Are complex numbers two dimensional or one dimensional?Geometric interpretation of the multiplication of complex numbers?4 dimensional numbersWhy are complex numbers considered to be numbers?Total order on complex numbersComplex Number on Cartesian Coordinate System QuestionWhy do we represent complex numbers as the sum of real and imaginary parts?How to plot complex numbers one-dimensionally?Transformation not defined for two different dimensionsWhy are complex numbers presented in a 2d panel (graph)?Multiplication of Phasors as Complex Numbers and Vectors
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Are complex numbers two dimensional or one dimensional?
Geometric interpretation of the multiplication of complex numbers?4 dimensional numbersWhy are complex numbers considered to be numbers?Total order on complex numbersComplex Number on Cartesian Coordinate System QuestionWhy do we represent complex numbers as the sum of real and imaginary parts?How to plot complex numbers one-dimensionally?Transformation not defined for two different dimensionsWhy are complex numbers presented in a 2d panel (graph)?Multiplication of Phasors as Complex Numbers and Vectors
$begingroup$
Complex numbers are represented as:
z = x + yi
This gives the impression that complex numbers are a real component plus an imaginary component. However, when doing math with complex numbers, they are represented as 2-D vectors like in this picture:
Complex Vector
Are complex numbers one or two dimensional?
If they are one dimensional, then why do we do math with them and represent them as vectors?
If they are two dimensional, then why are they represented as one number: "x + yi" instead of a coordinate pair: "(x, yi)"
I know this has been asked a zillion times, but most of the answers I've found online don't explain the subject well. The best explanation I've found so far was here on Quora: https://www.quora.com/Are-complex-numbers-2-dimensional
linear-algebra complex-numbers
edited 2 hours ago
Ethan Bolker
48.1k556123
asked 2 hours ago
Random ProgrammerRandom Programmer
161
New contributor
Random Programmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Complex numbers are represented as:
z = x + yi
This gives the impression that complex numbers are a real component plus an imaginary component. However, when doing math with complex numbers, they are represented as 2-D vectors like in this picture:
Complex Vector
Are complex numbers one or two dimensional?
If they are one dimensional, then why do we do math with them and represent them as vectors?
If they are two dimensional, then why are they represented as one number: "x + yi" instead of a coordinate pair: "(x, yi)"
I know this has been asked a zillion times, but most of the answers I've found online don't explain the subject well. The best explanation I've found so far was here on Quora: https://www.quora.com/Are-complex-numbers-2-dimensional
linear-algebra complex-numbers
edited 2 hours ago
Ethan Bolker
48.1k556123
asked 2 hours ago
Random ProgrammerRandom Programmer
161
New contributor
Random Programmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
It depends on the basis field. It is 1-dimensional with respect to itself $mathbbC$. But it is 2-dimensional wrt $mathbbR$
$endgroup$
– Julian Mejia
2 hours ago
1
$begingroup$
What distinguishes a complex number from a typical $2D$ real vector is the fact that you can multiply complex numbers to get another complex number: $(a+bi)(c+di) = (ac-bd)+(ab+bc)i$
$endgroup$
– Henry
2 hours ago
3
$begingroup$
As a perspective on what people have said, sometimes we can regard mathematical objects like the complex numbers as things in themselves which have an inherent unity. Other times we can decompose them into pieces in an interesting way. Sometimes we can do both, and quite often that shows that something interesting is going on.
$endgroup$
– Mark Bennet
1 hour ago
add a comment |
$begingroup$
Complex numbers are represented as:
z = x + yi
This gives the impression that complex numbers are a real component plus an imaginary component. However, when doing math with complex numbers, they are represented as 2-D vectors like in this picture:
Complex Vector
Are complex numbers one or two dimensional?
If they are one dimensional, then why do we do math with them and represent them as vectors?
If they are two dimensional, then why are they represented as one number: "x + yi" instead of a coordinate pair: "(x, yi)"
I know this has been asked a zillion times, but most of the answers I've found online don't explain the subject well. The best explanation I've found so far was here on Quora: https://www.quora.com/Are-complex-numbers-2-dimensional
linear-algebra complex-numbers
edited 2 hours ago
Ethan Bolker
48.1k556123
asked 2 hours ago
Random ProgrammerRandom Programmer
161
New contributor
Random Programmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Complex numbers are represented as:
z = x + yi
This gives the impression that complex numbers are a real component plus an imaginary component. However, when doing math with complex numbers, they are represented as 2-D vectors like in this picture:
Complex Vector
Are complex numbers one or two dimensional?
If they are one dimensional, then why do we do math with them and represent them as vectors?
If they are two dimensional, then why are they represented as one number: "x + yi" instead of a coordinate pair: "(x, yi)"
I know this has been asked a zillion times, but most of the answers I've found online don't explain the subject well. The best explanation I've found so far was here on Quora: https://www.quora.com/Are-complex-numbers-2-dimensional
linear-algebra complex-numbers
linear-algebra complex-numbers
edited 2 hours ago
Ethan Bolker
48.1k556123
asked 2 hours ago
Random ProgrammerRandom Programmer
161
New contributor
Random Programmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Ethan Bolker
48.1k556123
asked 2 hours ago
Random ProgrammerRandom Programmer
161
New contributor
Random Programmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Ethan Bolker
48.1k556123
edited 2 hours ago
Ethan Bolker
48.1k556123
edited 2 hours ago
Ethan Bolker
48.1k556123
48.1k556123
asked 2 hours ago
Random ProgrammerRandom Programmer
161
New contributor
Random Programmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Random ProgrammerRandom Programmer
161
asked 2 hours ago
Random ProgrammerRandom Programmer
161
161
New contributor
Random Programmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Random Programmer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
It depends on the basis field. It is 1-dimensional with respect to itself $mathbbC$. But it is 2-dimensional wrt $mathbbR$
$endgroup$
– Julian Mejia
2 hours ago
1
$begingroup$
What distinguishes a complex number from a typical $2D$ real vector is the fact that you can multiply complex numbers to get another complex number: $(a+bi)(c+di) = (ac-bd)+(ab+bc)i$
$endgroup$
– Henry
2 hours ago
3
$begingroup$
As a perspective on what people have said, sometimes we can regard mathematical objects like the complex numbers as things in themselves which have an inherent unity. Other times we can decompose them into pieces in an interesting way. Sometimes we can do both, and quite often that shows that something interesting is going on.
$endgroup$
– Mark Bennet
1 hour ago
add a comment |
3
$begingroup$
It depends on the basis field. It is 1-dimensional with respect to itself $mathbbC$. But it is 2-dimensional wrt $mathbbR$
$endgroup$
– Julian Mejia
2 hours ago
1
$begingroup$
What distinguishes a complex number from a typical $2D$ real vector is the fact that you can multiply complex numbers to get another complex number: $(a+bi)(c+di) = (ac-bd)+(ab+bc)i$
$endgroup$
– Henry
2 hours ago
3
$begingroup$
As a perspective on what people have said, sometimes we can regard mathematical objects like the complex numbers as things in themselves which have an inherent unity. Other times we can decompose them into pieces in an interesting way. Sometimes we can do both, and quite often that shows that something interesting is going on.
$endgroup$
– Mark Bennet
1 hour ago
3
3
$begingroup$
It depends on the basis field. It is 1-dimensional with respect to itself $mathbbC$. But it is 2-dimensional wrt $mathbbR$
$endgroup$
– Julian Mejia
2 hours ago
$begingroup$
It depends on the basis field. It is 1-dimensional with respect to itself $mathbbC$. But it is 2-dimensional wrt $mathbbR$
$endgroup$
– Julian Mejia
2 hours ago
1
1
$begingroup$
What distinguishes a complex number from a typical $2D$ real vector is the fact that you can multiply complex numbers to get another complex number: $(a+bi)(c+di) = (ac-bd)+(ab+bc)i$
$endgroup$
– Henry
2 hours ago
$begingroup$
What distinguishes a complex number from a typical $2D$ real vector is the fact that you can multiply complex numbers to get another complex number: $(a+bi)(c+di) = (ac-bd)+(ab+bc)i$
$endgroup$
– Henry
2 hours ago
3
3
$begingroup$
As a perspective on what people have said, sometimes we can regard mathematical objects like the complex numbers as things in themselves which have an inherent unity. Other times we can decompose them into pieces in an interesting way. Sometimes we can do both, and quite often that shows that something interesting is going on.
$endgroup$
– Mark Bennet
1 hour ago
$begingroup$
As a perspective on what people have said, sometimes we can regard mathematical objects like the complex numbers as things in themselves which have an inherent unity. Other times we can decompose them into pieces in an interesting way. Sometimes we can do both, and quite often that shows that something interesting is going on.
$endgroup$
– Mark Bennet
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can consider $mathbbC$ as a $mathbbC$vector space in this case complex numbers is a $1$-dimensional vector space over $mathbbC$.
You can consider $mathbbC$ as a $mathbbR$vector space in this case complex numbers is a $2$-dimensional vector space over $mathbbR$.
answered 2 hours ago
Tsemo AristideTsemo Aristide
62k11447
$endgroup$
$begingroup$
(+1) Just what I was about to write.
$endgroup$
– José Carlos Santos
2 hours ago
1
$begingroup$
Yes, but if yo think of it as a vector space you have to be careful, because it is not an inner product space in the usual sense. That is, the "dot product" between tow complex "vectors" is not the usual complex multiplcation of the two numbers represented.
$endgroup$
– Mark Fischler
2 hours ago
add a comment |
$begingroup$
Quoting from https://en.m.wikipedia.org/wiki/Complex_number#Construction_as_ordered_pairs:
William Rowan Hamilton introduced the approach to define the set C of complex numbers as the set R^2 of ordered pairs (a, b) of real numbers, in which the following rules for addition and multiplication are imposed:
beginaligned(a,b)+(c,d)&=(a+c,b+d)\(a,b)cdot (c,d)&=(ac-bd,bc+ad).endaligned
It is then just a matter of notation to express (a, b) as a + bi.
answered 1 hour ago
ahartelahartel
134
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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active
oldest
votes
$begingroup$
You can consider $mathbbC$ as a $mathbbC$vector space in this case complex numbers is a $1$-dimensional vector space over $mathbbC$.
You can consider $mathbbC$ as a $mathbbR$vector space in this case complex numbers is a $2$-dimensional vector space over $mathbbR$.
answered 2 hours ago
Tsemo AristideTsemo Aristide
62k11447
$endgroup$
$begingroup$
(+1) Just what I was about to write.
$endgroup$
– José Carlos Santos
2 hours ago
1
$begingroup$
Yes, but if yo think of it as a vector space you have to be careful, because it is not an inner product space in the usual sense. That is, the "dot product" between tow complex "vectors" is not the usual complex multiplcation of the two numbers represented.
$endgroup$
– Mark Fischler
2 hours ago
add a comment |
$begingroup$
You can consider $mathbbC$ as a $mathbbC$vector space in this case complex numbers is a $1$-dimensional vector space over $mathbbC$.
You can consider $mathbbC$ as a $mathbbR$vector space in this case complex numbers is a $2$-dimensional vector space over $mathbbR$.
answered 2 hours ago
Tsemo AristideTsemo Aristide
62k11447
$endgroup$
$begingroup$
(+1) Just what I was about to write.
$endgroup$
– José Carlos Santos
2 hours ago
1
$begingroup$
Yes, but if yo think of it as a vector space you have to be careful, because it is not an inner product space in the usual sense. That is, the "dot product" between tow complex "vectors" is not the usual complex multiplcation of the two numbers represented.
$endgroup$
– Mark Fischler
2 hours ago
add a comment |
$begingroup$
You can consider $mathbbC$ as a $mathbbC$vector space in this case complex numbers is a $1$-dimensional vector space over $mathbbC$.
You can consider $mathbbC$ as a $mathbbR$vector space in this case complex numbers is a $2$-dimensional vector space over $mathbbR$.
answered 2 hours ago
Tsemo AristideTsemo Aristide
62k11447
$endgroup$
You can consider $mathbbC$ as a $mathbbC$vector space in this case complex numbers is a $1$-dimensional vector space over $mathbbC$.
You can consider $mathbbC$ as a $mathbbR$vector space in this case complex numbers is a $2$-dimensional vector space over $mathbbR$.
answered 2 hours ago
Tsemo AristideTsemo Aristide
62k11447
answered 2 hours ago
Tsemo AristideTsemo Aristide
62k11447
answered 2 hours ago
Tsemo AristideTsemo Aristide
62k11447
answered 2 hours ago
Tsemo AristideTsemo Aristide
62k11447
62k11447
$begingroup$
(+1) Just what I was about to write.
$endgroup$
– José Carlos Santos
2 hours ago
1
$begingroup$
Yes, but if yo think of it as a vector space you have to be careful, because it is not an inner product space in the usual sense. That is, the "dot product" between tow complex "vectors" is not the usual complex multiplcation of the two numbers represented.
$endgroup$
– Mark Fischler
2 hours ago
add a comment |
$begingroup$
(+1) Just what I was about to write.
$endgroup$
– José Carlos Santos
2 hours ago
1
$begingroup$
Yes, but if yo think of it as a vector space you have to be careful, because it is not an inner product space in the usual sense. That is, the "dot product" between tow complex "vectors" is not the usual complex multiplcation of the two numbers represented.
$endgroup$
– Mark Fischler
2 hours ago
$begingroup$
(+1) Just what I was about to write.
$endgroup$
– José Carlos Santos
2 hours ago
$begingroup$
(+1) Just what I was about to write.
$endgroup$
– José Carlos Santos
2 hours ago
1
1
$begingroup$
Yes, but if yo think of it as a vector space you have to be careful, because it is not an inner product space in the usual sense. That is, the "dot product" between tow complex "vectors" is not the usual complex multiplcation of the two numbers represented.
$endgroup$
– Mark Fischler
2 hours ago
$begingroup$
Yes, but if yo think of it as a vector space you have to be careful, because it is not an inner product space in the usual sense. That is, the "dot product" between tow complex "vectors" is not the usual complex multiplcation of the two numbers represented.
$endgroup$
– Mark Fischler
2 hours ago
add a comment |
$begingroup$
Quoting from https://en.m.wikipedia.org/wiki/Complex_number#Construction_as_ordered_pairs:
William Rowan Hamilton introduced the approach to define the set C of complex numbers as the set R^2 of ordered pairs (a, b) of real numbers, in which the following rules for addition and multiplication are imposed:
beginaligned(a,b)+(c,d)&=(a+c,b+d)\(a,b)cdot (c,d)&=(ac-bd,bc+ad).endaligned
It is then just a matter of notation to express (a, b) as a + bi.
answered 1 hour ago
ahartelahartel
134
$endgroup$
add a comment |
$begingroup$
Quoting from https://en.m.wikipedia.org/wiki/Complex_number#Construction_as_ordered_pairs:
William Rowan Hamilton introduced the approach to define the set C of complex numbers as the set R^2 of ordered pairs (a, b) of real numbers, in which the following rules for addition and multiplication are imposed:
beginaligned(a,b)+(c,d)&=(a+c,b+d)\(a,b)cdot (c,d)&=(ac-bd,bc+ad).endaligned
It is then just a matter of notation to express (a, b) as a + bi.
answered 1 hour ago
ahartelahartel
134
$endgroup$
add a comment |
$begingroup$
Quoting from https://en.m.wikipedia.org/wiki/Complex_number#Construction_as_ordered_pairs:
William Rowan Hamilton introduced the approach to define the set C of complex numbers as the set R^2 of ordered pairs (a, b) of real numbers, in which the following rules for addition and multiplication are imposed:
beginaligned(a,b)+(c,d)&=(a+c,b+d)\(a,b)cdot (c,d)&=(ac-bd,bc+ad).endaligned
It is then just a matter of notation to express (a, b) as a + bi.
answered 1 hour ago
ahartelahartel
134
$endgroup$
Quoting from https://en.m.wikipedia.org/wiki/Complex_number#Construction_as_ordered_pairs:
William Rowan Hamilton introduced the approach to define the set C of complex numbers as the set R^2 of ordered pairs (a, b) of real numbers, in which the following rules for addition and multiplication are imposed:
beginaligned(a,b)+(c,d)&=(a+c,b+d)\(a,b)cdot (c,d)&=(ac-bd,bc+ad).endaligned
It is then just a matter of notation to express (a, b) as a + bi.
answered 1 hour ago
ahartelahartel
134
answered 1 hour ago
ahartelahartel
134
answered 1 hour ago
ahartelahartel
134
answered 1 hour ago
ahartelahartel
134
134
add a comment |
add a comment |
Random Programmer is a new contributor. Be nice, and check out our Code of Conduct.
Random Programmer is a new contributor. Be nice, and check out our Code of Conduct.
Random Programmer is a new contributor. Be nice, and check out our Code of Conduct.
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3
$begingroup$
It depends on the basis field. It is 1-dimensional with respect to itself $mathbbC$. But it is 2-dimensional wrt $mathbbR$
$endgroup$
– Julian Mejia
2 hours ago
1
$begingroup$
What distinguishes a complex number from a typical $2D$ real vector is the fact that you can multiply complex numbers to get another complex number: $(a+bi)(c+di) = (ac-bd)+(ab+bc)i$
$endgroup$
– Henry
2 hours ago
3
$begingroup$
As a perspective on what people have said, sometimes we can regard mathematical objects like the complex numbers as things in themselves which have an inherent unity. Other times we can decompose them into pieces in an interesting way. Sometimes we can do both, and quite often that shows that something interesting is going on.
$endgroup$
– Mark Bennet
1 hour ago