imaginary number analysis

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Vikram50517

Joined Jan 4, 2020
81
Hello all! I have a doubt regarding i's nature.wkt i=√-1.but it also performs the "operation" of rotating a number 90° ,just like" - "sign which rotates the number by 180°.so is it a number or an operator
 

WBahn

Joined Mar 31, 2012
32,794
Multiplying a number by 2 makes the number twice as big, so is 2 a number or an operator?

The rotation occurs because you multiply the number by i.
 

MrChips

Joined Oct 2, 2009
34,766
I consider j to be an operator that rotates the vector by π/2.

Note that we use j instead of i.

If you apply the j operator twice, it rotates the vector by π.
 

MrAl

Joined Jun 17, 2014
13,702
Yes it can be either.
A number like 1+2*j can be regarded as a number, but if you multiply it by 'j' you get a rotation.

Look at the position of 1+2j in the complex plane. It occurs 1 to the right of zero on the real axis and 2 up on the imaginary axis. Draw a line from the point (0.0) to the point (1,2) and note the angle is about 63.435 degrees. Now multiply it by 'j', the result is -2+j. Now look for the position of -2+j in the same plane. It occurs at the position 2 to the left of zero on the real axis and 1 up on the imaginary axis. The angle is now 153.435 degrees, which is 63.435 plus 90 degrees. So multiplication by 'j' results in a rotation by 90 degrees.
Multiply again by 'j' and it will be rotated by 180, again and it will be rotated by 270 degrees, and once more and it will be back to the original 63.435 degrees. This is how j^4 becomes equal to 1.

But even simpler, if you look at the 'number' 1+2*j you see that the '1' is on the real axis and the '2' is multiplied by 'j' so it is 'rotated' 90 degrees from the real axis, and that is of course the imaginary part and the imaginary axis is considered 90 degrees from the real axis.
 

MrChips

Joined Oct 2, 2009
34,766
Here is another view.

1 + 1 = 2

Assume that each 1 represents a sine wave at the same frequency, same amplitude.
Then 1 + 1 would be a sine wave at twice the amplitude of each sine wave.

Now let us look at 1 + j.
1 is a sine wave
j is a sine wave phase shifted by π/2, i.e. a cosine wave.
The result of 1 + j is a sine wave phase shifted by π/4 with an amplitude of √2.
 

bogosort

Joined Sep 24, 2011
696
Hello all! I have a doubt regarding i's nature.wkt i=√-1.but it also performs the "operation" of rotating a number 90° ,just like" - "sign which rotates the number by 180°.so is it a number or an operator
Mathematical theorems are meaningful only within some context, i.e., some semantic model that provides meaning to the various objects of the theory. Unfortunately, the context is rarely made explicit.

For example, you may read on one site that \( \mathbb{C} \) is isomorphic to \( \mathbb{R}^2 \), and then on another site you may read that they are not isomorphic. In fact, both statements are true within a given context. If we are talking about vector spaces, we have a model that includes a number field, vector addition, scalar multiplication, and not much else. In such a context, \( \mathbb{C} \) is indeed isomorphic to \( \mathbb{R}^2 \), through the map:
\[ (a, b) \in \mathbb{R}^2 \qquad \Leftrightarrow \qquad a + ib \in \mathbb{C} \]
However, if we are talking about algebraic rings (a richer structure than vector spaces), then \( \mathbb{C} \) is not isomorphic to \( \mathbb{R}^2 \) as multiplication is not compatible between the two.

Likewise, what someone means by the symbol \( i \) depends entirely on the context in which it is being used. As an element in a set (a number), it is defined by a simple property:
\[ i^2 = -1 \]
Though this element is mostly famously associated with \( \mathbb{C} \) as the algebraic closure of \( \mathbb{R} \), it pops up in many other contexts. For example, \( i^2 = -1 \) for every \( i \) in any Boolean structure (algebras, rings, lattices). The quaternions \( \mathbb{H} \) have three such elements. I mention these to show that \( i \) as a number is not unique to \( \mathbb{C} \).

Now, when we talk about rotations, we are necessarily introducing richer structure, as now we need the concept of an angle, which means we need all the structure of an inner-product space, which brings with it all the structure of a normed space, which brings with it all the structure of a metric space, which brings with it all the structure of a vector space, which brings with it all the structure of a number field. Remember, the original context defining \( i \) above was a simple number field. In this far-richer context of a geometry, the object \( i \) can take on any role in the hierarchy of structures. At the bottom level, we can treat it as a simple element member (a number); at the top level, we can treat it as an operator. All representations are valid within their context; confusion only arises when we forget (or aren't told) that there is an implict hierarchy of mathematical structure in the given context.

In general, mathematics (and logic) becomes a lot clearer once you train yourself to think this way about all such objects.
 

ci139

Joined Jul 11, 2016
1,982
purely for a theoretical interest i - long ago - defined a complex functions that preserve the arbitrary argument including |φ|>2π . . . and then i noticed that the particular "use" for such is highly limited . . . if i remember right , also due most "std." functions do not distinct beyond the Argument , Principal_value (complex analysis) - Wikipedia
 

MrAl

Joined Jun 17, 2014
13,702
purely for a theoretical interest i - long ago - defined a complex functions that preserve the arbitrary argument including |φ|>2π . . . and then i noticed that the particular "use" for such is highly limited . . . if i remember right , also due most "std." functions do not distinct beyond the Argument , Principal_value (complex analysis) - Wikipedia
Long long time ago i used to belong to a chess club, and we used to play until 4 in the morning sometimes and at that time we were all getting so tired we could hardly play right. An old master told me and i quote, "You gotta know when to go home".

I am not completely sure of what you are saying but i think you are talking about angles that extend beyond 2*pi because the computation returns values that are only within the range 0 to 2pi.
So you really gotta know when to take another step and figure out if the angle is really beyond 2pi. Many times it behaves as if it was between 0 and 2pi anyway, but sometimes you do have to recognize when it is really going beyond 2pi (or less than -2pi) and make any adjustments if the application demands this. One example i can think of is a sort of delay line where the phase at the end of the line could theoretically run way beyond 2pi otherwise we would be too limited. But if you are faced with a circuit like that then you have to be able to recognize that the angle could be Th, Th+2pi, Th+4pi, etc., and assess if it makes any difference to even consider.

There is also a failure at the point (0,0) but looking at the application itself usually reveals what we have to do to get the right angle.
 
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ci139

Joined Jul 11, 2016
1,982
not completely sure
no , you input an arbitrary angle argument that gets translated to another complex value then that to another ... until the original is restored with it's original argument
. . . trying to find the thing . . . something
Def.: \(\displaystyle{y=inp\left({x}\right)=e^{-i·x^i}=e^{\frac{Sin\ Ln\left|{x}\right|}{e^{arg\left({x}\right)}}}·e^{-i·\frac{Cos\ Ln\left|{x}\right|}{e^{arg\left({x}\right)}}}\\ x=lng\left({y}\right)={\left({i\ ln\ y}\right)}^{-i}=e^{-arctan\frac{Ln\left|{y}\right|}{arg\left({y}\right)}}·e^{-i·\frac12·ln\left({Ln^2\left|{y}\right|+arg^2\left({y}\right)}\right)}}\)
then ( Update ! ▲ this thing preserves the "extended argument" ▲ )
\(\displaystyle{x=ln\left({inp\left({ln\left({inp\left({ln\left({inp\left({ln\left({inp\left({x}\right)}\right)}\right)}\right)}\right)}\right)}\right)}\right)}\)
and/or ( Update ! ▲▼ those things do not preserve the "extended argument" )
\(\displaystyle{\text{PS!}\ x≠lng\left({exp\left({lng\left({exp\left({lng\left({exp\left({lng\left({exp\left({x}\right)}\right)}\right)}\right)}\right)}\right)}\right)}\right)}\)
\(\displaystyle{\text{instead : }\ x=exp\left({lng\left({exp\left({lng\left({exp\left({lng\left({exp\left({lng\left({x}\right)}\right)}\right)}\right)}\right)}\right)}\right)}\right)}\)

↓ BASIC SIMPLE ↓
inp~exp.png inp~exp - 2.png
 

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