power of j

[hanhan]

How can you know this?

Thanks for help.

 

[misterT]

You multiply the first (in red circle) with:

-90 / -90

The above is same as 1. And when you multiply with one, you do not change the value.

You get:

1 (-90) / 1 (90 - 90)

which is:

1 (-90) / 1

which is

1 (-90)

I omitted the angle symbols. I hope that was readable.

 

[hanhan]

Thanks misterT! I understand your proof now.
Can you help me explain it considering j as an rotate operator? I want to understand what does this really mean?

 

[misterT]

There is a very long debate about wheter j is a rotate operator or not. Mathematicians call it imaginary unit or imaginary number, and they use the letter i to denote it. Electrical engineers use j and call it rotate operation.. I wont go into explaining why. I don't call it rotation operator.. because it is not.

If you are interested to read a thread from hell: https://www.electro-tech-online.com/threads/why-dont-imaginary-numbers-make-so-much-sense.121331/
But please.. do not reply to that thread.

This is much better reading:
https://mathworld.wolfram.com/i.html

 

[MrAl]

Hi,


There is a much simpler way to understand this and that is through the idea that the operator 'j' is the square root of minus 1: j=sqrt(-1).

This produces an imaginary number which we call simply "j" for short. But as we all know, when you multiply the square root of a number by itself you get the original number back. So this leads to the first identity:
j^2=-1

and that is simple enough to understand because we multiplied j*j and that meant we did sqrt(-1)*sqrt(-1) and just got -1 back again. No problem there.

Next comes multiplying that by j again, and this leads to:
j^3=(j*j)*j

and we already know that j*j equals -1 so we get:
j^3=-j

Next we multiply that by j again and we have:
j^4=(-j)*j

or written out slightly different we have:
j^4=(-1)*(j*j)

and we already know what j*j is, it is -1 again, so we get:
j^4=(-1)*(-1)

and of course that equals 1 so we just get j^4=1.

So now we create a little table:
j=sqrt(-1)=j^1=j
j^2=-1
j^3=-j
j^4=1

So what about j^5 then? Well, when we multiply j^4 times j we just get j again, so we see the table expands again:
j^1=j
j^2=-1
j^3=-j
j^4=1
j^5=j

and notice that this is the same as j^1 which is the first entry in the table. What happened here is the values have started to repeat, so we end up with the same repetition of the group:
{j,-1,-j,1}

and so we can immediately expand the table:
j^1=j
j^2=-1
j^3=-j
j^4=1
j^5=j
j^6=-1
j^7=-j
j^8=1

and this goes on forever, so that the value of any power of j is equal to the power of j that has the power minus 4, so we see:
j^(n+4)=j^(n)

and that causes a repeat of j, -1, -j, 1. So those are the only four values we can get.

Now when we look at something like 1/j, that is really j^(-1), and you can create a little table for minus j's too, or just use simple algebra. "j" can also be considered an algebraic constant so we can reduce it based on that too by multiplying the numerator and denominator both by j:
1/j=(j*1)/(j*j)

and again we know what j*1 is and what j*j is:
(j*1)/(j*j)=j/(-1)=-j

So you can use that idea instead of building a table for the negative powers of j.

Viewed as a rotation, multiplying times j means rotating counter clockwise by 90 degrees. Multiplying by j^2 means rotating counter clockwise by 180 degrees, and so on.
Note that as we rotate over and over by 90 degrees that makes the quantity real, then imaginary, then real again, then imaginary, etc.

 

[hanhan]

Thanks MrAl,
I understand your point but my mind feel confused about sqrt(-1). The mathematical j=sqrt(-1) and your proof is not confusing me. But then about the specific point, considering sqrt(-1).
What is really meaning of this? What is real world application of this?
The only way I have seen to prove that j^2 = -1 makes sense is using rotation.
Thanks.

 

[misterT]

Imaginary numbers are the solution to the problem "what is sqrt(-1)" how do I deal with that..

 

[MrAl]

Thanks MrAl,
I understand your point but my mind feel confused about sqrt(-1). The mathematical j=sqrt(-1) and your proof is not confusing me. But then about the specific point, considering sqrt(-1).
What is really meaning of this? What is real world application of this?
The only way I have seen to prove that j^2 = -1 makes sense is using rotation.
Thanks.

Hi again,


This very simple algebraic interpretation comes from the pattern that we get when we take 'i' up to a power and when we take sqrt(-1) up to a power, we get the same results, so 'i' can be interpreted as the square root of minus 1.

Algebraically, we have the simple identity:
x=(sqrt(x))^2

So when we substitute numbers for x we get:
1=(sqrt(1))^2
2=(sqrt(2))^2
3=(sqrt(3))^2
4=(sqrt(4))^2
N=(sqrt(N))^2

Note that the number 1, 2, 3, 4, N, etc., appears on both sides, and and when we do minus 1 we get the same thing back also:
-1=(sqrt(-1))^2

which really just follows the convention we established with the positive numbers. And when we take powers of sqrt(-1) we get:
(sqrt(-1))^1=sqrt(-1)
(sqrt(-1))^2=-1
(sqrt(-1))^3=-sqrt(-1)
(sqrt(-1))^4=1

and the pattern of results repeats. Comparing this to powers of 'i', we see the same pattern:
i^1=i
i^2=-1
i^3=-i
i^4=1

and note that if we substitute sqrt(-1) in for 'i' in the above four, we get the very same results when we apply the rules of algebra. So it starts to make sense that we can interpret 'i' as the square root of minus 1 in algebraic settings.

 

[hanhan]

Hi MrAl,
Your interpretation makes sense if ignores what does sqrt(-1) really mean. All your proofs are right but my mind tend to reject/ can't accept sqrt(-1). To me sqrt(-1) don't exist.
How about the understand of imaginahttps://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/

If we consider j as a rotate operator then j^2 = -1 (or j= sqrt(-1) ) makes sense.
In your explanation, you consider that sqrt(-1) existed and consider as at above level not a low level like what does sqrt(-1) really mean .
And sorry for my bad English.