I felt that the class could use some more exposure to complex numbers so I said some things again and then some other things about them. The point (1,0) in the plane is written 1 when you regard it as a complex number, and (0,1) is written i. The real line is the x-axis. I reminded the class that multiplication works by Multiplying Magnitudes and Adding Angles. I thought about the nth roots of unity, by way of illustration. I also explained that once you've adjoined a root of the quadratic polynomial z^2+1, you can factor any degree n polynomial as p(z) = a(z-r1)(z-r2)...(z-rn) , where a, r1, ... rn are complex numbers, namely the roots of p.
Then I pointed out that the expression e^t doesn't really instruct us to raise the number e to the t-th power; think about the case t=sqrt(2) to be convinced of this. Rather, raising to powers is defined in terms of the exponential function, which is really simply the solution to the initial value problem z'=z, z(0)=1. More generally, e^{at} is the solution to z'=az, z(0)=1.
This is for a real. I tried Wednesday to discuss e^{it}. Let's try to define this function as the solution to z'=iz, z(0)=1. This seems a little underhanded of me, since at first I led you to believe that solutions to ODEs should be real-valued, while this z can't be. But let it happen: look for a function z from the reals to the complexes. Since C is just the plane R^2, we know how to think about such a function: it parametrizes a curve in the plane; it's given by z(t)=(x(t),y(t)), or in terms of complex numbers, z(t) = x(t) + iy(t). Its derivative is the velocity vector, computed coordinatewise: z' = x' + iy'. To investigate the right-hand side of the ODE, observe that multiplication by i doesn't change the magnitude, and adds pi/2 to the angle: it rotates counterclockwise by 90 degrees. When you add the initial condition z(0)=1, you find a parametrization of the unit circle. In formulas,
z(t) = cox(t) + i sin(t) .
I checked that indeed, z'=iz and z(0)=1. Since we had agreed that the solution to this IVP would be e^{it}, we've recovered what I claimed on Wednesday, with better reason.
Similarly, if a+bi is an arbitrary complex number, e^{(a+bi)t} is the solution to z'=(a+bi)z, z(0)=1, which as you can check is
e^{(a+bi)t} = e^{at}(cos(bt) + i sin(bt)) .
I took a positive and sketched this curve: a spiral. I did not point out that if w1 and w2 are any complex numbers, then e^{w1} e^{w2 t} and e^{w1 + w2 t} are both solutions to the IVP z' = w2 z, z(0) = e^{w1}, and hence must be equal. With t = 0 this gives
e^{w1} e^{w2} = e^{w1 + w2}.
I returned to the second-order constant coefficient homogeneous linear ODE y" + py' + qy = 0, reminded the class of the significance of the coefficients q (the spring constant) and p (the damping term), that it's a good idea to look for exponential solutions y = e^{rt}, and that y is a solution exactly when r is a root of the characteristic polynomial r^2 + pr + q. I wrote the quadratic formula as
r = -p/2 +- sqrt{(p/2)^2-q}.
Take the case p^2 greater than 4q : distinct real roots r1 and r1, leading to exponential solutions e^{r1 x}, e^{r2 x}, and the general solution y = a e^{r1 x} + b e^{r2 x} . To justify calling this the general solution I should check that if r1 is different from r2 then e^{r1 x} and e^{r2 x} are linearly independent. I did this by assuming
a e^{r1 x} + be{r2 x} = 0,
setting x = 0 to get a + b = 0, then differentiating and setting x = 0 again to get r1 a + r2 b = 0. These force a = b = 0, and so linear independence.