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How are imaginary numbers used in real life?

Hye there! How are they applied in the real world? If possible, I would not like just like listing, but an explanation..I have to write quite a bit about them. I know they are used in electrical engineering with oscillating volage/current graphs. Where else, possibly outside electricity? Thanks, Any links would be very much appreciated where I can find more information. I'd like to stress again, I am looking for an explanation if possible. :)

Public Comments

1. I think electricity might be it for physical interpretations, but De Moirve's formula, that cis^n(θ)=cis(nθ), allows people to find multiple angle formulas quickly, by equating the real or imaginary parts of cos(nθ)+isin(nθ) to the expansion of (cos(θ)+isin(θ))^n. If you can think of any real-life applications of sines of triple angles, these'd be helpful.
2. Hi, Here is the answer from the website listed below: 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. I had to delete part of the answer due to length. See website for entire article. I hope that helps!! :-)