9. Partial Fractions

b2. Finding Coefficients - Quadratic

a. Complex Methods (Optional)

When we discussed finding the coefficients for terms with a linear denominators, \(x-a\), we found that plugging in the "obvious" value \(x=a\) gave the numerator of the linear denominator if it was non-repeating and the numerator of the highest power if it was repeating. Repeatedly differentiating and plugging in the obvious value gave the other coefficients.

However, when trying to find the coefficients for terms with a quadratic denominators, \((x-b)^2+c^2\), plugging in the "obvious" value \(x=b\) only reduced these factors to \(c^2\) and did not completely eliminate all factors other than the desired \(B(x-b)+C\) factor. However, the quadratic denominator can actually be factored into linear factors, if we allow complex numbers: \[ (x-b)^2+c^2=(x-b+c\,{\rm i})(x-b-c\,{\rm i}) \] (Check this by multiplying out the right side and using \({\rm i}^2=-1\).) So \(x=b\) is not really the correct "obvious" value to plug in; \(x=b\pm c\,{\rm i}\) is!

Before discussing finding the coefficients for quadratic terms using complex methods, we need to review some facts about complex numbers.

Given the complex number \(z=x+y\,{\rm i}\),

the length or magnitude of the complex number is: \[ |z|=\sqrt{x^2+y^2} \] Note: This is the distance from the point \((x,y)\) to the origin in the \(xy\)-plane, or the length of the vector \(\left< x,y \right>\).
the complex conjugate of the complex number is: \[ \bar z=x-y\,{\rm i} \] Note: To construct a complex conjugate, just replace all \({\rm i}\)'s by \(-{\rm i}\)'s.
Note: The product of a complex number and its conjugate is the square of its magnitude: \[ z\bar z=\bar z z=|z|^2 \] To see this just multiply it out: \[ z\bar z=(x+y\,{\rm i})(x-y\,{\rm i})=x^2+y^2=|z|^2 \]
the inverse of the complex number is: \[ z^{-1}=\dfrac{\bar z}{|z|^2} \] To see it really is the multiplicative inverse, we multiply: \[\begin{aligned} z^{-1}z=\dfrac{\bar z}{|z|^2}z=\dfrac{|z|^2}{|z|^2}=1 \end{aligned}\]

So if you are willing to use complex numbers, the procedure is to clear the denominator and then follow this method:

Find the partial fraction expansion for: \[ \dfrac{16x+128}{\left[(x-3)^2+4\right]\left[(x-1)^2+4\right]^2} =\dfrac{A(x-3)+B}{(x-3)^2+4}+\dfrac{C(x-1)+D}{(x-1)^2+4} +\dfrac{E(x-1)+F}{\left[(x-1)^2+4\right]^2} \]

This has one non-repeated and one repeated quadratic factor in the denominator. Since there are \(6\) coefficients, it would be very difficult to find the coefficients using the techniques of the previous page. So we here use the complex methods.

As usual, we first clear the denominator: \[\begin{aligned} 16x+128 =&[A(x-3)+B]\left[(x-1)^2+4\right]^2 \\ &+[C(x-1)+D]\left[(x-3)^2+4\right]\left[(x-1)^2+4\right] \\ &+[E(x-1)+F]\left[(x-3)^2+4\right] \qquad \qquad \text{(*)} \end{aligned}\] The roots of \((x-3)^2+4\) are \(x=3\pm2\,{\rm i}\). The roots of \((x-1)^2+4\) are \(x=1\pm2\,{\rm i}\). So we plug in one of each of these:

\(x=3+2\,{\rm i}\): \[\begin{aligned} 48+32\,{\rm i}+128 &=[A(2\,{\rm i})+B][(2+2\,{\rm i})^2+4]^2 \\ &\Longrightarrow \qquad B+2A\,{\rm i}=-1-2\,{\rm i} \\ &\Longrightarrow \qquad A=-1, \quad B=-1 \end{aligned}\]

How do we solve \( 176+32\,{\rm i} =[A(2\,{\rm i})+B][(2+2\,{\rm i})^2+4]^2 \)?

We first compute that: \[\begin{aligned} [(2+2\,{\rm i})^2+4]^2 &=[4+8\,{\rm i}+4\,{\rm i}^2+4]^2 =[4+8\,{\rm i}]^2 \\ &=16+64\,{\rm i}+64\,{\rm i}^2 =-48+64\,{\rm i} \end{aligned}\] Then the inverse (reciprocal) of \(-48+64\,{\rm i}\) is \[ (-48+64\,{\rm i})^{-1} =\dfrac{-48-64\,{\rm i}}{48^2+64^2} =\dfrac{-1}{400}(3+4\,{\rm i}) \] So we multiply both sides of our equation by \(\dfrac{-1}{400}(3+4\,{\rm i})\) to get: \[\begin{aligned} [A(2\,{\rm i})+B] &=\dfrac{-1}{400}(3+4\,{\rm i})(176+32\,{\rm i}) \\ &=\dfrac{-1}{400}(528+96\,{\rm i}+704\,{\rm i}+128\,{\rm i}^2) \\ &=\dfrac{-1}{400}(400+800\,{\rm i}) =-1-2\,{\rm i} \end{aligned}\] We finally equate real and imaginary parts to get \(A=-1\) and \(B=-1\).

\(x=1+2\,{\rm i}\): \[\begin{aligned} 16+32\,{\rm i}+128 &=[E(2\,{\rm i})+F][(-2+2\,{\rm i})^2+4] \\ &\Longrightarrow \qquad F+2E\,{\rm i}=4+16\,{\rm i} \\ &\Longrightarrow \qquad E=8, \quad F=4 \end{aligned}\]

How do we solve \( 144+32\,{\rm i} =[E(2\,{\rm i})+F][(-2+2\,{\rm i})^2+4] \)?

We first compute that: \[ [(-2+2\,{\rm i})^2+4] =[4-8\,{\rm i}+4\,{\rm i}^2+4] =4-8\,{\rm i} \] Then the inverse (reciprocal) of \(4-8\,{\rm i}\) is \[ (4-8\,{\rm i})^{-1} =\dfrac{4+8\,{\rm i}}{4^2+8^2} =\dfrac{1}{20}(1+2\,{\rm i}) \] So we multiply both sides of our equation by \(\dfrac{1}{20}(1+2\,{\rm i})\) to get: \[\begin{aligned} [E(2\,{\rm i})+F] &=\dfrac{1}{20}(1+2\,{\rm i})(144+32\,{\rm i}) \\ &=\dfrac{1}{20}(144+32\,{\rm i}+288\,{\rm i}+64\,{\rm i}^2) \\ &=\dfrac{1}{20}(80+320\,{\rm i}) =4+16\,{\rm i} \end{aligned}\] We finally equate real and imaginary parts to get \(E=8\) and \(F=4\).

Notice that this gave us both coefficients in the numerator of \((x-3)^2+4\) and both coefficients in the numerator of the highest power of \((x-1)^2+4\).

To find the last \(2\) coefficients, we differentate (*) to get: \[\begin{aligned} 16=&A\left[(x-1)^2+4\right]^2+[A(x-3)+B]2\left[(x-1)^2+4\right]2(x-1) \\ &+C\left[(x-3)^2+4\right]\left[(x-1)^2+4\right] \\ &+[C(x-1)+D]\left[(x-3)^2+4\right]2(x-1) \\ &+[C(x-1)+D]2(x-3)\left[(x-1)^2+4\right] \\ &+E\left[(x-3)^2+4\right]+[E(x-1)+F]2(x-3) \qquad \qquad \text{(**)} \end{aligned}\] Notice we used the product rule and did not simplify!
We now plug in \(x=1+2\,{\rm i}\) and use the values of \(E\) and \(F\):

\(x=1+2\,{\rm i}\): \[\begin{aligned} 16&=[C(2\,{\rm i})+D]\left[(-2+2\,{\rm i})^2+4\right]2(2\,{\rm i}) \\ &\quad+E[(-2+2\,{\rm i})^2+4]+[E(2\,{\rm i})+F]2(-2+2\,{\rm i}) \\ &=[C(2\,{\rm i})+D][32+16\,{\rm i}] \\ &\quad+E[4-8\,{\rm i}]+[E(2\,{\rm i})+F](-4+4\,{\rm i}) \\ &=[C(2\,{\rm i})+D][32+16\,{\rm i}]-48-112\,{\rm i} \\ &\Longrightarrow \qquad D+2C\,{\rm i}=3+2\,{\rm i} \\ &\Longrightarrow \qquad C=1, \quad D=3 \end{aligned}\]

How do we solve \( 16=[C(2\,{\rm i})+D][32+16\,{\rm i}]-48-112\,{\rm i} \)?

We first rewrite the equation as \[ [D+2C\,{\rm i}][32+16\,{\rm i}]=64+112\,{\rm i} \] Then the inverse (reciprocal) of \(32+16\,{\rm i}\) is \[ (32+16\,{\rm i})^{-1} =\dfrac{32-16\,{\rm i}}{32^2+16^2} =\dfrac{1}{80}(2-{\rm i}) \] So we multiply both sides of our equation by \(\dfrac{1}{80}(2-{\rm i})\) to get: \[\begin{aligned} D+2C\,{\rm i} &=\dfrac{1}{80}(2-{\rm i})(64+112\,{\rm i}) \\ &=\dfrac{1}{80}(128+224\,{\rm i}-64\,{\rm i}-112{\rm i}^2) \\ &=\dfrac{1}{80}(240+160\,{\rm i}) =3+2\,{\rm i} \end{aligned}\] We finally equate real and imaginary parts to get \(C=1\) and \(D=3\).

So the partial fraction expansion is \[ \dfrac{16x+128}{\left[(x-3)^2+4\right]\left[(x-1)^2+4\right]^2} =\dfrac{-(x-3)-1}{(x-3)^2+4}+\dfrac{(x-1)+3}{(x-1)^2+4} +\dfrac{8(x-1)+4}{\left[(x-1)^2+4\right]^2} \]

It's amazing how much easier complex numbers made this process. Well, it wasn't really easy! But it was still much easier then solving \(6\) equations for \(6\) unknowns.

You can also practice finding the Coefficients in a Partial Fraction Expansion by using the following Maplet (requires Maple on the computer where this is executed):

Partial Fractions: Finding Coefficients Rate It

We are now ready to do the integrals.