(a+bi^{2}.a-bi^{2})i=ai+bi^{3}.ai-bi^{3}

ai(ai-bi^{3})-bi^{3}(ai+bi^{3})

a^{2}i^{2}-a^{2}bi^{3}-a^{2}bi^{3}-b^{2}i^{3}

a^{2}(-1)-a^{2}bi(-i)-a^{2}b(-i)-b^{2}(-i)

a^{2}+a^{2}b+a^2b+b^{2}

From here is uncertainty… do I solve this way…

(a^{2}+ab)=b^2 and (a^{2}b-b^{2})=a

So, a+b^{2}=ab^{2}
…..
Or should it be done this way:
a^{8}+b^{4}
Or neither way… please tell me where I am going wrong

Your notation is confusing - parentheses would help. Does "." mean multiply? Finally what are you looking for?

Your notation is confusing - parentheses would help. Does "." mean multiply? Finally what are you looking for?

My notation may very well be confusing. Please ask me where to make it more lucid. And much short less, then it should be, what do you think i may be possibly looking for

nO''.'' does not mean multiply... where did you see this?

I don't mean to be so... agressive towards you, however, rthis is basic algebra. Basic even in the sense i do not even know if it is true, so my knowlegde is heavily limited.

I will ask thes question again: Is the math correct or not?

Only until someone has proposed it be correct, will i divulge its final goal...

... if there is no final goal, then is the math, nevertheless, still holding true, in algebraic function and discipline?

No one will answer??? Interesting

Seriously. WTF is your question? I don't get it.

And what's with the "."?

, Sorry, and mind my language, BUT WHAT THE FUCK DO YOU MEAN ''.'' ?????

(a+bi^{2}.a-bi^{2})i=ai+bi^{3}.ai-bi^{3}

I think this line is where people are having the most confusion, right at the beginning. On the left hand side, are you trying to say
(a+bi^2)(a-bi^2) or (a+bi)^2(a-bi)^2 or what?

Also, is each new line supposed to be equal to the expression in the previous line?

Reiku: It's your "dot". If you refuse to explain what it means, how can anyone offer you the help you asked for? Why is it a secret?

Reiku, also can you tell me, what you're looking in this task? What is the task about?

Oh i see.... sorry. Yes, the dot is multiplication. I cannot allow brackets in this sense.

Oh i see.... sorry. Yes, the dot is multiplication. I cannot allow brackets in this sense.

Well, then your first line is wrong.

Why... whats the difference... don't they produce the smae answers:

(a+bi)(a-bi)=a^{2}-b^{2}i^{2}

as with

a+bi.a-bi=a^{2}-b^{2}i^{2}

Cheers

Even though the normal representation of the equations of conjugates are usually oresented with the former, i am not following the rigor of a textbook; with that said, it still can be seen this other way surely? In theory no?

Or must it be done like this

((a+bi)(a-bi))i

Is that better?
The goals that i can reduce the equation is that i can make a final positive answer, and in this case, it would be most appropriate to have a, rather than ab^2 or something like that.

Please don't ask why... lol

(a+bi^{2}.a-bi^{2})i=ai+bi^{3}.ai-bi^{3}

ai(ai-bi^{3})-bi^{3}(ai+bi^{3})


Ok, I think I see what your problem is.
Assuming your first line is meant to read: (a+bi^2)(a-bi^2)i=(ai+bi^3)(ai-bi^3), this would be false. Your logic is equivalent to saying (x*y)i=(x*i)(y*i) when the proper result is (x*y)i=(x*i)y=x(y*i). Thus your first line should read:
(a+bi^2)(a-bi^2)i=(ai+bi^3)(a-bi^2) or equivalently,
(a+bi^2)(a-bi^2)i=(a+bi^2)(ai-bi^3).

You also make a mistake going from the first line to the second. Your statement in this step is:
(ai+bi^3)(ai-bi^3)=ai(ai-bi^3)-bi^3(ai+bi^3)
when it should actually say
(ai+bi^3)(ai-bi^3)=ai(ai+bi^3)-bi^3(ai+bi^3)
or equivalently, (ai+bi^3)(ai-bi^3)=ai(ai-bi^3)+bi^3(ai-bi^3).

I would honestly recommend that wherever you see i^2, you should replace it with -1 as soon as possible. Assuming you do your calculations correctly, it won't make a difference to your final answer and it will simply the math substantially.

Right, so i have been working with the variable i, before resolving inside the brackets? Is this right?

Thank you very much :) Its been doing my simple mind horrors

I will sit down and calculate it, but if you can see this all laid out in your head (seemingly you are quite a whizz with numbers,) can i achieve by reduction a single value of a... and i would be willing to settle with -a as well...

Judging by the final calculation, in your post, can i still achieve the goal in other words, when i reach =a...

Or must it be done like this

((a+bi)(a-bi))i

Is that better?
The goals that i can reduce the equation is that i can make a final positive answer, and in this case, it would be most appropriate to have a, rather than ab^2 or something like that.

Please don't ask why... lol

This will give you i(a^2 + b^2).

That's all you can get.

Wait a wee min friend... I see you are rearranging what i need....

I cannot allow a single calculation to create (a+bi^{3})(a-bi^{2}), for i need i to raise the the i variables in both conjugates... this is why i introfuced the dot, and put the i outside a single bracket... if this is not allowed mathematically, what about what i said before...

((a+bi)(a-bi))i

????

Judging by the final calculation, in your post, can i still achieve the goal in other words, when i reach =a...

Well if you're trying to show that a+b^2=ab^2, then you won't get this result when you do the calculations properly. Your calculation makes no assumptions about what the actual values of a and b are, so it should be true for any values you choose. Yet clearly there are many values you could choose for a and b so that a+b^2 \neq ab^2.

As for your question about "resolving the i into the brackets", I'm not quite sure what you mean but I'll take a guess. Real and complex numbers have the exact same algebraic properties. So when doing algebra, the calculations are the same when you're dividing, multiplying, adding and subtracting. In general, (ac)(bc)=abc^2 which is not the same as abc.

Oops... just got that ben, cheers

so why not (a+bi.a-bi)i, the way i presented it before?

the variable i, and stress MUST be multipled by all the variables in BOTH the brackets containing the conguates... HELP MEEEEEE!!!!!

Wait a wee min friend... I see you are rearranging what i need....

I cannot allow a single calculation to create (a+bi^{3})(a-bi^{2}), for i need i to raise the the i variables in both conjugates... this is why i introfuced the dot, and put the i outside a single bracket... if this is not allowed mathematically, what about what i said before...

((a+bi)(a-bi))i

????

To get an expression like (ai+bi^3)(ai-bi^3), you could take (a+bi^2)(a-bi^2)i^2 which is also the same as (-1)(a+bi^2)(a-bi^2).

AHHHHHHH YOU'RE A GOD-SEND.... NOW WHY COULDN'T I SEE THAT????? Lol

I must be retarded lol

AHHHHHHH YOUR A GOD-SEND.... NOW WHY COULDN'T I SEE THAT????? Lol

I must be retarded lol

Things don't usually come easy the first time around, but the harder you think about stuff when you're learning it, the better you'll understand it for next time ;)

Here's the sad part. i'm not working from any referenced question. This is my own investigation. It just shows even i don't understand what i am trying to do... quite shameful actually

When I started out working with things like complex numbers, I was deriving all kinds of crazy results that I knew couldn't be true. There are a lot of subtleties and you have to be careful with your calculations, but it's great practice even when you're making mistakes. My biggest insights usually come when I figure out where my mistakes are, then everything else makes that much more sense. No pain, no gain.

* Why should one ''immedately'' replace i^2 with -1... is it an invariant rule...? Because as far as i understood algebra, you can deal with the variables any way you like, so long as you deduce them correctly in the end????

A new question...

With the final value ab^{2}, plug in -a, so that ab^2(-a)... now

In Einstinian notation, you can can trace AB=BA... using the subvariables of i and j... can i and j, seen though used as a reference of trace, be allowed to have actual values? real or complex?

And if not, mathematically, can we not attempt to add some value to them? It is under my impression, math makes the most out of what it has, and even breaks its own rules from time to time...

I work out, that if:

ab^2(-a), if i say a_{i} and b_{j}

So that (ii)j=i(ji)=-j

then

ab^{2}(-a)

=(a_{i}b_{j})-a_{i}

and

(-a_{i}.a_{i})b_j

so that (ii)j=i(ij)= - j...

?????

I say, ''break its own rules,'' and as an example, just consider quarternions...

* Why should one ''immedately'' replace i^2 with -1... is it an invariant rule...? Because as far as i understood algebra, you can deal with the variables any way you like, so long as you deduce them correctly in the end????

One of the nicest properties of algebraic fields (a field is a special kind of set) like the sets of real and complex numbers is that in any calculation you do, the order in which you add, subtract, multiply and divide doesn't matter, you still get the same result at the end. So you can leave i^2 as is and set it to -1 at the very end of the calculation, or you can set i^2=-1 right at the very beginning, and you'll get the same answer. I prefer the second option because it simplifies your calculations and makes them less messy, reducing the chances of making mistakes in your algebra.

A new question...

With the final value ab^{2}, plug in -a, so that ab^2(-a)... now

In Einstinian notation, you can can trace AB=BA... using the subvariables of i and j... can i and j, seen though used as a reference of trace, be allowed to have actual values? real or complex?


I'm having a tough time understanding your question. Are A and B supposed to be matrices? Are you talking about the trace of the matrix product AB as in (using the Einstein summation convention) A_{ij}B_{ji}?

And a third question... a counterpart of the original set of equation,

a^{2} + b^{2} + c^{2} + tdi^{2} - Tdi^{2}

I solved the real part of the equation by allowing i^{2} = i *k^{2} so that the result is

a^{2}+ b^{2} + c^{2} - i^{2}*k^{2}^{2} = 0
a^{2} + b^{2} +c^{2} + k2^{2} =0

This is right, is it not?

yes friend. Matrix notation... however, i was speculating whether we can remove the idea of the matrix, and give i and j actual values... .... ....

Well if you mean A_{ij} to be the entry of matrix A located in row i, column j, then i and j can only be positive integers. After all, it doesn't make any sense to say "this entry is located in row 0.5, column sqrt{3}.

As for constructing the analogue of a matrix but allowing i and j to take on any values you want, you could just define a 2-variable function f(i,j) mapping the pair (i,j) onto whatever you like, with both i and j being allowed to take on complex values if that's what you want.

I said, forgive me again, remove the notion of matrix algebra, and make them as sources that have a real value... imaginary, or complex...

But i am guessing, from the last post, i certainly can?

But i am guessing, from the last post, i certainly can?

You certainly can, but a lot of nice properties and results of matrices won't carry over to this analogue. Depends what you're trying to do.

Thanks
What about the xtra math?

The result, i would rather mail you. If i mention my intentions here, this thread will be moved to psuedoscience, without hesitation.

Well now that I have 20 posts, you can send me a private message if you like. As for the extra math, I'm afraid I can't follow at all what you're doing or why, or what all your variables mean. I need you to carefully explain what it is you're trying to do at each step, and I need you to be very clear about what variables you're using and what you're doing with them. You seem to be treating a and b as vectors, matrices and scalars all at the same time, which is adding to the confusion.

Right, correct me if i am wrong..

.. as you do lol

.. But i am finding something a bit odd. You say,

to get an expression like

(a+bi^{3})(a-bi^{3})

You can take


(a+bi^{2})(a-bi^{2})i^{2}

butttttt, i was always informed that:

i^{5}=i^{2}.i^{2}.i

Because

i^{2+2+1}

So,

(a+bi^{2})(a-bi^{2})i^{2}

wouldn't that lead to exponentially reducing it as

(ai^{2}+bi^{4}), working with just one side for now,

because i^{2}.bi^{2}=bi^{4}...

Just a bit confused about this right now.

Right, correct me if i am wrong..

.. as you do lol

.. But i am finding something a bit odd. You say,

to get an expression like

(a+bi^{3})(a-bi^{3})

You can take


(a+bi^{2})(a-bi^{2})i^{2}

butttttt, i was always informed that:

i^{5}=i^{2}.i^{2}.i

Because

i^{2+2+1}

So,

(a+bi^{2})(a-bi^{2})i^{2}

wouldn't that lead to exponentially reducing it as

(ai^{2}+bi^{4}), working with just one side for now,

because i^{2}.bi^{2}=bi^{4}...

Just a bit confused about this right now.

Or is it that you are replacing i^{2} with -1, so that...

(a+bi^{2})(a-bi^{2})(-1)

which would give you

(ai+bi^{3})(ai-bi^{3})

?

And even if the latter is the way to get the result, why can't it be solved in the way i shown previously? What makes one more correct than the other?

I know it was directed at CpT, but if anyone has any insights, it would be appreciated. Both ways seem logical... but if one of them have an error, i would like to know what it is.

Right, correct me if i am wrong..

.. as you do lol

.. But i am finding something a bit odd. You say,

to get an expression like

(a+bi^{3})(a-bi^{3})

Careful, I typed (ai+bi^3)(ai-bi^3) which is different from what you have above.




You can take

(a+bi^{2})(a-bi^{2})i^{2}

butttttt, i was always informed that:

i^{5}=i^{2}.i^{2}.i

Because

i^{2+2+1}

So,

(a+bi^{2})(a-bi^{2})i^{2}

wouldn't that lead to exponentially reducing it as

(ai^{2}+bi^{4}), working with just one side for now,

because i^{2}.bi^{2}=bi^{4}...

Just a bit confused about this right now.

You can distribute the i's however you want, as long as each i is distributed into only one set of brackets, no more. So the expression
(a+bi^2)(a-bi^2)i^2 could be equivalently written as
(ai+bi^3)(ai-bi^3)=(ai^2+bi^4)(a-bi^2)=(a+bi^2)(ai^2-bi^4), these expressions are all the same and you can convert one form of the expression into any other form by algebraic manipulation. You can plug in i^2=-1 into any of these expressions at any point in your calculations, and your final answer is the same.

For instance, you can set
i^3=(i^2)i=(-1)i=-i,\ i^4=(i^2)(i^2)=(-1)(-1)=1,\ldots and so on. You might want to get a good textbook on functions of a complex variable if you're new to these things. Just curious, what level is your math background? I could help you out a lot better if I knew more about your abilities.

I thought so ... thank you very much :)

Remember i was saying i was requiring, when you reduced the conjugates (a+bi)(a-bi), i required a single answer, preferably squared a^{2}, i sat down, and found a simpler way of deducing it. Tell me if these assumptions are correct:

(a+bi)(a-bi)=a^{2}-b^{2}+2iab
=2iab=(-1)ab=ab=a^{2}

??

Remember i was saying i was requiring, when you reduced the conjugates (a+bi)(a-bi), i required a single answer, preferably squared a^{2}, i sat down, and found a simpler way of deducing it. Tell me if these assumptions are correct:

(a+bi)(a-bi)=a^{2}-b^{2}+2iab
=2iab=(-1)ab=ab=a^{2}

??

No this is wrong.

(a+bi)(a-bi)=a^{2}+b^{2}+iab-iab=a^2+b^2

Well, Dr Wolf has

(a+ib)(a-ib)=a^{2}-b^{2}+2iab

I could just ask him, but i want to see if anyone knows first...

Oh Ben. Don't even answer me back. I am misreading what is said. The answer is of course a^2+b^2. Cheers.

Is this reduced correctly?

(a+bi^{2})(a-bi^{2})i^{2}=(a+bi^{3})(a-bi^{3})
a(a+bi^{3})-bi^{3}(a-bi^{3})=a^{2}+abi^{3}-abi^{3}-bi^{6}
a^{2}+ab-ab+b=a^{3}b-ab^{2}=a

I'm afraid not. Starting with (a+bi^2)(a-bi^2)i^2, you should end up with the expression b^2-a^2.

Ok. I am guessing i have made an error along the way. I'll try and work out what i have done wrong.

It pains me, but here goes: given (a+bi^2)(a- bi^2)i^2 you should proceed as follows.....

the usual rule is to work inside parentheses first. Noticing that i^2 = -1 you will find that

(a+bi^2)(a- bi^2) = (a-b)(a+b) = a^2 +ab -ba -b^2 = a^2 - b^2 (assuming commutativity)

So, again remembering that i^2 = -1 you must have that (a^2 - b^2)i^2 = -(a^2 - b^2) = -a^2 + b^2 = b^2 - a^2 just like on the tin that CaptBork offered you.

P.S. This help was offered on the strict condition that you stop describing yourself as a physics student. No person, in any country, would be admitted to any course in physics without this very basic mathematical knowledge.

I sware on my mothers grave i am a physics student. I am also studying biology and chemistry now.

So not a physics undergraduate but a high school student whose topics include physics. I took chemistry in high school but I'm not a chemist and never was.

Oh, so i am lying about my age now as well am i.

If you need to know, when i finish this, i will get my National Diploma in physics. The chemistry and biology are subjects just started.

Know what, fuck you alphanumeric. I thought good... no sign of him, i hoped that your computer had crashed with a bug, or something worse. Then you have to come back, with your filthy attitude toward me.

Well suck it up, i finished highschool seven years ago.

testhrdeqibf

Know what, fuck you alphanumeric. I thought good... no sign of him, i hoped that your computer had crashed with a bug, or something worse. Then you have to come back, with your filthy attitude toward me.Actually, I was at a string theory conference for a week. People like Ed Witten were there too. Given I was listening to about 8 hours of algebraic topology and differential geometry a day (and only understanding small parts of it) I didn't feel like coming onto these forums during that time.

That and I was typing up a draft of a paper.

And I didn't say you were lying about your age. You're just inconsistent when it comes to sayingwhat you actually do. It's uni, then it's pre-uni. You're taught by lecturers then it's teachers. You've got a degree, then you don't. You did physics at university, then you didn't.

Perhaps if you answered me when I asked you to say precisely what you're doing and what you've done in education things would be clearer. But you seem to avoid answering.