College Math Teaching

October 9, 2013

Fun with complex numbers and particular solutions

I was fooling around with y''+by'+cy = e^{at}cos(bt) and thought about how to use complex numbers in the case when e^{at}cos(bt) is not a solution to the related homogenous equation. It then hit me: it is really quite simple.

First notes the following: e^{rt}cos(st) = \frac{1}{2}(e^{(r + si)t} + e^{(r - si)t}) and e^{rt}sin(st) = \frac{1}{2i}(e^{(r + si)t} - e^{(r - si)t}).

Then it is a routine exercise to see the following: given that z = r+si, \bar{z} = r-si are NOT solutions to p(m)= m^2 + bm + c = 0 p(m) is the characteristic equation of the differential equation. Then: attempt y_p = Ae^{zt} + Be^{\bar{z}t} Put into the differential equation to see y''_p + by'_p + cy_p = A(z^2+bz+c)e^{zt} + B(\bar{z}^2 + b\bar{z} + c)e^{\bar{z}t} .

Then: if the forcing function is e^{rt}cos(st) , a particular solution is y_p = Ae^{zt} + Be^{\bar{z}t} where A = \frac{1}{2(p(z))}, B = \frac{1}{2(p(\bar{z}))}. If the forcing function is e^{rt}cos(st) , a particular solution is y_p = Ae^{zt} - Be^{\bar{z}t} where A = \frac{1}{2i(p(z))}, B = \frac{1}{2i(p(\bar{z}))}.

That isn’t profound, but it does lead to the charming exercise: if z, \bar{z} are NOT roots to the quadratic with real coefficients p(x), then \frac{1}{p(z)} + \frac{1}{p(\bar{z})} is real as is \frac{i}{p(z)} - \frac{i}{p(\bar{z})} .

Let’s check this out: \frac{1}{p(z)} + \frac{1}{p(\bar{z})} = \frac{p(\bar{z})+p(z)}{p(z)p(\bar{z})}. Now look at the numerator and the denominator separately. The denominator: p(z)p(\bar{z})= (z^2 + bz +c)(\bar{z}^2 + b\bar{z} + c) = (z^2 \bar(z)^2) + b(b (z \bar{z}) + (z(z \bar{z}) +\bar{z}(z \bar{z})) + c (\bar{z} + z) + (z^2 + \bar{z}^2)) + (c^2) Now note that every term inside a parenthesis is real.

The numerator: z^2 + \bar{z}^2 + b (z + \bar{z}) + 2c is clearly real.

What about \frac{i}{p(z)} - \frac{i}{p(\bar{z})} ? We need to only check the numerator: i (z^2 - \bar{z}^2 + b(z - \bar{z}) + c-c) is indeed real.

Yeah, this is elementary but this might appear as an exercise for my next complex variables class.

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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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