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+c2, plugging in the "obvious" value x=b only
reduced these factors to c2 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+c2=(x−b+ci)(x−b−ci)
(Check this by multiplying out the right side and using i2=−1.)
So x=b is not really the correct "obvious" value to plug in;
x=b±ci 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+yi,
the length or
magnitude of the complex number is:
∣z∣=x2+y2
Note: This is the distance from the point (x,y) to the origin in
the xy-plane, or the length of the vector ⟨x,y⟩.
the complex conjugate of the complex number is:
zˉ=x−yi
Note: To construct a complex conjugate, just replace all i's
by −i's.
Note: The product of a complex number and its conjugate is the square of its magnitude:
zzˉ=zˉz=∣z∣2
To see this just multiply it out:
zzˉ=(x+yi)(x−yi)=x2+y2=∣z∣2
the inverse of the complex number is:
z−1=∣z∣2zˉ
To see it really is the multiplicative inverse, we multiply:
z−1z=∣z∣2zˉz=∣z∣2∣z∣2=1
So if you are willing to use complex numbers, the procedure is to clear
the denominator and then follow this method:
Plug in the Obvious Complex Values of x, Differentiate and Repeat
Find the partial fraction expansion for:
[(x−3)2+4][(x−1)2+4]216x+128=(x−3)2+4A(x−3)+B+(x−1)2+4C(x−1)+D+[(x−1)2+4]2E(x−1)+F
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:
16x+128=[A(x−3)+B][(x−1)2+4]2+[C(x−1)+D][(x−3)2+4][(x−1)2+4]+[E(x−1)+F][(x−3)2+4](*)
The roots of (x−3)2+4 are x=3±2i.
The roots of (x−1)2+4 are x=1±2i.
So we plug in one of each of these:
We first compute that:
[(2+2i)2+4]2=[4+8i+4i2+4]2=[4+8i]2=16+64i+64i2=−48+64i
Then the inverse (reciprocal) of −48+64i is
(−48+64i)−1=482+642−48−64i=400−1(3+4i)
So we multiply both sides of our equation by
400−1(3+4i) to get:
[A(2i)+B]=400−1(3+4i)(176+32i)=400−1(528+96i+704i+128i2)=400−1(400+800i)=−1−2i
We finally equate real and imaginary parts to get A=−1 and B=−1.
We first compute that:
[(−2+2i)2+4]=[4−8i+4i2+4]=4−8i
Then the inverse (reciprocal) of 4−8i is
(4−8i)−1=42+824+8i=201(1+2i)
So we multiply both sides of our equation by
201(1+2i) to get:
[E(2i)+F]=201(1+2i)(144+32i)=201(144+32i+288i+64i2)=201(80+320i)=4+16i
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:
16=A[(x−1)2+4]2+[A(x−3)+B]2[(x−1)2+4]2(x−1)+C[(x−3)2+4][(x−1)2+4]+[C(x−1)+D][(x−3)2+4]2(x−1)+[C(x−1)+D]2(x−3)[(x−1)2+4]+E[(x−3)2+4]+[E(x−1)+F]2(x−3)(**)Notice we used the product rule and did not simplify!
We now plug in x=1+2i and use the values of E and F:
We first rewrite the equation as
[D+2Ci][32+16i]=64+112i
Then the inverse (reciprocal) of 32+16i is
(32+16i)−1=322+16232−16i=801(2−i)
So we multiply both sides of our equation by
801(2−i) to get:
D+2Ci=801(2−i)(64+112i)=801(128+224i−64i−112i2)=801(240+160i)=3+2i
We finally equate real and imaginary parts to get C=1 and D=3.
So the partial fraction expansion is
[(x−3)2+4][(x−1)2+4]216x+128=(x−3)2+4−(x−3)−1+(x−1)2+4(x−1)+3+[(x−1)2+4]28(x−1)+4
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):