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.

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