what are imaginary numbers?why we use iota and omega?

Question

iota and iota squre,iota cube,omega ,omega squre,omega cube?

Answer 1

imaginary numbers :

eleventeen
tenteen
threeteen
fiveteen
oneteen
twoteen

Answer 2

Imaginary numbers are, broadly speaking, numbers that are not real--they do not have a location on a real number line. However, they can be defined by a PROPERTY that they have.

I assume that by "iota" you mean i, which is commonly used to represent a number that, if you square it, you get -1. That is the DEFINITION of i: i^2 = -1. (I believe we use "i" because it stands for "imaginary", but don't quote me.)

Because of this property, powers of i follow an orderly pattern:

i^1 = i
i^2 = -1
i^3 = (i^2)(i) = (-1)(i) = -i
i^4 = (i^2)(i^2) = (-1)(-1) = 1
i^5 = (i^4)(i) = (1)(i) = i
i^6 = (i^5)(i) = (i)(i) = -1

... and so forth. Since i^4 = 1, we could say that i is a FOURTH ROOT of 1.

As for omega... I belive omega (which kind of looks like a "w") is used to represent a CUBE root of 1. In other words, w^3 = 1.

Obviously 1 itself is a cube root of 1, but there are two others as well, w and w^2. But w can be re-written using i as follows:

w = -1/2 + (sqrt(3)/2) i

Using that definition, you can verify for yourself that if you cube w, you get 1.

So i is a fourth root of 1, and w is a cube (third) root of 1.

(As for why omega us used, rather than some other symbol, your guess is as good as mine.)

Answer 3

By the very definition of IMAGINARY numbers...They are IMAGINARY!!!!....why do we need imaginary numbers??? well because they can have real consequences!!!....for example iota squared is a real number ie -1.......thats why we need "imaginary numbers".......
iota and and omega are chosen completely for convience sake...iota[i]...stands for imaginary...omega of course means something that is not well known........

Answer 4

Mathematicians wanted to solve equations like x^2+1=0. Since there is no real number that can solve this, mathematicians said that one of the solutions is i. This leads of course to the other solution being -i. Although you may use iota (ι), I believe that it is more standard to use i as the square root of -1. This can also be known as a primitive 4th root of unity.

Omega (ω) and zeta (ζ) are both used often for the same thing. That is a primitive n-th root of unity. This can be given as ω=ζ=exp(2πi/n).

What is a primitive n-th root of unity in theory?
Well, let's first define an n-th root of unity.

An n-th root of unity is a number α (usually complex and not real, but can be real) such that α^n=1. Since 1^n=1, and (-1)^n=1 when n is even, 1 is always an n-th root of unity, and -1 is an n-th root of unity when n is even. Also ω^n=(exp(2πi/n))^n = exp(2πi/n•n) = exp(2πi) = 1, so ω is also an n-th root of unity (this is good, since we would want a primitive n-th root of unity to be an n-th root of unity).

So what is a primitive n-th root of unity? A primitive n-th root of unity ω is an n-th root of unity and moreover all n-th roots of unity can be given by powers of ω.

For n=3, we have ω=exp(2πi/3) = cos(2π/3)+i•sin(2π/3) = -1/2+i•âˆš3
and we have ω^2=exp(2πi/3)^2 = exp(4πi/3) = cos(π/3)+i•sin(π/3) = -1/2-i•âˆš3.
You will also notice that ω^3=exp(2πi/3)^3 = exp(6πi/3) = exp(2πi) = 1, and (ω^2)^3 = (^3)^2 = 1^2 = 1. Thus ω, ω^2, and ω^3=1 are the 3rd roots of unity, and as stated they are all powers of ω.

Also notice that for n=4, we have ω=exp(2πi/4) = exp(πi/2) = cos(π/2)+i•sin(π/2) =i. Thus i is a primitive 4th root of unity. This can be checked by i^2=-1, so i^3=-1•i=-i and i^4=(-1)^2=1. These four numbers are all solutions of x^4=1, so that shows that all 4th roots of unity can be represented as powers of i. Thus i IS a primitive 4th root of unity.

Answer 5

sqrt(-1) is not a real number. So we call it imaginary. We use the symbol 'i' to denote it. we can now easily write the sqrt.s of -ve numbers in terms of i
Using this we can write sqrt(-4) as 2i or -2i

Answer 6

Those are not imaginary numbers. You need to talk to your math and history professors and have them answer those questions.