Imaginary numbers

PinkFloyd
Science 08 Jul '08 22:29
1. 08 Jul '08 22:29
I vaguely remember this term from high school--either adv. algebra or trig. Anybody care to remind me what they are? And is infinity an imaginary number?
2. flexmore
Quack Quack Quack !
08 Jul '08 23:39
Originally posted by PinkFloyd
I vaguely remember this term from high school--either adv. algebra or trig. Anybody care to remind me what they are? And is infinity an imaginary number?
If you just use normal numbers then you cannot find the square root of negative numbers ... imaginary numbers are tacked on so that you can
3. sonhouse
Fast and Curious
09 Jul '08 00:381 edit
Originally posted by flexmore
If you just use normal numbers then you cannot find the square root of negative numbers ... imaginary numbers are tacked on so that you can
Like the square root of minus one. They call it "i" for want of a better name. It just is a symbol that means whatever the square root of minus one really is, we don't know exactly what, but we are going to call it 'i'.
So you can have numbers like 6+ 3i which is an imaginary number, the 3i part is imaginary. They are complex numbers. Its based on the following rules:
i^0=+1
i^1=+1
i^2=-1
i^3=-i
i^4=+1
i^5=i
i^6=-1
i^7=i
i^8=+1 and the pattern repeats.
4. 09 Jul '08 05:44
Originally posted by PinkFloyd
I vaguely remember this term from high school--either adv. algebra or trig. Anybody care to remind me what they are? And is infinity an imaginary number?
Try to solve the equation x^2=-1. There are no solution with ordinary numbers but if you introduce a number "i" with the property that i^2 = -1 the equation is solvable.

With the aid of "i" every equation of the form ax^2+bx+c=d is solvable. Even every equation with x^n where n can be any natural number can be solved.

There is a natural blockage for thinking that s^2=-1 is solvable. By disregard this blockage you can understand the world of complex numbers involving "i". Mathematics is thereafter one degree funnier.
5. 09 Jul '08 08:17
Originally posted by sonhouse
It just is a symbol that means whatever the square root of minus one really is, we don't know exactly what, but we are going to call it 'i'.
There is no "whatever the square root of minus one really is" to it - that sort of implies that i lies somewhere on the real number line, which it doesn't. We know precisely what the square root of minus one is - it is the square root of minus one. There is no other way of writing it! (other than, say, 0.5*sqrt(-4), but that's just being petty.)
6. sonhouse
Fast and Curious
09 Jul '08 09:30
Originally posted by Swlabr
There is no "whatever the square root of minus one really is" to it - that sort of implies that i lies somewhere on the real number line, which it doesn't. We know precisely what the square root of minus one is - it is the square root of minus one. There is no other way of writing it! (other than, say, 0.5*sqrt(-4), but that's just being petty.)
That is a mere description, not any kind of deep understanding. We call this unknown quantity i and have logic that proves it works, thats about all we can say about it. Not exactly a fundamental understanding.
7. flexmore
Quack Quack Quack !
09 Jul '08 11:501 edit
Originally posted by Swlabr
There is no "whatever the square root of minus one really is" to it - that sort of implies that i lies somewhere on the real number line, which it doesn't. We know precisely what the square root of minus one is - it is the square root of minus one. There is no other way of writing it! (other than, say, 0.5*sqrt(-4), but that's just being petty.)
it does not imply that it lies on the real number line ... the real number line is just a play toy like the other contrived number sets.

they keep tacking on thinking of things they might like to do:
add .. and they end up with the positive integers,
subtract .. and they end up with all the integers positive and negative,
divide ... and end up with all the rationals
get to all limit points. . and have the reals,
square root ... get the complex numbers
8. ChronicLeaky
Don't Fear Me
09 Jul '08 15:29
Find the ring of integers in the field Q[x]/(x^2+1), then find a generator for the (multiplicative) group of units in this ring. The real multiples of this generator are the imaginary numbers.
9. 09 Jul '08 16:32
Originally posted by ChronicLeaky
Find the ring of integers in the field Q[x]/(x^2+1), then find a generator for the (multiplicative) group of units in this ring. The real multiples of this generator are the imaginary numbers.
What does the Q mean in your formula? I'd like to try it.
10. 09 Jul '08 16:34
Originally posted by FabianFnas
Try to solve the equation x^2=-1. There are no solution with ordinary numbers but if you introduce a number "i" with the property that i^2 = -1 the equation is solvable.

With the aid of "i" every equation of the form ax^2+bx+c=d is solvable. Even every equation with x^n where n can be any natural number can be solved.

There is a natural blockage for ...[text shortened]... the world of complex numbers involving "i". Mathematics is thereafter one degree funnier.
So, other than a curiosity, or to make fun mathematical puzzles, imaginary numbers have no real use in mathematics?
11. sonhouse
Fast and Curious
09 Jul '08 20:47
Originally posted by PinkFloyd
So, other than a curiosity, or to make fun mathematical puzzles, imaginary numbers have no real use in mathematics?
They have real value in electronics. I don't know how much basic electricity you have taken but if you have a straight DC circuit, say a battery and a light bulb, if you have a ten volt battery and a light that will draw one amp at ten volts, that represents a ten watt bulb. You just multiply the amps times the volts and you get the power in watts.
With AC circuits, it's not so easy. For ordinary household wiring, you have say, 110 volts, and 60 hertz or 60 cycles per second. So with AC you can't just multiply the current times the voltage, especially with what are called inductive loads. A non-inductive load is like a light bulb, regular filament bulb, not fluorescent bulbs, and with a regular filament bulb you get a pretty good estimate of the power drawn, just multiply the volts times the amps, 110 volts, 1 amp, is about 110 watts.
But in an inductive load, that is to say one where there is a big coil of wire involved, like the windings in an electric motor, things get complicated because the voltage and the current peaks don't happen at the same time but a different times around that 60 cycle per second frequency. They 'Lead' or "Lag' each other and so the power consumed is a bit trickier to calculate. That's where imaginary numbers come into play. The calculation goes into vectors now and the imaginary numbers are part of that calculation. I don't think you want a full tour of that, you can't cover that in paragraph but suffice it to say, without imaginary numbers you would have a hard time trying to figure out just how much power a given electric motor will use up. That also works for the much higher frequencies of radio and TV signals, same idea but MUCH faster cycle times.
12. 09 Jul '08 22:27
Originally posted by PinkFloyd
So, other than a curiosity, or to make fun mathematical puzzles, imaginary numbers have no real use in mathematics?
"imaginary numbers have no real use in mathematics?"

This was intended to be funny, wasn't it? 😀

Real use for imaginary number, or imaginary use for real numbers!

I think there are a lot of areas in mathematics where complex numbers (of the form a+bi) are used. Partly because it helps calculations. Some trig problems are very simple when you know complex numbers.

Everything you can do with reals, like derivatives, integrals, series, and stuff, you can do with with complex numbers, more or less as easy. But everything you can do with complex numbers, you cannot do with reals. Like sqrt(-5), ln(-3), etc...

Fun, fun, really!
13. 10 Jul '08 00:07
Originally posted by FabianFnas
"imaginary numbers have no real use in mathematics?"

This was intended to be funny, wasn't it? 😀

Real use for imaginary number, or imaginary use for real numbers!

I think there are a lot of areas in mathematics where complex numbers (of the form a+bi) are used. Partly because it helps calculations. Some trig problems are very simple when you kno ...[text shortened]... complex numbers, you cannot do with reals. Like sqrt(-5), ln(-3), etc...

Fun, fun, really!
No it was a real question; expecting a sober, non-sarcastic answer.
14. 10 Jul '08 07:441 edit
No sarcasm intended. I just thought your wordings was quite humourus. Perhaps I should blame my lack of linguistic skills.
15. 10 Jul '08 07:59
Originally posted by FabianFnas
No sarcasm intended. I just thought your wordings was quite humourus. Perhaps I should blame my lack of linguistic skills.
You're not the only one to find the wordings fun - maybe it's a nordic thingy then. Personally I can't wait to get to use complex numbers in math, our next course should include that. Too bad we don't get to use complex numbers in physics until in university 🙁

On another note what use do graphs with complex numbers have in electricity calculations? I always thought they used imaginary numbers just to make calculations much simpler and thus they would have no point in representing things with them in graphs.