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Complex numbers

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I am trying to understsand complex numbers, And I can imagine it is a vector, but why needs the y coordinate (imaginary part) to be multiplied by i. Why don''t one just use a normal vector for it. Is there some paper on the net that explains how they got the idea to use that number My Site

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Guest Anonymous Poster
The idea probably came from equations like x² + 2 = 0. Then they found that complex numbers are useful for a lot of things, eg wave equations. Complex numbers are very different from 2D vectors though.

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A complex number is the combination of a standard number and an imaginary number. Thus why you have the i component. If you didnt have the imaginary part, then it would just be any normal number.

Complex numbers are often represented as pairs or vectors, just as a means of storage: so 2 + 3i is often stored as (2, 3). But you cant forget the + sign, a complex number is truely single entity.

The question you should be asking is whats so magical about imaginary numbers and how they interact with real numbers. For example, go look up the source for calculating (x + iy)^n. Most implementations use the radian coordinate system and sin/cos to calculate complex pow().

On a related note, because complex numbers are stored as pairs, you can use the x,y coordinate field to generate fractals etc.

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You cannot write it as a normal vector, because that complex number is referring to a position on the Argand diagram, not on the XY plane. The Y axis of the Argand diagram is the imaginary number line, whereas the X is the real number line.

You can write complex numbers in numerous forms. There''s x + iy. Then ther is r(cos THETA + i sin THETA) which is also equivalent to re^i(THETA). r in the latter two cases is the modulus of the complex number, which is just the length of the complex number when drawn on the Argand diagram, and THETa is the angle that is made with the real number line.

Using these, you get the equation e^i(PI) + 1 = 0, which is beautiful, apparently :-)

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The idea of looking at complex numbers like you do a 2D vector is really just notation. I guess if you''re really hung up on the i, then try thinking of it like this:

complex number = a*r + b*i

where
r=real=1
i=imaginary=sqrt(-1)

and voila it''s in the same form as 2D vector notation of
a*i + b*j = (a,b)

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hm. I am getting to understanding how to work with them, but I still don't see how they came on the idea to make a number who's square is -1, and why they use it for complex numbers. Is there a reason for this, or should I just believe it.

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[edited by - Quasar3D on February 2, 2003 6:31:21 PM]

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it was not an idea to do like this.. it was automagically.. as everything by math, its just based on the basic rules...

they though about this:

2 = sqrt(4)
a*2 = sqrt(a*a*4)

now.. as they had more and more complex equations, and realised they need to find square roots of negative numbers:

i*2 = sqrt(-4)
i*2 = sqrt(-1*4)
i*2 = sqrt(-1)*2
i = sqrt(-1)

and voilà. the whole mis(t)ery is gone

(okay, there is much more than this.. but the basic idea on why i is what it is is now given..)

"take a look around" - limp bizkit
www.google.com

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quote:
Original post by quasar3d
hm. I am getting to understanding how to work with them, but I still don''t see how they came on the idea to make a number who''s square is -1, and why they use it for complex numbers. Is there a reason for this, or should I just believe it.

My Site

[edited by - Quasar3D on February 2, 2003 6:31:21 PM]


They came about because mathematicians found that there was a massive hole in their theories when it came to trying to solve x^2 + 2 = 0.

Try to solve it without complex numbers.

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Well, complex numbers just make things easier to work with. You don''t have to keep track of the sign of numbers you''re dealing with. They always have a squareroot no matter if they are positive or negative so you have lass cases to deal with. Besides, every equation has a solution in complex numbers so you can always be sure that a solution exists.


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Guest Anonymous Poster
Every equation has a solution in complex numbers? What about z*0 = 1?

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Hmm, I think that z*0 = 1 is a false statement, z*0 = 0 and 0 is not equal to 1. So it isn''t really an equation.
Correction: saying every equation, I meant every polynomial. So every polynomial has a root, thus a solution.


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z*0 = 1 can be made more easy: z*0 = 0 => 0 = 1. simply a false statement, no variable is in there. no equation

"take a look around" - limp bizkit
www.google.com

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Yup, that''s the "Fundamental Theorem of Algebra", originally proved by Gauss (and just to show it wasn''t a fluke, he provided another 9 alternative proofs; none of which would stand up to today''s standards of rigour). The proof is a very hard theorem of complex analysis (which is also very hard).

Every polynomial with complex coefficients has a complex root.

That says that the field of complex numbers is ''algebraically closed''. Normally we have to invent new kinds of numbers to solve equations...

x + 1 = 0 -> invent negative numbers (x = -1)
2x = 1 -> invent rational numbers ( x = 1/2 )
x^2 = 2 -> invent irrational numbers ( x = sqrt(2) )
x^2 = -1 -> invent complex numbers ( x = i )

but what the theorem shows is that for polynomials, we don''t have to invent super-complex numbers like quaternions or anything to solve them. That''s what makes complex numbers so great.

"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley

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Guest Anonymous Poster
In that case x² = -1 (x real) is no equation either. Since x² >= 0 => 0 >= -1. simply a false statement, no variable is in there. no equation......

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quote:
Original post by Anonymous Poster
In that case x² = -1 (x real) is no equation either. Since x² >= 0 => 0 >= -1. simply a false statement, no variable is in there. no equation......



Of course, once you start dealing with complex numbers, relations like greater-than and less-than lose their meaning...


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Guest Anonymous Poster
I guess you didn''t see the (x real) then.

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x^2 >= 0 isn''t a true statement with x complex.
Any complex number greater than another complex number isn''t valid at all (complex numbers have no ordering).

z^2 = -1 is a valid statement with z complex though. z = +/- sqrt(-1) are the 2 solutions, i.e. z = i or z = -i.

"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley

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Guest Anonymous Poster
I guess you didn''t see the (x real) either.

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I guess you didn''t see that in my original post I didn''t say x was real at any point.


"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley

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Guest Anonymous Poster
I guess you didn''t realize that I never said anything about your original post. I was answering davepermen...

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daveperman didn''t mention >= either. He did use =>, as in "implies". 0z = 1 (in a ring) implies 0 = 1, that''s false except in a trivial ring (where there is only one element x = 0 [additive identity] = 1 [multiplicative identity] and we define x+x = x*x = x). Trivial rings are super-uninteresting though.

x^2 = -1 is a perfectly valid statement if we are considering complex numbers (or quaternions, in which case solutions are +/- i,j or k). x^2 >= 0 is only valid if we have ordering of elements.

"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley

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Guest Anonymous Poster
You just don''t get it, do you? daveperman argued that z*0 = 1 is no equation since it implies a simple false statement. I answered that in that case x² = -1 (where x is taken to be real) is no equation either, since it implies 0 >= -1. In fact, ANY equation not having a solution implies a simple false statement like the above, and consequently is not an equation at all according to daveperman. However, it is common practice to speak of equations having no solution, therefore I question his argument. If this doesn''t make sense to you, that''s perfectly fine with me. Don''t worry about it.

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you don''t get it, don''t you.. all an equation uses is =, no >, no <, nothing.. and base math forces 0*x = 0 => your thing is 0 = 1, wich is just wrong.

x^2 = -1 is not possible to simplify to something like 0 = 1. but it is possible to "simplify" it like that

x^2 = -1

-x^2 = 1

(-1)*x^2 = 1

sqrt(-1)^2*x^2 = 1
(sqrt(-1)*x)^2 = 1

and that sqrt(-1) has to be found. its not possible to find in the real numbers, so you can argue that its impossible to solve..

but when ever we had a number, describeable by math, but not to write out, people found some way to name it.

in this case, its called i


as long as there where no imaginary or complex numbers, yes, there was no solution. as long as there where no negative numbers, there was no solution for x + 1 = 0 as well.

thats why it got expanded.

anyways, you''re just an ap bitching around.. who cares

"take a look around" - limp bizkit
www.google.com

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The eager anonymous poster is right. "0*z=1" IS an equation. It's just that the theorem that every equation involving complex numbers has a solution is just false. The theorem is (as someone has already noted) that every non-constant polynomial with complex coeficients has a complex root, so "0*z=1" doesn't satisfy the hypothesis.

I also wanted to mention that having equations with no solutions is no "massive hole" in a theory. A better way to see it is as follows:
We start with natural numbers {0,1,2,3,4,...} (0 is optional). We can add and multiply, but we have a "massive hole": There is no number such that 5+n=2. We want to find a number that can be the answer to "2-5", but we don't have one. We fix this by creating integers, which include naturals but also have {-1,-2,...}. Now we can add, substract and multiply.
Now we can't divide 1/2. We create rationals, which include integers and some other numbers. Now we can add, substract, multiply and divide.
But there are sequences, like {3, 3.1, 3.14, 3.141, 3.1415, 3.14159, ...} that have the Cauchy property but have no limit. We create real numbers, that include the limits of all Cauchy sequences.
Now we can't find roots for some non-constant polynomials. We create complex numbers to fix this.

Note that we could have skipped the part with the limits and try to find numbers which include the rationals and have solutions for every polynomial. Complex numbers are not the only field with that property. We could also complete the rationals so that we have limits to all Cauchy sequences under a different definition of absolute value, which gives birth to p-adic numbers.

This is basically how I think of different sets of numbers. Complex numbers turn out to be very useful. As 2D-vectors, they have the advantage that the multiplication makes it very easy to rotate around the origin. Just multiply by (cos(alpha) + i sin(alpha)).



[edited by - alvaro on February 3, 2003 2:04:54 PM]

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