# 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.