18.03 Class 10, Wed. Feb. 24, 1999


Constant coefficient homogeneous linear second order ODEs, Complex Numbers

I handed out a second part to PS2 (which contains a little misprint: y" + py' + q should of course be y" + py' +qy. The PS is due *Monday.*

After reminding the class of the significance of each term in the general form of a linear second order ODE, I focused on the constant coefficient homogeneous case,

y" + by' + cy = 0.

y" + y = 0 has periodic solutions (undamped spring). The example y" - y = 0 (for which e^x is one solution) leads to the ansatz y = e^{rx}. There remains a mystery about how this ansatz will accomodate the periodic solutions as well.

The ansatz works as long as r is a root of the "characteristic polynomial"

p(r) = r^2 + br + c.

This gives us two linearly independent (though I didn't check this) solutions as long as there are two distinct real roots. We are left with two problems: 1. only one real root (repeated root), and 2. no real roots. The solution to this problem is: complex numbers.

I described the Argand plane and the multiplication forced on us by i^2 = -1. I defined complex conjugation, pointed out the formulas for Re(z) and Im(z), explained that conjugation respects addition and multiplication, and showed that z z-bar = |z|^2 and |zw| = |z||w|.

Then I talked about the unit circle, where z-bar and z-inverse conincide. I defined

cis(theta) = cos(theta) + i sin(theta),

checked using trig addition formulas that it behaves exponentially: you multiply by adding angles. Then I used this as an excuse to define

e^{i theta} = cis(theta)

and then e^z for general complex z accordingly. Then

e^{z+w) = e^z e^w , e^0 = 1.

Polar coordiates take the form z = |z|e^{i theta}.

I pointed out that exponentiation and conjugation commute, and used this to see the formulas for cos and sin in terms of exponentials of purely imaginary numbers.


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