Physics 263: Problem Set #9

1. (BTM 5.3.2.) These are a bit tedious, but once you slug through them, you should be expert at converting to polar form and manipulating polar form. I always start by finding the modulus r from sqrt(z z*). Then I draw a picture with the x and y coordinates, so that I can see roughly what the angle should be (e.g., what quadrant it is in). Often the picture is enough to read off the angle, because it is one of the common ones (like pi/3). If not, then the triangle gives theta = arctan(y/x), but make sure the sign is correct for the quadrant the angle is in. Once you have z1 and z2 in polar form, the product and quotient are simple. Most of the answers are in the back of the book, so no excuses! (This also means you have to show your work.)
   (BTM 5.3.3.) Convert back to Cartesian using r e^{i theta} = r(cos theta + i sin theta), add and then convert back to polar. Just use a calculator or MATLAB for the conversion.
2. (BTM 5.3.4.) These are very quick once you convert to exponential form. Just do the arithmetic and show the two sides of the equation give the same answer.
3. (BTM 5.3.6.) This is very quick and elegant. Just substitute n=4, do the binomial expansion, and equate real and imaginary parts! Similarly, equate real and imaginary parts for the two sides of the given equation, using e^{i A} = cos A + i sin A repeatedly.
4. (BTM 5.4.4.) The first part is just plug-and-chug to get familiar with the formulas of section 5.4. To get the last part, find the formula for |Z| and minimize with respect to omega. The answer in terms of L and C should be familiar from your past study of circuits.
5. (BTM 5.4.6.) Combining three impedances in parallel is the obvious generalization (familiar from resistors) of equation (5.4.24). The individual impedances for each element follow from equation (5.4.16); do you see how? The answer is in the back.
6. (BTM 5.4.7.) To solve this one, you need to find the full solution, including the transient solution. The steady-state solution is no problem, but the transient solution has to be chosen to satisfy the initial conditions. You can make an ansatz that looks like the solution in the back or an ansatz that has a complex amplitude and frequency.