The haps in SF

Leif Johnson — 14 Oct 2003, 19:10

imaginary numbers

today in class we started going over imaginary numbers. i'm not sure i've ever thought much about this topic, just more or less accepted it and moved on. now, don't get me wrong ; i'm a fan of the patented alex rosefielde smile-and-nod methodology. when faced with potentially upsetting new situations, it lets one pass along and not get too tied up on things that are probably not that important in the long run. but now, looking back a little on imaginary numbers, i started thinking that i need to get to a deeper understanding : just accepting the definition isn't really helping explain the concept to other people.

in particular, prompted by persistent questions from a couple of students in class, i started thinking about what an imaginary number represents. perplexed students in class kept pressuring the teacher to explain “which number %% i %% represents.” his best answer, which is more or less the one i would have given, is that %% i %% is not a number that we would recognize. it's a different sort of number.

coming from a math background, surely this “which number” deal sounds at first like a nonsensical question : after all, %% i %% squared is defined to be negative one, plain and simple. but what, really, does an imaginary number represent ? i wondered for a bit in class and abruptly came across a seemingly simple example : a parabola with no real roots, i.e. one that doesn't touch the %% x %%-axis. clearly such functions have imaginary roots, but can't those roots somehow be displayed on the %% x %%-%% y %% coordinate plane ? perhaps something like the magnitude of a vector from the origin to the base of the parabola ...

consider %% f(x) = x^2 + 1 %%. the roots of this equation are %% i %% and %% -i %%. or a more contrived example : %% g(x) = (x - 1)^2 + 1 = x^2 - 2x + 2 %%. this parabola's base lies at the point %% (1, 1) %%, and its roots are %% 1 + i %% and %% 1 - i %%. how about the parabola whose base is at %% (2, 5) %% : %% h(x) = (x - 2)^2 + 5 = x^2 - 4x + 9 %%. the roots of %% h(x) %% are %% 2 + \sqrt{5}i %% and %% 2 - \sqrt{5}i %%.

in these simple cases, the roots, though complex, contain the real numbers that locate the base of the parabola. interesting. is this always true ? i fear a symbolic solution is needed. not going to happen here in front of my computer though, stupid html.


last night i booked it over to the aikido dojo on my bike, just barely getting there in time to sneak on the matt (after class had started though). practice was excellent, though : lots of bokken work and concentration on moving from the center and such. it's really challenging to do this, and quite difficult to explain in words.

at the end of class we got out these dodgeballs and did a leg exercise against the wall : lean up against the wall, ball between back and wall, knees and feet about shoulder width apart, so your knees and toes point straight forward. lower your torso so your knees are at a 90 degree angle, then raise up 45 degrees, and lower back to 90 degrees. pretty tough on the thighs, but it's good to get some strength training back on these weak, weak upper legs of mine. the knees held up well, i really just need to get in better shape.

it's funny : in high school, being in shape was just one of those additional benefits of exercising, and the only real threat of injury was from exercising itself. now being in shape is becoming a necessity for avoiding injury during an otherwise exercise–less existence. hmm.


i wonder if i can get my monthly food budget to under $ 100 a month. i just got back from this one store just off of pacific on the east side of stockton, where i bought a dozen eggs, 6 bananas, an onion, a tub of tofu (made in san francisco), a bag of snow peas, and a can of coconut milk : $ 4.75 total.