Real Partial Fraction Decomposition

Summary: Euler’s method (§199–§210) for resolving a real proper rational function into a sum of real partial fractions whose denominators are real linear factors or real quadratic trinomial factors $p^2-2pqz\cos \varphi +q^2{z}^2$. Extends the §39–§46 algorithm to handle the complex-root case without introducing complex numerators.

Sources: chapter12

Last updated: 2026-05-04

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Setup

Let $M/N$ be a proper real rational function (see improper-rational-function if it is improper). By the §28–§32 theorem (the fundamental-theorem-of-algebra), $N$ factors over $\mathbb{R}$ into real linear factors and real quadratic factors. Each real linear factor is handled by the Chapter 2 algorithm. Each real quadratic factor — the case Chapter 12 addresses — is normalized to the canonical form

$p^2-2pqz\cos \varphi +q^2{z}^2$

with complex roots $z=(p/q)(\cos \varphi \pm i\sin \varphi )$ (source: chapter12, §200).

Distinct quadratic factors (§200–§205)

For one such factor of $N$, the corresponding partial fraction has the form

$\frac{P+Qz}{p^2-2pqz\cos \varphi +q^2{z}^2}$

with two unknown real coefficients $P$ and $Q$. The numerator is exactly first degree: a higher degree would leave behind a polynomial part (which should already have been removed) (source: chapter12, §200).

Derivation (§201)

Write $N=(p^2-2pqz\cos \varphi +q^2{z}^2)Z$. Subtracting $(P+Qz)/(\text{trinomial})$ from $M/N$ leaves a fraction whose numerator $M-PZ-QZz$ must be divisible by the trinomial — i.e. it must vanish at the trinomial’s two roots. Setting $f=p/q$, De Moivre gives at the roots

$z^n=f^n(\cos n\varphi \pm i\sin n\varphi ).$

(source: chapter12, §201).

Two real equations (§202)

Substituting both roots into $M=PZ+QZz$ and expanding the cosines/sines gives, in the abbreviations

$$\begin{array}{rl}R& =A+Bf\cos \varphi +C{f}^2\cos 2\varphi +D{f}^3\cos 3\varphi +\cdots ,\\ r& =Bf\sin \varphi +C{f}^2\sin 2\varphi +D{f}^3\sin 3\varphi +\cdots ,\\ S& =\alpha +\beta f\cos \varphi +\gamma f^2\cos 2\varphi +\delta f^3\cos 3\varphi +\cdots ,\\ s& =\beta f\sin \varphi +\gamma f^2\sin 2\varphi +\delta f^3\sin 3\varphi +\cdots ,\\ T& =\alpha f\cos \varphi +\beta f^2\cos 2\varphi +\gamma f^3\cos 3\varphi +\cdots ,\\ t& =\alpha f\sin \varphi +\beta f^2\sin 2\varphi +\gamma f^3\sin 3\varphi +\cdots ,\end{array}$$

the single complex equation $R\pm ri=(PS\pm Psi)+(QT\pm Qti)$. Separating real and imaginary parts:

$R=PS+QT,r=Ps+Qt.$

(source: chapter12, §202–§203). Notice the pattern: $R,r$ come from substituting $z^n\to f^n\cos n\varphi$ (resp. $f^n\sin n\varphi$) in $M$; $S,s$ come from the same substitution in $Z$; $T,t$ from the same substitution in $Zz$ (which shifts every exponent by one).

Closed-form solution (§203)

The two-equation system gives

$\boxed{P=\frac{Rt-rT}{St-sT},Q=\frac{Rs-rS}{sT-St}.}$

Once $P$ and $Q$ are computed, the partial fraction $(P+Qz)/(p^2-2pqz\cos \varphi +q^2{z}^2)$ is determined. Subtracting it from $M/N$ leaves a “complementary fraction” whose denominator is $Z$, which can be decomposed by the same rule applied to its own quadratic factors (and by the Chapter 2 algorithm for any linear factors).

Streamlined formula (§204–§205)

A direct calculation from the definitions gives

$T=f(S\cos \varphi -s\sin \varphi ),t=f(S\sin \varphi +s\cos \varphi ),$

so $T$ and $t$ are not independent computations — they are determined by $S,s$. Substituting into the §203 system yields

$St-sT=(S^2+s^2)f\sin \varphi ,$

$Rt-rT=(RS+rs)f\sin \varphi +(Rs-rS)f\cos \varphi .$

Plugging back gives the compact form

$\frac{P+Qz}{p^2-2pqz\cos \varphi +q^2{z}^2}=\frac{(RS+rs)p\sin \varphi +(Rs-rS)(p\cos \varphi -qz)}{(p^2-2pqz\cos \varphi +q^2{z}^2)(S^2+s^2)p\sin \varphi }.$

Only the four scalars $R,r,S,s$ are needed — half the trigonometric multiples of §203 (source: chapter12, §204–§205).

Worked Example I (§203)

Decompose $\frac{z^2}{(1-z+z^2)(1+z^4)}$ using the factor $1-z+z^2$.

Compare $1-z+z^2$ to $p^2-2pqz\cos \varphi +q^2{z}^2$: $p=1$, $q=1$, $\cos \varphi =1/2$, hence $\varphi =\pi /3$. Here $M=z^2$ and $Z=1+z^4$, $f=1$.

• From $M=z^2$: only the $z^2$-coefficient $C=1$ contributes. $R=\cos (2\pi /3)=-1/2$, $r=\sin (2\pi /3)=\sqrt{3}/2$.
• From $Z=1+z^4$: the $z^0$-coefficient $\alpha =1$ and the $z^4$-coefficient $ϵ=1$ contribute. $S=1+\cos (4\pi /3)=1/2$, $s=\sin (4\pi /3)=-\sqrt{3}/2$.
• From $Zz$: $T=\cos (\pi /3)+\cos (5\pi /3)=1/2+1/2=1$, $t=\sin (\pi /3)+\sin (5\pi /3)=\sqrt{3}/2-\sqrt{3}/2=0$.

Then $P=(Rt-rT)/(St-sT)=(0-\sqrt{3}/2)/(0-(-\sqrt{3}/2))=-1$ and $Q=0$. The partial fraction is

$\frac{-1}{1-z+z^2}.$

Subtracting from the original gives the complementary fraction $(1+z+z^2)/(1+z^4)$ (using $1+z^2+z^4=(1+z+z^2)(1-z+z^2)$). The complement still has denominator $1+z^4=(1+\sqrt{2}z+z^2)(1-\sqrt{2}z+z^2)$, two more trinomial factors with $\varphi =\pi /4$, decomposed identically in Example II (source: chapter12, §203).

Worked Example III (§203, abridged)

Decompose $\frac{1+2z+z^2}{(1-\frac{8}{5}z+z^2)(1+2z+3z^2)}$. The factor $1-\frac{8}{5}z+z^2$ has $p=1$, $q=1$, $\cos \varphi =4/5$. This is not a fractional part of a right angle, so $\sin \varphi$, $\cos 2\varphi ,\sin 2\varphi ,\cos 3\varphi ,\sin 3\varphi$ must be computed by the addition formulas: $\sin \varphi =3/5$, $\cos 2\varphi =7/25$, $\sin 2\varphi =24/25$, $\cos 3\varphi =-44/125$, $\sin 3\varphi =117/125$. With $M=1+2z+z^2$ and $Z=1+2z+3z^2$ (and $f=1$), the formulas give $R=72/25$, $r=54/25$, $S=86/25$, $s=102/25$, $T=38/125$, $t=666/125$, and finally $P=153/178$, $Q=-45/178$. The partial fraction is

$\frac{9(17-5z)/178}{1-\frac{8}{5}z+z^2}.$

A symmetric computation for the other factor $1+2z+3z^2$ (now $f=-1/\sqrt{3}$, $\cos \varphi =1/\sqrt{3}$) yields $\frac{5(5+27z)/178}{1+2z+3z^2}$ (source: chapter12, §203).

Repeated quadratic factors (§206–§210)

If $(p^2-2pqz\cos \varphi +q^2{z}^2{)}^k$ divides $N$, the algorithm above degenerates: after substituting the trinomial’s roots, both $M-PZ-QZz$ and $Z$ vanish, and the §203 system becomes $0=0$ (source: chapter12, §206). A separate iterative procedure is needed.

Tower of partial fractions (§206)

Write $N=(p^2-2pqz\cos \varphi +q^2{z}^2{)}^{k}Z$ where $Z$ contains no further power of this trinomial. The contribution to the partial-fraction decomposition is the tower

$\frac{U+uz}{(p^2-2pqz\cos \varphi +q^2{z}^2{)}^k}+\frac{V+vz}{(p^2-2pqz\cos \varphi +q^2{z}^2{)}^{k-1}}+\cdots +\frac{X+xz}{p^2-2pqz\cos \varphi +q^2{z}^2}.$

(source: chapter12, §209). Each numerator pair has two unknowns; together there are $2k$ unknowns to determine.

One step at a time (§207–§209)

Each numerator pair is found by a single application of a fixed formula. For the top numerator $(U+uz)$:

• Substitute $z^n\to (p/q{)}^n\cos n\varphi$ in $M$ to obtain the scalar $Y$, and in $Z$ to obtain $N$.
• Substitute $z^n\to (p/q{)}^n\sin n\varphi$ in $M$ to obtain $y$, and in $Z$ to obtain $n$.
• Then

$U=\frac{YN+yn}{N^2+n^2}+\frac{Yn-yN}{N^2+n^2}\cdot \frac{\cos \varphi }{\sin \varphi },u=-\frac{Yn-yN}{N^2+n^2}\cdot \frac{q}{p\sin \varphi }.$

(source: chapter12, §207). After $U,u$ are known, define the next polynomial

$F=\frac{M-(U+uz)Z}{p^2-2pqz\cos \varphi +q^2{z}^2}.$

The numerator is divisible by the trinomial (this is what the formulas for $U,u$ guarantee), so $F$ is a polynomial. Apply the same formula to $F$ in place of $M$: new substituted values $P,p$ (real and imaginary substitutions of $F$) replace $Y,y$; the values $N,n$ are unchanged because $Z$ has not changed. This produces $V,v$. Then $G=(F-(V+vz)Z)/(\text{trinomial})$, then $W,w$, then $H,X,x$, and so on for $k$ rounds (source: chapter12, §208–§209).

Complementary fraction (§210)

The sequence of polynomials $F,G,H,I,K,\dots$ produced in the iteration is exactly what is needed for the complementary fraction with denominator $Z$. After all $k$ numerators in the tower have been extracted, the next polynomial in the sequence (call it the last one) is the numerator of the complement: for $k=1$, $F/Z$; for $k=2$, $G/Z$; for $k=3$, $H/Z$; and so on. The complement, having denominator $Z$ which contains no further power of this trinomial, can itself be expressed in partial fractions by the §200–§205 rule applied to its own quadratic factors (source: chapter12, §210).

Worked example (§209)

Decompose $\frac{z-z^3}{(1+z^2{)}^4(1+z^4)}$. The repeated factor $(1+z^2{)}^4$ has $p=1$, $q=1$, $\cos \varphi =0$, $\varphi =\pi /2$, $\sin \varphi =1$, $f=1$. Take $M=z-z^3$, $Z=1+z^4$.

• At $z=i$: $M$ evaluates to $i-i^3=2i$, so $Y=0$, $y=2$. $Z$ evaluates to $1+i^4=2$, so $N=2$, $n=0$.
• $U=(YN+yn)/(N^2+n^2)+(Yn-yN)\cos \varphi /((N^2+n^2)\sin \varphi )=0+0=0$.
• $u=-(Yn-yN)\cdot q/((N^2+n^2)p\sin \varphi )=-(-4)/(4\cdot 1\cdot 1)=1$.

So $U+uz=z$. The first partial fraction is $\frac{z}{(1+z^2{)}^4}$.

Compute $F=(z-z^3-z(1+z^4))/(1+z^2)=-z^3$.

Repeat with $F=-z^3$: at $z=i$, $-i^3=i$, so the new $Y=0$, new $y=1$. $N,n$ unchanged ($N=2,n=0$). Then $V=0$, $v=-(0-2)/(4)=1/2$, so $V+vz=z/2$ and the second partial fraction is $\frac{z}{2(1+z^2{)}^3}$.

Compute $G=(-z^3-(z/2)(1+z^2))/(1+z^2)=-z/2-z^3/2\cdot$ wait — $G=(-z^3-z/2-z^5/2)/(1+z^2)=-z/2-z^3/2$.

Repeat: at $z=i$, $G=-i/2-i^3/2=-i/2+i/2=0$, so $W=w=0$. Third partial fraction is $0$.

Compute $H=(G-0)/(1+z^2)=(-z/2-z^3/2)/(1+z^2)=-z/2$.

Repeat: at $z=i$, $H=-i/2$, so the substituted values from $H$ are $0$ (real) and $-1/2$ (imaginary). Then $X=0+(0\cdot 0-(-1/2)\cdot 2)\cdot 0/(4\cdot 1)=0$ and $x=-(0\cdot 0-(-1/2)\cdot 2)\cdot 1/(4\cdot 1\cdot 1)=-1/4$. Fourth partial fraction is $-\frac{z}{4(1+z^2)}$ (source: chapter12, §209).

The complementary fraction has numerator $I=(H-(X+xz)Z)/(1+z^2)=-z/4+z^3/4$, divided by $1+z^4$:

$\frac{-z+z^3}{4(1+z^4)}.$

Putting everything together (source: chapter12, §209):

$\frac{z-z^3}{(1+z^2{)}^4(1+z^4)}=\frac{z}{(1+z^2{)}^4}+\frac{z}{2(1+z^2{)}^3}-\frac{z}{4(1+z^2)}+\frac{-z+z^3}{4(1+z^4)}.$

The remaining $1/(1+z^4)$ piece could itself be decomposed into trinomial fractions by the §200–§205 rule (Example II of §203 does exactly this for the cousin denominator $(1+\sqrt{2}z+z^2)(1-\sqrt{2}z+z^2)$).

General procedure

To decompose an arbitrary real proper rational function $M/N$:

1. If $M/N$ is improper, extract the polynomial part by division; see improper-rational-function.
2. Factor $N$ over $\mathbb{R}$ into real linear and real quadratic factors (see Chapter 9).
3. For each real linear factor (distinct or repeated), use the Chapter 2 algorithm.
4. For each distinct real quadratic trinomial factor, use §200–§205 to produce a single partial fraction $(P+Qz)/(\text{trinomial})$.
5. For each repeated real quadratic factor $(\text{trinomial}{)}^k$, use the §207–§209 iterative algorithm to produce the tower of $k$ partial fractions; the iteration’s last polynomial is the numerator of the complementary fraction.
6. Sum all partial fractions (plus the polynomial part, if any). The result equals $M/N$ in fully real partial-fraction form.

Why this matters

A real partial-fraction decomposition is what makes rational functions tractable when working over $\mathbb{R}$:

• Integration. Real linear pieces $A/(p-qz)$ integrate to logarithms; real quadratic pieces $(P+Qz)/(p^2-2pqz\cos \varphi +q^2{z}^2)$ integrate to a logarithm plus an arctangent (after completing the square). Without §200–§205, one would either work with complex partial fractions (and then re-pair them at the end) or simply not have a method.
• Series expansion. Each real quadratic piece is, by the §62–§70 recurrent-series machinery, a power series whose coefficients satisfy a trigonometric recurrence $a_{n+2}=2(\cos \varphi )a_{n+1}-a_n$. Real partial fractions therefore expose the trigonometric content of any real rational function’s series expansion.

Related pages

• partial-fraction-decomposition
• improper-rational-function
• trinomial-factor
• factoring-polynomials
• fundamental-theorem-of-algebra
• de-moivre-formula
• trigonometric-addition-formulas
• trigonometric-recurrent-progression
• recurrent-series
• chapter-2-on-the-transformation-of-functions
• chapter-9-on-trinomial-factors
• chapter-12-on-the-development-of-real-rational-functions