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Factorising complex numbers
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4 years ago, when I learned about factorising and complex numbers, me and my friend worked on factorising complex numbers.
For example, $ 4+2i= 3-(-1)+2i = 3-i^2+2i = -(i^2-2i-3) = -(i-3)(i+1) $
The goal was to represent $a+bi$ with product of same form. where $a$ and $b$ are integer.
Another example is, $ 8+i= 8i^4+i=i(8i^3+1)=i(2i+1)(4i^2-2i+1)=i(2i+1)(-2i-3)=-i(2i+1)(2i+3) $
I showed my teacher, and she said it's useless.
Now I think of it, I don't know why I did this and it looks like same thing just in different form.
Is there any research already done on this or can there be any use of it?
Apparently,
$(n+2)+ni=-(i-(n+1))(i+1)$
$m^3+n^3i=-i(mi+n)(mni+(m^2-n^2))$
complex-numbers factoring
asked Jul 26 at 14:03
Pizzaroot
1056
add a comment |Â
up vote
7
down vote
favorite
4 years ago, when I learned about factorising and complex numbers, me and my friend worked on factorising complex numbers.
For example, $ 4+2i= 3-(-1)+2i = 3-i^2+2i = -(i^2-2i-3) = -(i-3)(i+1) $
The goal was to represent $a+bi$ with product of same form. where $a$ and $b$ are integer.
Another example is, $ 8+i= 8i^4+i=i(8i^3+1)=i(2i+1)(4i^2-2i+1)=i(2i+1)(-2i-3)=-i(2i+1)(2i+3) $
I showed my teacher, and she said it's useless.
Now I think of it, I don't know why I did this and it looks like same thing just in different form.
Is there any research already done on this or can there be any use of it?
Apparently,
$(n+2)+ni=-(i-(n+1))(i+1)$
$m^3+n^3i=-i(mi+n)(mni+(m^2-n^2))$
complex-numbers factoring
asked Jul 26 at 14:03
Pizzaroot
1056
3
Yes, there's a lot of work done on this. Look up Gaussian Integers.
– Lord Shark the Unknown
Jul 26 at 14:07
5
And by no means is it useless!
– Lubin
Jul 26 at 14:11
There's research already done, but that does not mean there can't be any use of the work you do on this, when you've read up on the existing work.
– Henrik
Jul 26 at 14:13
1
I was thinking @Henrik, that knowing how to factor Gaussian integers is useful in what might look like other parts of mathematics.
– Lubin
Jul 26 at 14:15
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
4 years ago, when I learned about factorising and complex numbers, me and my friend worked on factorising complex numbers.
For example, $ 4+2i= 3-(-1)+2i = 3-i^2+2i = -(i^2-2i-3) = -(i-3)(i+1) $
The goal was to represent $a+bi$ with product of same form. where $a$ and $b$ are integer.
Another example is, $ 8+i= 8i^4+i=i(8i^3+1)=i(2i+1)(4i^2-2i+1)=i(2i+1)(-2i-3)=-i(2i+1)(2i+3) $
I showed my teacher, and she said it's useless.
Now I think of it, I don't know why I did this and it looks like same thing just in different form.
Is there any research already done on this or can there be any use of it?
Apparently,
$(n+2)+ni=-(i-(n+1))(i+1)$
$m^3+n^3i=-i(mi+n)(mni+(m^2-n^2))$
complex-numbers factoring
asked Jul 26 at 14:03
Pizzaroot
1056
4 years ago, when I learned about factorising and complex numbers, me and my friend worked on factorising complex numbers.
For example, $ 4+2i= 3-(-1)+2i = 3-i^2+2i = -(i^2-2i-3) = -(i-3)(i+1) $
The goal was to represent $a+bi$ with product of same form. where $a$ and $b$ are integer.
Another example is, $ 8+i= 8i^4+i=i(8i^3+1)=i(2i+1)(4i^2-2i+1)=i(2i+1)(-2i-3)=-i(2i+1)(2i+3) $
I showed my teacher, and she said it's useless.
Now I think of it, I don't know why I did this and it looks like same thing just in different form.
Is there any research already done on this or can there be any use of it?
Apparently,
$(n+2)+ni=-(i-(n+1))(i+1)$
$m^3+n^3i=-i(mi+n)(mni+(m^2-n^2))$
complex-numbers factoring
asked Jul 26 at 14:03
Pizzaroot
1056
asked Jul 26 at 14:03
Pizzaroot
1056
asked Jul 26 at 14:03
Pizzaroot
1056
asked Jul 26 at 14:03
Pizzaroot
1056
1056
3
Yes, there's a lot of work done on this. Look up Gaussian Integers.
– Lord Shark the Unknown
Jul 26 at 14:07
5
And by no means is it useless!
– Lubin
Jul 26 at 14:11
There's research already done, but that does not mean there can't be any use of the work you do on this, when you've read up on the existing work.
– Henrik
Jul 26 at 14:13
1
I was thinking @Henrik, that knowing how to factor Gaussian integers is useful in what might look like other parts of mathematics.
– Lubin
Jul 26 at 14:15
add a comment |Â
3
Yes, there's a lot of work done on this. Look up Gaussian Integers.
– Lord Shark the Unknown
Jul 26 at 14:07
5
And by no means is it useless!
– Lubin
Jul 26 at 14:11
There's research already done, but that does not mean there can't be any use of the work you do on this, when you've read up on the existing work.
– Henrik
Jul 26 at 14:13
1
I was thinking @Henrik, that knowing how to factor Gaussian integers is useful in what might look like other parts of mathematics.
– Lubin
Jul 26 at 14:15
3
3
Yes, there's a lot of work done on this. Look up Gaussian Integers.
– Lord Shark the Unknown
Jul 26 at 14:07
Yes, there's a lot of work done on this. Look up Gaussian Integers.
– Lord Shark the Unknown
Jul 26 at 14:07
5
5
And by no means is it useless!
– Lubin
Jul 26 at 14:11
And by no means is it useless!
– Lubin
Jul 26 at 14:11
There's research already done, but that does not mean there can't be any use of the work you do on this, when you've read up on the existing work.
– Henrik
Jul 26 at 14:13
There's research already done, but that does not mean there can't be any use of the work you do on this, when you've read up on the existing work.
– Henrik
Jul 26 at 14:13
1
1
I was thinking @Henrik, that knowing how to factor Gaussian integers is useful in what might look like other parts of mathematics.
– Lubin
Jul 26 at 14:15
I was thinking @Henrik, that knowing how to factor Gaussian integers is useful in what might look like other parts of mathematics.
– Lubin
Jul 26 at 14:15
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
You might want to look at Gaussian integers.
answered Jul 26 at 14:07
Kusma
1,097111
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
You might want to look at Gaussian integers.
answered Jul 26 at 14:07
Kusma
1,097111
add a comment |Â
up vote
5
down vote
accepted
You might want to look at Gaussian integers.
answered Jul 26 at 14:07
Kusma
1,097111
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
You might want to look at Gaussian integers.
answered Jul 26 at 14:07
Kusma
1,097111
You might want to look at Gaussian integers.
answered Jul 26 at 14:07
Kusma
1,097111
answered Jul 26 at 14:07
Kusma
1,097111
answered Jul 26 at 14:07
Kusma
1,097111
answered Jul 26 at 14:07
Kusma
1,097111
1,097111
add a comment |Â
add a comment |Â
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3
Yes, there's a lot of work done on this. Look up Gaussian Integers.
– Lord Shark the Unknown
Jul 26 at 14:07
5
And by no means is it useless!
– Lubin
Jul 26 at 14:11
There's research already done, but that does not mean there can't be any use of the work you do on this, when you've read up on the existing work.
– Henrik
Jul 26 at 14:13
1
I was thinking @Henrik, that knowing how to factor Gaussian integers is useful in what might look like other parts of mathematics.
– Lubin
Jul 26 at 14:15