Using Residues (Complex Analysis) to compute partial fractions

Dear Everybody, I am wondering how to compute the partial fraction decomposition of the following rational function: ##f(z)=\frac{z+2}{(z+1)^2(z^2+1)}.##

I understand how to do the simple poles of the function and how it is related to the decomposition's constants, i.e. ##f(z)=\frac{A_1}{z+i}+\frac{A_2}{z-i}+\frac{B_1}{z+1}+\frac{B_2}{(z+1)^2}##. Thus, I know that ##A_1=-\frac{i+2}{4}## and ##A_2=-\frac{-i+2}{4}.## But how do I computes double pole in terms of the residue, if possible? How I can write out ##B_1, B_2.##

Setting [itex]z = 0[/itex] gives [tex]
2 = i(A_2 - A_1) + B_1 + B_2.[/tex] Setting [itex]z = 1[/itex] gives [tex]
\frac 38 = \frac{A_1 + A_2 + i(A_2-A_1)}{2} + \frac {B_1}2 + \frac{B_2}4.[/tex] Once [itex]A_1[/itex] and [itex]A_2[/itex] are known, this system can be solved for [itex]B_1[/itex] and [itex]B_2[/itex].

pasmith said:

Setting [itex]z = 0[/itex] gives [tex]
2 = i(A_2 - A_1) + B_1 + B_2.[/tex] Setting [itex]z = 1[/itex] gives [tex]
\frac 38 = \frac{A_1 + A_2 + i(A_2-A_1)}{2} + \frac {B_1}2 + \frac{B_2}4.[/tex] Once [itex]A_1[/itex] and [itex]A_2[/itex] are known, this system can be solved for [itex]B_1[/itex] and [itex]B_2[/itex].

ok. thanks for your input. But, I think you may not answer my question because I am wondering how to do the partial fraction decomposition with respect to residue theory in complex analysis the multiple roots case?

For ##B_2##, define

\begin{align*}
g(z) = (z+1) f(z)
\end{align*}

Then

\begin{align*}
\text{Res} [g(z_0=-1)] = \text{Res} [\frac{B_2}{z+1}] = B_2
\end{align*}

You understand why:

\begin{align*}
\text{Res} [f(z_0=-1)] = \text{Res} [\frac{B_1}{z+1}] = B_1 ?
\end{align*}

where you would use the general formula for the residue of an ##nth-##order pole

\begin{align*}
\text{Res} [f(z_0)] = \frac{1}{(n-1)!} \lim_{z \rightarrow z_0} \left( \dfrac{d^{n-1}}{dz^{n-1}} [(z-z_0)^n f(z)] \right) .
\end{align*}

Are you sure your answers to ##A_1## and ##A_2## are right? So you are equating

\begin{align*}
\text{Res} [f(z_0=-i)] = \text{Res} [\frac{A_1}{z+i}] = A_1
\end{align*}

and then using

\begin{align*}
\text{Res} [f(z_0=-i)] = \lim_{z \rightarrow -i} [(z+i) f(z)]
\end{align*}

but I get ##A_1 = - \dfrac{-i+2}{4}##.

cbarker1 said:

ok. thanks for your input. But, I think you may not answer my question because I am wondering how to do the partial fraction decomposition with respect to residue theory in complex analysis the multiple roots case?

I was led astray by your statement [tex]
f(z) = \frac{A_1}{z + i} + \frac{A_2}{z-i} + \frac{B_1}{z+1} + \frac{B_2}{(z + 1)^2}[/tex] and therefore assumed you meant a partial fraction decomposition. But you meant to calculate the residue of [itex]f[/itex] at -1 instead. A function has only one residue at a root of its denominator, regardless of the multiplicity of that root.

But to compute the Laurent series of [itex]f[/itex] about [itex]-1[/itex] is fairly straightforward, if you do not know the formula quoted by @julian: [tex]
\begin{split}
f(z) &= \frac{z + 2}{(z+1)^2(z^2+1)} \\
&= \frac{(z + 1) + 1}{(z+1)^2(z^2+1)} \\
&=\frac{1}{z^2 + 1}\left(\frac1{(z+1)} + \frac{1}{(z+1)^2}\right).
\end{split}[/tex] The common factor [itex]g(z) = (z^2 + 1)^{-1}[/itex] is analytic at [itex]z = -1[/itex], so it has a Taylor series which converges in some open neighbourhood of it. So [tex]\begin{split}
f(z) &= \sum_{n=0}^\infty \frac{g^{(n)}(-1)(z + 1)^n}{n!}\left(\frac1{(z+1)} + \frac{1}{(z+1)^2}\right) \\
&= \frac{g(-1)}{(z + 1)^2} + \sum_{n=-1}^\infty\left(\frac{g^{(n+1)}(-1)}{(n+1)!} +
\frac{g^{(n+2)}(-1)}{(n+2)!}\right)(z + 1)^n
\end{split}[/tex] and the residue is [itex]g(-1) + g'(-1)[/itex].