# Using Residues (Complex Analysis) to compute partial fractions

• I
• cbarker1
In summary, to compute the partial fraction decomposition of a rational function, one can use the method of setting z = 0 and z = 1 to find the constants. In the case of multiple roots, one can use residue theory to find the constants B1 and B2. Additionally, the Laurent series of the function about the root can also be used to find the residue.
cbarker1
Gold Member
MHB
TL;DR Summary
How to compute partial fractions decompose when one the factors is a root of multiplicity in the residue method?
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 $z = 0$ gives $$2 = i(A_2 - A_1) + B_1 + B_2.$$ Setting $z = 1$ gives $$\frac 38 = \frac{A_1 + A_2 + i(A_2-A_1)}{2} + \frac {B_1}2 + \frac{B_2}4.$$ Once $A_1$ and $A_2$ are known, this system can be solved for $B_1$ and $B_2$.

cbarker1
pasmith said:
Setting $z = 0$ gives $$2 = i(A_2 - A_1) + B_1 + B_2.$$ Setting $z = 1$ gives $$\frac 38 = \frac{A_1 + A_2 + i(A_2-A_1)}{2} + \frac {B_1}2 + \frac{B_2}4.$$ Once $A_1$ and $A_2$ are known, this system can be solved for $B_1$ and $B_2$.

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}##.

Last edited:
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 $$f(z) = \frac{A_1}{z + i} + \frac{A_2}{z-i} + \frac{B_1}{z+1} + \frac{B_2}{(z + 1)^2}$$ and therefore assumed you meant a partial fraction decomposition. But you meant to calculate the residue of $f$ 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 $f$ about $-1$ is fairly straightforward, if you do not know the formula quoted by @julian: $$\begin{split} f(z) &= \frac{z + 2}{(z+1)^2(z^2+1)} \\ &= \frac{(z + 1) + 1}{(z+1)^2(z^2+1)} \\ &=\frac{1}{z^2 + 1}\left(\frac1{(z+1)} + \frac{1}{(z+1)^2}\right). \end{split}$$ The common factor $g(z) = (z^2 + 1)^{-1}$ is analytic at $z = -1$, so it has a Taylor series which converges in some open neighbourhood of it. So $$\begin{split} f(z) &= \sum_{n=0}^\infty \frac{g^{(n)}(-1)(z + 1)^n}{n!}\left(\frac1{(z+1)} + \frac{1}{(z+1)^2}\right) \\ &= \frac{g(-1)}{(z + 1)^2} + \sum_{n=-1}^\infty\left(\frac{g^{(n+1)}(-1)}{(n+1)!} + \frac{g^{(n+2)}(-1)}{(n+2)!}\right)(z + 1)^n \end{split}$$ and the residue is $g(-1) + g'(-1)$.

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