# College Math Teaching

## September 13, 2013

### Partial Fractions Expansion: sometimes complex numbers can save time.

Filed under: calculus, differential equations, integrals — Tags: , — collegemathteaching @ 8:56 pm

One sometimes needs to do a partial fraction expansion when one is integrating or when one is doing Laplace transforms. Most people know the standard methods: either gather terms and compare coefficients, or use selected (real) values for $x$.

But sometimes, (NOT all of the time), one can speed things up by using complex numbers.

Here is an example: expand $\frac{1}{(x+1)(x^2 + 1)}$.

Solution: set this up as $\frac{1}{(x+1)(x^2 + 1)}=\frac{A}{(x+1)}+\frac{Bx+C}{(x^2 + 1)}$.
Now clear denominators to obtain $1 = A(x^2+1)+ (Bx+C)(x+1)$.

Setting $x = -1$ yields $1 = 2A$ which means $A = \frac{1}{2}$

(Yes, I know that we used a number not in the domain of the original fraction…but why can we get away with that? :-))

Now set $x = i$ we obtain $1= (Bi+C)(i+1)=(B+C)i +C-B$. By comparing real and imaginary parts, we obtain $-B = C$ and then $C = \frac{1}{2}, B = \frac{-1}{2}$.

Here is a second, more complicated case. Expand $\frac{1}{x^3 -1} = \frac{A}{x-1}+ \frac{Bx + C}{x^2+x + 1}$.

Clear denominators again to obtain $1 = A(x^2+x+1)+ (Bx+C)(x-1)$. Trying $x = 1$ yields $A = \frac{1}{3}$.

Now use a primitive complex 3’rd root of unity: $x = e^{\frac{2\pi i}{3}}$; this causes the first term to vanish. The second term becomes immediately:
$B(e^{\frac{4\pi i}{3}}-e^{\frac{2\pi i}{3}}) + C(e^{\frac{2\pi i}{3}}-1)$ which simplifies to: $\sqrt{3}i(B + \frac{1}{2}C) - \frac{3}{2}C = 1$.
Comparing real and imaginary parts again: $C = \frac{2}{3}$ and $B = -\frac{1}{3}$.

Caveat: one has to be very comfortable with complex arithmetic to use this method, but some engineers and physicists are.