Applications of Imaginary NumbersDate: 10/14/97 at 18:40:14 From: Beatka Zakrzewski Subject: Imaginary Numbers in the work force (applications) Dear Dr. Math, I know that you have already received tens of questions about imaginary numbers but I can't seem to find a straight answer to mine. Where are imaginary numbers used today in real life, as in the work force or other areas that use math? I have heard that they are used in electrical technologies - if so, could you specifically tell me in which areas? Thank you. Sincerely, Beatka Zakrzewski Date: 10/14/97 at 19:03:06 From: Doctor Tom Subject: Re: Imaginary Numbers in the work force (applications) Hi Beatka, I answered almost exactly the same question for a teacher a couple of years ago and saved my reply. I'll send it to you. Unfortunately, as you'll see from the description, they are often used with more sophisticated math than you know yet, but if you ignore the details below and concentrate on the problem descriptions, they may make some sense to you. I teach Algebra II and Trigonometry in Gilroy High School, Gilroy, CA. The students' ages are 15-17. Their question is: Who uses imaginary numbers in the real world? The students learn to calculate with i and complex numbers of the form a + bi. The book indicates that electrical engineers use imaginary numbers, but use j instead of i. What do they *do* with the numbers? And does anyone else actually use them? First off, electrical engineers use "j" because "i" is almost always used to mean "current" in their equations. Where most of us would write "a + bi", they'd write "a + bj", but it means the same thing. Electrical engineers often have to solve what are called "differential equations," which are a bit hard to explain without knowing a bit about calculus. Basically, a differential equation relates functions to their rates of growth. The solution to a differential equation is usually a function, not a number. As a specific example, suppose you have a snowplow that keeps piling up more and more snow in front of it so that the farther it goes, the heavier the load it is pushing, and the heavier the load, the slower it goes, and the slower it goes the slower the pile of snow in front of it grows. You can (with a differential equation) relate the amount of snow at a given time t (call it A(t)) to the velocity of the plow, and the equations can be solved to give the function A(t) at all times t. But often, it's easier to solve differential equations in the domain of complex numbers because the equations are a lot nicer, but you know that the solution you care about is just the real part of the solution. It's difficult to give an example without some calculus. I can, however, show you a nice example that may make it clear that working in the domain of complex numbers is easier in some cases than working strictly in the reals. If you're studying the complex numbers and trigonometry at the same time, in theory you can follow the next steps, but if you can't, don't worry; just look at the messiness of the calculation - I'm trying to show how going to the domain of complex numbers can drastically simplify a problem. 3 Try to work out the following: (cos x + i*sin x) . Multiply this out and reduce it to the simplest form. I get this: 3 2 2 3 cos x + 3*i*cos x * sin x - 3*cos x * sin x - i*sin x. (It's a nice exercise in algebra, complex numbers, and trigonometry.) Let's just look at the real part (you can do the same sort of thing with the imaginary part): 3 2 cos x - 3*cos x * sin x 2 2 2 = cos x * ( cos x - sin x) - 2*cos x * sin x 2 2 Now, since cos(2x) = cos x - sin x, and sin(2x) = 2*sin x * cos x, we can re-write the equation above as: cos x * cos(2x) - sin x * sin(2x). Now, cos(3x) = cos(x + 2x) = cos x * cos(2x) - sin x * sin(2x), so the real part is just cos(3x). With a similar amount of ugly calculation, we can get that the imaginary part is i*sin(3x). So the answer is: cos(3x) + i*sin(3x). 2 3 I assume you understand exponents = x = x * x, x = x * x * x, and so on. Well, it turns out that there's a special number called e which is equal (approximately) to 2.71828182845... which satisfies the following equation (in the complex numbers: ix e = cos(x) + i*sin(x) So the original problem I stated was to find the cube of the number on the left: ix 3 i(3x) (e ) = e = cos(3x) + i*sin(3x), so you're done in a single step. Another important application of complex numbers to the real world is in physics. In quantum mechanics, one cannot say with precision where a particle is. You can only give an probability distribution of its position in space. And the only way to calculate the distributions is using complex variables. Unfortunately, this is even harder to explain, but such calculations have to be done for almost any calculations about nuclear reactions. Finally, something that may not be precisely an "application," but one that you can easily experiment with, is that a certain class of complex numbers behave as rotation operators. For example, draw the usual real and imaginary axes, and plot any point on it (say 3 + 5i) Multiply this number by i, and you get (-5 + 3i). If you plot this new point, you'll find that it is the original point rotated about the origin by 90 degrees counter- clockwise. This works for ANY complex number. Multiply by i, and you'll rotate it by 90 degrees. Now, take any complex number, and multiply it by cos(45) + i*sin(45) (in degrees), which is about (.707 + .707*i). This rotates points clockwise by 45 degrees. And there's nothing special about 45 degrees. Multiply any complex number by cos(x) + i*sin(x), and you'll rotate the number about the origin by an angle x. In the same way, adding a fixed complex number is equivalent to a translation, and multiplying by a real number expands or contracts the values. By combinations of rotations, translations, and expansions/ shrinkages, you can do most 2-dimensional computer graphics operations on objects in a plane. Unfortunately, in 3 dimensions, it's not so easy, and the easy way involves something called matrix multiplication. I know none of these is easy to understand, but if you want to do them eventually, you've got to have a solid foundation in the basic operations on complex numbers - just as you can't do algebra until you learn to add and subtract real numbers. -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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