frac{B_{2}}{\left( z+1\right) ^{2}}- \frac{C_{1}}{z+1}.\tag{2} \end{equation*}
4. We can proceed likewise to above, multiplying by $\left( z+1\right) ^{2}$ and let $z\rightarrow -1$ \begin{eqnarray*} \lim_{z\rightarrow -1}-\frac{3}{4\left( z-1\right) ^{2}} &=&\lim_{z \rightarrow -1}\frac{B_{1}\left( z+1\right) ^{2}}{\left( z-1\right) ^{2}}+ \frac{C_{1}\left( z+1\right) ^{2}}{z-1}+B_{2}-\frac{C_{1}\left( z+1\right) ^{2}}{z+1} \\ -\frac{3}{16} &=&0+B_{2}. \end{eqnarray*}
5. Similarly, if we multiply by $\left( z-1\right) ^{2}$ and let $z\rightarrow 1 $, we find $B_{1}=-\frac{3}{16}=B_{2}$ \begin{equation*} -\frac{3}{4\left( z+1\right) ^{2}\left( z-1\right) ^{2}}=-\frac{3}{16\left( z-1\right) ^{2}}+\frac{C_{1}}{z-1}-\frac{3}{16\left( z+1\right) ^{2}}-\frac{ C_{1}}{z+1}. \end{equation*}
6. Setting e.g. $z=0$ on both sides, and solving for $C_{1}$ we finally get $C_{1}= \frac{3}{16}=-C_{2}$. Thus \begin{align} \frac{1}{\left( z-1\right) ^{3}(z+1)^{3}}&=\frac{1}{8\left( z-1\right) ^{3}}- \frac{3}{16\left( z-1\right) ^{2}}+\frac{3}{16\left( z-1\right) }\\&-\frac{1}{ 8\left( z+1\right) ^{3}}-\frac{3}{16\left( z+1\right) ^{2}}-\frac{3}{ 16\left( z+1\right) }.\tag{3} \end{align}