What is the purpose of imaginary numbers and how do they work?

[NanakiXIII]

Sorry if this isn't the right forum, I didn't know so I just went to general.

Could someone explain how this i (imaginary numbers) thing works? I know i is supposed to be a number which is the sqrt of a negative number, which isn't supposed to exist, but what's its use? And yeah...really any information would be appreciated.

 

[gravenewworld]

i=sqrt(-1)


why can't it exist? complex numbers have applications everywhere, especially in engineering. Trigonometric identities are proved using complex numbers, not with trig.

 

[recon]

NanakiXIII maybe you should have googled it. They've got tonnes of information about it - mostly elementary. I assume that is what you're looking for.

 

[matt grime]

It's purely a formalism, as is the rest of mathematics. Complex numbers are numbers written as:

x+iy

where x,y are real numbers (that is, possibly those numbers you think of as being decimal expansions, they aren't but you're used to working with them as if they were, and it's no different with complex numbers: remember the rules and manipulate accordingly)

x is called the real component, y the imaginary component
the rules are:

(x+iy)+(u+iv) = (x+u) + i(y+v)

(x+iy)(u+iv)= (xu-yv)+i(xv+yu)

that is i is a symbol with the property that when ever you see i^2, you can replace it with -1, and just do your algebra as if it were what you're used to.

its utility is that it allows you to solve lots of physical problems. but mathematically it is just a formal object that satisfies some rules, as with everything else.

another example of this formalism you use without realizing is:

what is 1/2?
correct answer, it is the object such that, when multiplied by 2 gives 1, it is *not* 1 divided by 2. you use this idea all the time, just no one tells you you are, so if you can manipulate fractions then you can manipulate complex numbers.

all these things are just inventions (they really do not exist at all in any physical sense) that are of use for us.

we have the whole numbers, 1,2,3... good for counting sheep, and half a sheep is no good so we don't need halves and fractions. but we then might want to count things that can be divided up so we invent fractions to describe them, then we realize that there are things, like the length of th diagonal of a square of side 1 unit that aren't fractions of that unit so we invent real numbers, (I missed out negative ones, sorry), then we realize that we can't square root all things, so we invent square roots of negative numbers, and lo and behold these things describe the current flow in an electric circuit, or the oscillation of a spring. ok, that might seem strange, what does a current of 3i amps feel like, but we aren't meauring the current in imaginary numbers, we use it to describe the way currents flow, and we read off the imaginary components (which are real numbers don't forget) when we want to find the numerical

if you don't like the i thing then there's another way:


just take all pairs of real numbers (x,y) and define and arithemtic on them:

(x,y)+(u,v) = (x+y,u+v)

(x,y)(u,v)=(xu-yv,yu+xv)

this contains a copy of the real numbers in it: (x,0)

and has the property that (0,1)(0,1)=(-1,0)

 

[NanakiXIII]

I meant it doesn't exist as in sqrt(-1) = error on my calculator. And I've no idea what the rest you said means. I'm not a math wiz, just a curious 14-year-old.

 

[NanakiXIII]

Sorry, those replies weren't there when I started typing mine.

I did google and read all about it on Wikipedia but I still don't really understand.

Where did that u come from? And this might be a stupid question, but what does the comma mean in those equations?

 

[Integral]

Have you ever played Battleship? The game where you call the coordinates of a grid like A-1 or B-10 to look for your opponents ships. The notation (x,y) is simply that A-1 would be (A,1). So a complex number is represented as a pair of numbers, where the first number is called the Real part with 1 represented as (1,0) or 1 + 0i. The second number is the imaginary part where i = sqrt(-1)=(0,1) or 0 + i. This is the starting part of a very complex (pun intended!) and powerful branch of Mathematics. Though we call the number imaginary it has some very real uses.

Edit: A sufficently sophisocated scientific calculator should return (0,1) for sqrt(-1)... you need a better calculator!

 

[NanakiXIII]

Thanks. I don't understand complex numbers either, but I'll just wait 'til I get that in school...I don't understand any of the pages I've read. sqrt(-1)=(1,0)? Then what exactly are the 1 and 0? And what is the relationship between such a pair of numbers? They're not co-ordinates, are they?

But so basically, i just equals sqrt(-1), and you need that for certian things?

I've a TI-83 Plus calculator, which is what school told me to get, so. How do you use i on a calculator? Because whatever I type, the i just remains the letter i. Except for some i^x equations.

 

[matt grime]

The u didn't "come" from anywhere.

Let's do a quick potted history so you know what the terms are I'm using.

The natural numbers are the ones you use for counting:
0,1,2,3...

the integers are what you get when you add in negative whole numbers:

...-3,-2,-1,0,1,2,3...

the rationals are the ones of the form:

a/b

for a and b integers.

the greeks used to think that with these you could describe all the things that needed to be measured in some linear sense: distance, time, area, and so on.

but then they proved that in a right angled triangle that:

h^2=a^2+b^2

h the length of the hypoteneuse, a, b the lengths of the other sides.

and some one showed that if you've two smaller sides of length 1 unit that the hypoteneuse is sqrt(2) and that isn't writable as a rational.

so if you like, in the number line, ie on a ruler, there are bits between all the rationals. adding these in, numbers like pi, and the square root of 2, and so on we get the real numbers.

so each time we're coming up against some restriction and we're finding a way to extend our number system 'til we have what mathematicians call the real numbers.

in the real numbers x^2 is always positive, so the inverse, the square root, doesn't give an answer inside the reals for the square root of -1, or any other negative number. but this shouldn't worry us too much, after all, there were no solutions to x+1=0 in the naturals, so we added the negatives, but then there were no solutions to 2x=1 in the integers so we added the rationals, and then there were no roots to x^2=2 in the rationals so we added in again to get the reals, thus we can add to the reals so we can solve x^2=-1.

now at each stage in the extensions above, we were creating the number line: the real numbers come with an ordering, a < or > and we could *represent* them as all being on a line. obviously we can't add IN to this line.
so we *define* i to be a symbol such that squaring it gives -1, and magically, every root of every polynomial, can be expressed as some real number plus i times some imaginary number.

that is in some sense all the numbers of the form

x+iy

for x and y real are all you need to find write down the zeroes of any polynomial.

the difference is you can't picture these new numbers on the number line.

so we often draw them as a plane.

the (x,y) means the coordinate of some point in the plane.

if you're trying to imagine what an imaginary number looks like, then I've got a question for you: what does a real number look like? it doesn't look like anything; these things don't actually have a physical existence. all that we have is a set of things and rules for adding multipliying and dividing and so on, that we can use to model stuff in the real world.

 

[NanakiXIII]

I understand that it's just an extension, but it's harder to comprehend. To me, a real number looks like a point on a line. To me the system of numbers is 1-dimensional.

What is a polynomial? And what exactly do you mean by a plane?

 

[TenaliRaman]

A polynomial is an equation of the form,
f(x)=a_n*x^n+a_(n-1)*x^(n-1)+...+a_0*x^0

where a_i is a with subscript i and all such a_i's are constants
and x^n is x to the power of n

A plane ...
draw two number lines , one horizontal and one vertical
i can then place a point anywhere and pin point its position using a number from the horizontal number line and one from the vertical number line
The collection of such points becomes a plane.
**It would be great to have a figure at this point but maybe later.**

-- AI
Disclaimer : what i gave is a very non-technical definition ... a very proper definition can be found googling for the phrase 'plane mathworld' or 'plane wikipedia' or ...

 

[mathwonk]

my teacher taught it to uas like this; make up a new number system as follows:

a "complex number" is a pair of real numbers (a,b) where both a and b are real numbers, integers if you like.

then we add them componentwise, i.e. (a,b) + (c,d) = (a+c, b+d), but we multiply in a funnier way: (a,b)(c,d) = (ac-bd, ad+bc).

Then notice that the complex numbers of form (a,0) behave just loike the usual numbers, i.e. (a,0)+(c,0) = (a+c,0), and (a,0)(c,0) = (ac,0). so its the same as before but wioth an unnecesdsary 0 hanging around thatw e could ignore.

But notice also that then (1,0) behaves like the old number 1, since (1,0)(c,0) = (c,0) and also (1,0)(c,d) = (c,d). i.e. (1,0) even behaves like 1 when multiplying by new numbers too.

so (1,0) is our new "1". and when restricted to the subsystem (a,0) of old numbers it really is the old number 1.


But now we have a new number we did not have before, namely (0,1).

now square this guy and see what you get! i.e. work out (0,1)(0,1) = (??)

it should come out that this new number has a squatre that equals one of our old numbers, namely the old number -1.

so complex numbers are a way of enlarghing our number system so that one of the enw numbers thrown in happoens to have its square equal to minus the old number 1.


So there is still no square root of minus one in the oldsystem, but if your calculator also handles these new guys it can do this calculation. in fact you can probably program it to do this if it handles vectors, i.e. pairs of numbers.

actually i learned this in high school when i was 16, from a wonderful book called principles of mathematics by carl allendoerfer and cletus oakley. and then we studied it again in college in my freshman math class from a famous algebraist, john tate.

 

[NanakiXIII]

Thanks, both of you. I've an image in my head now. Please tell me if I'm thinking this right.

The complex numbers system is a 2-dimensional system, like a graph where you determine a place with x an y, and in this case x, the horizontal line, is the old number system, meaning (x, 0) is where you'd have plainly x in the old system.

Am I right this far? I think so.

And this new system obviously has different rules, as mathwonk showed.

(The following is just for me sort of)

(0,1)(0,1) = (0*0-1*1, 0*1-1*0) = (-1, 0) or plain -1 on the old system.

Explains i^2=-1, so I'm thinking I'm still on track.


Now back to what matt grime posted (which suddenly makes a lot more sense). Complex numbers are written as x+iy. I'm not sure I understand this yet. Isn't i itself a complex number?


Thanks for all the replies so far, I think I'm starting to understand.

 

[arildno]

Your image is excellent!
Further, in the vector notation i=(0,1)
A general complex number (x,y) (x and y themselves ordinary reals) can be written as:
(x,y)=(x,0)+(0,y)=x(1,0)+(0,1)y=x*1+iy=x+iy

 

[NanakiXIII]

Good. I understand then.

I see. But like a complex number (5,5) for example, you'd write that 5+5i then?

But if we're going to have a 2-dimensional system, wouldn't there be systems with 3 or more dimensions? I don't know what they'd be used for, but yeah.

 

[Gonzolo]

Good. I understand then.

I see. But like a complex number (5,5) for example, you'd write that 5+5i then?

But if we're going to have a 2-dimensional system, wouldn't there be systems with 3 or more dimensions? I don't know what they'd be used for, but yeah.

Yes, if you're talking about complex numbers (5, 5) would be 5+5i . In general x + iy <=> (x, y).

As for three-dimensionnal complex number systems, look up "quaternions". Not sure what they are used for either. I believed some have used them in advanced quantum theories. Regular complex numbers are immensely more common.

 

[NanakiXIII]

Thanks. I looked up quaternions and got to that and hypercomplex numbers, octonions, things like that. Seems complex numbers are the simple things. Oh well.

 

[matt grime]

Thanks, both of you. I've an image in my head now. Please tell me if I'm thinking this right.

The complex numbers system is a 2-dimensional system, like a graph where you determine a place with x an y, and in this case x, the horizontal line, is the old number system, meaning (x, 0) is where you'd have plainly x in the old system.

Am I right this far? I think so.

And this new system obviously has different rules, as mathwonk showed.

(The following is just for me sort of)

(0,1)(0,1) = (0*0-1*1, 0*1-1*0) = (-1, 0) or plain -1 on the old system.

Explains i^2=-1, so I'm thinking I'm still on track.


Now back to what matt grime posted (which suddenly makes a lot more sense). Complex numbers are written as x+iy. I'm not sure I understand this yet. Isn't i itself a complex number?


Thanks for all the replies so far, I think I'm starting to understand.

yep, you're getting a good idea now. sorry, thought that if you were asking about complex numbers you'd know about planes and polynomials.

i, is like a place holder: there is a complex number: 0+1i or (0,1) if you like that by the analogy with the real numbers and algebra in general we write as i, because 1 times x equals x, and that is a rule we want in our extended system.

 

[matt grime]

So, sorry, to double post but just remembered what I'm trying to say:

you';re happy with the plane description as the pairs of numbes (x,y) with the multiplication we've defined.

now, let's identify the pairs (x,y) with the symbols x+iy where i is a symbol that squares to give -1, and where we can perform addition: x+iy + u+iv =x+u +(y+v)i

and multiplication like we do for ordinary numbers:

(x+iy)*(u+iv)= xu +xvi+uyi+yvi^2

but of course i^2 equals -1 by definition so that expression is the same as

xu-yv + i(uy+xv)

which is what we'd identified with the pair (xu-yv,uy+xv)

so you see given two pairs (x,y) and (u,v) we can do the identification, multiply there, and go back to the pair expression OR we could have just multiplied the two pairs, but both of these methods give the same answer.

so you can either think in terms of symbols (u,v) or as expressions u+iy and they are essentially the same.

 

[geometer]

Good. I understand then.

I see. But like a complex number (5,5) for example, you'd write that 5+5i then?

But if we're going to have a 2-dimensional system, wouldn't there be systems with 3 or more dimensions? I don't know what they'd be used for, but yeah.

That's right. And yes, there are multi-dimensional systems, as many dimensions as you want. For example, in the real world, we have 4 dimensions; 3 of space (length, width, and height) and one of time. Events in spacetime are located with respect to some beginning point (origin) with four numbers (t, x, y, z).

 

[DarkEternal]

I like to think of it like this. You have any vector, Ax+By, where x and y are unit vectors that are perpendicular to each other. This expression contains three pieces of information: the magnitudes A and B, and the fact that x and y are defined to be perpendicular. With this information you can locate points. Let's try to generalize this to a form A + Bi. So we still have A and B, so let's define i to be a 90 degree counterclockwise rotation - this gives us the same amount of information as a vector. That is, if you think of the number A as as a line segment pointing to the right with length A, then Ai would be a line segment pointing up with length A. So what is Ai^2? That would be a line segment pointing to the left with length A (180 degree rotation), or one pointing to the right with -A. so Ai^2=-A -> i^2=-1.

 

[robphy]

Here's another representation of the complex numbers.

$$a+ib=\left( \begin{array}{cc} a & -b \\ b & a \\ \end{array} \right)$$

Addition and subtraction are obvious.
Multiplication:

$$\begin{align*} (a+ib)(c+id)&=\left( \begin{array}{cc} a & -b \\ b & a \\ \end{array} \right)\left( \begin{array}{cc} c & -d \\ d & c \\ \end{array} \right)\\ (ac+i^2 bd)+i(ad+bc) &= \left( \begin{array}{cc} ac-bd & -ad-bc \\ bc+ad & -bd+ac \\ \end{array}\right)\\ (ac-bd)+i(ad+bc) &= \left( \begin{array}{cc} ac-bd & -(ad+bc) \\ ad+bc & ac-bd \\ \end{array} \right) \end{align*}$$

 

[mathwonk]

There are no three dimensional systems but there are 4 and 8 dimensional ones, each losing a few properties of the previous ones.

quaternions were invented by hamilton, and have form a+bi+cj+dk where a,b,c,d are all real, and you multiply by noting that i^2 = j^2 = k^2 = -1, and ij = k, jk = i, ki = j, and

now here is the interesting part, ji = -k, kj = -i, ik = -j. so they are no loinger commutative.

maxwell used them as a system combining real numbers a, and three dimensional vectors bi+cj+dk, and used them in his fundamental physics book on electricity and magnetism.

nowadays people separate out the vectors form the numbers.

 

[NanakiXIII]

Thanks for your replies, everyone.

 

[NanakiXIII]

Just getting back to this post for a quick question. Could someone give me the rules of addition, multiplying, etc. with complex numbers in the form of (x,y)? I can only seem to find the rules for x+yi.

 

[Muzza]

But it's the same thing... For example, a + bi + x + yi = a + x + (b + y)i, so (a, b) + (x, y) = (a + x, b + y).

 

[NanakiXIII]

So... (a,b)-(x,y)=(a-x,v-y)
The problem is, I don't fully understand, for example, how to divide using x+yi. That's why I wanted the rules for (x,y), because I find them simpler.

 

[mathwonk]

I already gave you the rules for components in the very beginning (on p.1):

I.e Recall:

"my teacher taught it to uas like this; make up a new number system as follows:

a "complex number" is a pair of real numbers (a,b) where both a and b are real numbers, integers if you like.

then we add them componentwise, i.e. (a,b) + (c,d) = (a+c, b+d), but we multiply in a funnier way: (a,b)(c,d) = (ac-bd, ad+bc)."

What you want is dividing, which is a little tricky. That was homework exewrcise in my first calcl class: i.e. find numbers x,y such that (a,b).(x,y) = (1,0), assuming a,b, are not both zero. Or equivalently find x,y, such that (a+bi)(x+iy) = 1. You get two linear equations in two unknowns and you have to solve them. I had trouble because all the letters confused me: I forgot that a,b are constants and only x,y, are unknowns, so i solved it for a=3 and b= 5, or something, and then guessed the general answer.

But there is also a shorter way. Just assume such a number exists, i.e. it would be

x+ iy = 1/[a+bi]. Now try to write this number without any i's in the bottom, i.e. so that it looks like x+iy. Remember that i = sqrt(-1), so this is 1/[a+bsqrt(-1)]. Now remember how to "rationalize a denominator" to remove square roots.

 

[NanakiXIII]

I'm not sure what rationalizing a denominator is, but removings square roots I think one would do by squaring it.

I'm really not sure what to do with your explenation. I'm really just looking for something like the multiplying rule: (a,b)(c,d) = (ac-bd, ad+bc). (a,b)/(c,d) = ?

Isn't there such a rule?

 

[Gonzolo]

So... (a,b)-(x,y)=(a-x,v-y)
The problem is, I don't fully understand, for example, how to divide using x+yi...

$$\frac{a + ib}{c + id} = \frac{a + ib}{c + id}(\frac{c - id}{c - id}) = \frac{ac + bd + i(bc - ad)}{c^2 + d^2} = \frac{ac + bd}{c^2 + d^2} + i\frac{(bc - ad)}{c^2 + d^2}$$

 

[NanakiXIII]

Uhh thanks, I've seen that...

Someone told me this: (a,b)/(c,d) = ((ac+bd)/(c^2+d^2), (bc-ad)i/(c^2+d^2))

Any truth to it?

 

[Gonzolo]

It is exactly what I just demonstrated to you! Look carefully. (you don't need the i in the parenthesis, it is understood that it is there in that type of notation)

 

[NanakiXIII]

It is exactly what I just demonstrated to you!

...Is that a yes?

 

[TenaliRaman]

...Is that a yes?

yes its a yes but without the i...

(a,b)/(c,d)
Multiply numerator and denominator with (c,-d)
[(a,b)(c,-d)]/[(c,d)(c,-d)]
= (ac+bd, -ad+bc)/(c^2+d^2,0)

The denominator is real and can be absorbed in the numerator and therefore,
(ac+bd, -ad+bc)/(c^2+d^2,0)
=([(ac+bd)/(c^2+d^2)],[(-ad+bc) /(c^2+d^2)])

-- AI

 

[Gonzolo]

...Is that a yes?

If you want to learn how complex numbers work, if you are smart enough to find this site and ask such an intelligent question, if you understand how to multiply them, and if you understand how to do to add $$\frac{3}{5}+\frac{2}{3}$$, then I believe you are smart enough to understand what I posted. Rewrite it carefully, doing the steps I skipped with a pencil and paper.

It is the next step in your understanding of complex numbers. When you know this, you'll be ready for other things.

Note that the first thing I did was to multiply by 1. Yes, $$\frac{c-id}{c-id} = 1$$, exactly as $$\frac{13}{13} = \frac{c}{c}=1$$. When you divide complex numbers, the first thing you have to do is multiply by 1. If the denominator is c + id, you must multiply it by c - id, this gets the i out of the denominator.

The rest is regular algebra, taking into account the definition of i. Now Hut! Hut! Hut!