Complex numbers and their matrix form.












7












$begingroup$


I have a line starting at the origin, and i extend it to a point $(a,b)$ in the plane. This thing can be called a vector and be represented as $(a,b), [atext{ }b]^T$ (column vector) or by $amathbf{i}+bmathbf{j}$, where $(mathbf{i},mathbf{j})$ is the stardard basis in $mathbb{R}^2$ Or it could be seen as a visual representation of a complex number where $(a,b)=a+bi,$ where $i=sqrt{-1}$.



So I want to rotate this vector $(a,b)$ $90$ degrees counter clockwise, so i know I can use my trusty matrix for rotations
$begin{bmatrix} cos(90) & -sin(90) \ sin(90) & cos(90)\ end{bmatrix}$=$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}$ and we find that
$$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}begin{bmatrix} a \ b\ end{bmatrix}=begin{bmatrix} -b \ a\ end{bmatrix}$$
Or, I could choose the complex multiplication way and say,
$i(a+bi)=ai+bi^2=ai-b=-b+ai$



So we all know that, but what are some of the advantages and disadvantages to having two things that are completely identical operation in different systems?






linear-algebra matrices complex-numbers rotations







share|cite|improve this question











$endgroup$












  • $begingroup$
    For this particular example, one obvious difference is computation. To use the rotation matrix, you needed 4 trigonometric computations, 4 scalar multiplications and 2 additions. But in the other method, you swapped the entries and changed the sign of one. In general, for any degree of rotation, both will have same amount of computation.
    $endgroup$
    – dineshdileep
    Sep 19 '13 at 11:07










  • $begingroup$
    Not clear what does the author mean. Multiplication of complex numbers gets two numbers (elements of the field ℂ) and yields a product from the same set. Euclidean 2-vectors are another thing. Although you can “multiply” two rotations and obtain another rotation, there is no multiplication on 2-vectors that gives 2-vectors.
    $endgroup$
    – Incnis Mrsi
    Nov 2 '14 at 7:09


















7












$begingroup$


I have a line starting at the origin, and i extend it to a point $(a,b)$ in the plane. This thing can be called a vector and be represented as $(a,b), [atext{ }b]^T$ (column vector) or by $amathbf{i}+bmathbf{j}$, where $(mathbf{i},mathbf{j})$ is the stardard basis in $mathbb{R}^2$ Or it could be seen as a visual representation of a complex number where $(a,b)=a+bi,$ where $i=sqrt{-1}$.



So I want to rotate this vector $(a,b)$ $90$ degrees counter clockwise, so i know I can use my trusty matrix for rotations
$begin{bmatrix} cos(90) & -sin(90) \ sin(90) & cos(90)\ end{bmatrix}$=$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}$ and we find that
$$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}begin{bmatrix} a \ b\ end{bmatrix}=begin{bmatrix} -b \ a\ end{bmatrix}$$
Or, I could choose the complex multiplication way and say,
$i(a+bi)=ai+bi^2=ai-b=-b+ai$



So we all know that, but what are some of the advantages and disadvantages to having two things that are completely identical operation in different systems?






linear-algebra matrices complex-numbers rotations







share|cite|improve this question











$endgroup$












  • $begingroup$
    For this particular example, one obvious difference is computation. To use the rotation matrix, you needed 4 trigonometric computations, 4 scalar multiplications and 2 additions. But in the other method, you swapped the entries and changed the sign of one. In general, for any degree of rotation, both will have same amount of computation.
    $endgroup$
    – dineshdileep
    Sep 19 '13 at 11:07










  • $begingroup$
    Not clear what does the author mean. Multiplication of complex numbers gets two numbers (elements of the field ℂ) and yields a product from the same set. Euclidean 2-vectors are another thing. Although you can “multiply” two rotations and obtain another rotation, there is no multiplication on 2-vectors that gives 2-vectors.
    $endgroup$
    – Incnis Mrsi
    Nov 2 '14 at 7:09
















7












7








7


5



$begingroup$


I have a line starting at the origin, and i extend it to a point $(a,b)$ in the plane. This thing can be called a vector and be represented as $(a,b), [atext{ }b]^T$ (column vector) or by $amathbf{i}+bmathbf{j}$, where $(mathbf{i},mathbf{j})$ is the stardard basis in $mathbb{R}^2$ Or it could be seen as a visual representation of a complex number where $(a,b)=a+bi,$ where $i=sqrt{-1}$.



So I want to rotate this vector $(a,b)$ $90$ degrees counter clockwise, so i know I can use my trusty matrix for rotations
$begin{bmatrix} cos(90) & -sin(90) \ sin(90) & cos(90)\ end{bmatrix}$=$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}$ and we find that
$$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}begin{bmatrix} a \ b\ end{bmatrix}=begin{bmatrix} -b \ a\ end{bmatrix}$$
Or, I could choose the complex multiplication way and say,
$i(a+bi)=ai+bi^2=ai-b=-b+ai$



So we all know that, but what are some of the advantages and disadvantages to having two things that are completely identical operation in different systems?






linear-algebra matrices complex-numbers rotations







share|cite|improve this question











$endgroup$




I have a line starting at the origin, and i extend it to a point $(a,b)$ in the plane. This thing can be called a vector and be represented as $(a,b), [atext{ }b]^T$ (column vector) or by $amathbf{i}+bmathbf{j}$, where $(mathbf{i},mathbf{j})$ is the stardard basis in $mathbb{R}^2$ Or it could be seen as a visual representation of a complex number where $(a,b)=a+bi,$ where $i=sqrt{-1}$.



So I want to rotate this vector $(a,b)$ $90$ degrees counter clockwise, so i know I can use my trusty matrix for rotations
$begin{bmatrix} cos(90) & -sin(90) \ sin(90) & cos(90)\ end{bmatrix}$=$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}$ and we find that
$$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}begin{bmatrix} a \ b\ end{bmatrix}=begin{bmatrix} -b \ a\ end{bmatrix}$$
Or, I could choose the complex multiplication way and say,
$i(a+bi)=ai+bi^2=ai-b=-b+ai$



So we all know that, but what are some of the advantages and disadvantages to having two things that are completely identical operation in different systems?






linear-algebra matrices complex-numbers rotations




linear-algebra matrices complex-numbers rotations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 0:16









dantopa

6,46942243




6,46942243










asked Sep 19 '13 at 2:37









Irish M PowersIrish M Powers

1155




1155












  • $begingroup$
    For this particular example, one obvious difference is computation. To use the rotation matrix, you needed 4 trigonometric computations, 4 scalar multiplications and 2 additions. But in the other method, you swapped the entries and changed the sign of one. In general, for any degree of rotation, both will have same amount of computation.
    $endgroup$
    – dineshdileep
    Sep 19 '13 at 11:07










  • $begingroup$
    Not clear what does the author mean. Multiplication of complex numbers gets two numbers (elements of the field ℂ) and yields a product from the same set. Euclidean 2-vectors are another thing. Although you can “multiply” two rotations and obtain another rotation, there is no multiplication on 2-vectors that gives 2-vectors.
    $endgroup$
    – Incnis Mrsi
    Nov 2 '14 at 7:09




















  • $begingroup$
    For this particular example, one obvious difference is computation. To use the rotation matrix, you needed 4 trigonometric computations, 4 scalar multiplications and 2 additions. But in the other method, you swapped the entries and changed the sign of one. In general, for any degree of rotation, both will have same amount of computation.
    $endgroup$
    – dineshdileep
    Sep 19 '13 at 11:07










  • $begingroup$
    Not clear what does the author mean. Multiplication of complex numbers gets two numbers (elements of the field ℂ) and yields a product from the same set. Euclidean 2-vectors are another thing. Although you can “multiply” two rotations and obtain another rotation, there is no multiplication on 2-vectors that gives 2-vectors.
    $endgroup$
    – Incnis Mrsi
    Nov 2 '14 at 7:09


















$begingroup$
For this particular example, one obvious difference is computation. To use the rotation matrix, you needed 4 trigonometric computations, 4 scalar multiplications and 2 additions. But in the other method, you swapped the entries and changed the sign of one. In general, for any degree of rotation, both will have same amount of computation.
$endgroup$
– dineshdileep
Sep 19 '13 at 11:07




$begingroup$
For this particular example, one obvious difference is computation. To use the rotation matrix, you needed 4 trigonometric computations, 4 scalar multiplications and 2 additions. But in the other method, you swapped the entries and changed the sign of one. In general, for any degree of rotation, both will have same amount of computation.
$endgroup$
– dineshdileep
Sep 19 '13 at 11:07












$begingroup$
Not clear what does the author mean. Multiplication of complex numbers gets two numbers (elements of the field ℂ) and yields a product from the same set. Euclidean 2-vectors are another thing. Although you can “multiply” two rotations and obtain another rotation, there is no multiplication on 2-vectors that gives 2-vectors.
$endgroup$
– Incnis Mrsi
Nov 2 '14 at 7:09






$begingroup$
Not clear what does the author mean. Multiplication of complex numbers gets two numbers (elements of the field ℂ) and yields a product from the same set. Euclidean 2-vectors are another thing. Although you can “multiply” two rotations and obtain another rotation, there is no multiplication on 2-vectors that gives 2-vectors.
$endgroup$
– Incnis Mrsi
Nov 2 '14 at 7:09












1 Answer
1






active

oldest

votes


















2












$begingroup$

There is a homeomorphism between the the complex numbers



$$
color{blue}{a} + color{red}{b}i
$$



and the rotation matrices



$$
left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{a} \
end{array}
right)
$$




Let the Cartesian forms be
$$
z_{1} = color{blue}{a} + color{red}{b}i, quad
z_{2} = color{blue}{c} + color{red}{d}i
$$
and the matrix forms
$$
z_{1} = left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{c} \
end{array}
right),
quad
z_{2} = left(
begin{array}{rc}
color{blue}{c}& color{red}{d} \
- color{red}{d} & color{blue}{c}\
end{array}
right)
$$



Addition



$$
begin{align}
%
z_{1} +
z_{2} &=
(color{blue}{a}+color{blue}{c}) + (color{red}{b} + color{red}{d} )i \[3pt]
%
&=
left(
begin{array}{rc}
color{blue}{a}+color{blue}{c}& color{red}{b}+ color{red}{d} \
-color{red}{b}- color{red}{d} & color{blue}{a}+color{blue}{c}\
end{array}
right)
end{align}
$$



Multiplication



$$
begin{align}
z_{1}
z_{2} &=
(color{blue}{ac}- color{red}{bd}) +
(color{red}{b}color{blue}{c}+color{blue}{a}color{red}{d})i\[3pt]
%
&=
%
left(
begin{array}{rc}
color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} & color{red}{b}color{blue}{c}+color{blue}{a} color{red}{d} \
- color{red}{b} color{blue}{c}-color{blue}{a} color{red}{d} & color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} \
end{array}
right)
end{align}
$$



Inversion



$$
begin{align}
%
frac{1}{z} &= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1} left( color{blue}{a} - color{red}{b} i right) \[3pt]
%
&= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1}
left(
begin{array}{cr}
color{blue}{a} & - color{red}{b} \
color{red}{b} & color{blue}{a} \
end{array}
right)
end{align}
$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology.
    $endgroup$
    – Starfall
    Apr 24 '17 at 4:04










  • $begingroup$
    @Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion.
    $endgroup$
    – dantopa
    Apr 24 '17 at 4:21











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f498132%2fcomplex-numbers-and-their-matrix-form%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

There is a homeomorphism between the the complex numbers



$$
color{blue}{a} + color{red}{b}i
$$



and the rotation matrices



$$
left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{a} \
end{array}
right)
$$




Let the Cartesian forms be
$$
z_{1} = color{blue}{a} + color{red}{b}i, quad
z_{2} = color{blue}{c} + color{red}{d}i
$$
and the matrix forms
$$
z_{1} = left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{c} \
end{array}
right),
quad
z_{2} = left(
begin{array}{rc}
color{blue}{c}& color{red}{d} \
- color{red}{d} & color{blue}{c}\
end{array}
right)
$$



Addition



$$
begin{align}
%
z_{1} +
z_{2} &=
(color{blue}{a}+color{blue}{c}) + (color{red}{b} + color{red}{d} )i \[3pt]
%
&=
left(
begin{array}{rc}
color{blue}{a}+color{blue}{c}& color{red}{b}+ color{red}{d} \
-color{red}{b}- color{red}{d} & color{blue}{a}+color{blue}{c}\
end{array}
right)
end{align}
$$



Multiplication



$$
begin{align}
z_{1}
z_{2} &=
(color{blue}{ac}- color{red}{bd}) +
(color{red}{b}color{blue}{c}+color{blue}{a}color{red}{d})i\[3pt]
%
&=
%
left(
begin{array}{rc}
color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} & color{red}{b}color{blue}{c}+color{blue}{a} color{red}{d} \
- color{red}{b} color{blue}{c}-color{blue}{a} color{red}{d} & color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} \
end{array}
right)
end{align}
$$



Inversion



$$
begin{align}
%
frac{1}{z} &= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1} left( color{blue}{a} - color{red}{b} i right) \[3pt]
%
&= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1}
left(
begin{array}{cr}
color{blue}{a} & - color{red}{b} \
color{red}{b} & color{blue}{a} \
end{array}
right)
end{align}
$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology.
    $endgroup$
    – Starfall
    Apr 24 '17 at 4:04










  • $begingroup$
    @Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion.
    $endgroup$
    – dantopa
    Apr 24 '17 at 4:21
















2












$begingroup$

There is a homeomorphism between the the complex numbers



$$
color{blue}{a} + color{red}{b}i
$$



and the rotation matrices



$$
left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{a} \
end{array}
right)
$$




Let the Cartesian forms be
$$
z_{1} = color{blue}{a} + color{red}{b}i, quad
z_{2} = color{blue}{c} + color{red}{d}i
$$
and the matrix forms
$$
z_{1} = left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{c} \
end{array}
right),
quad
z_{2} = left(
begin{array}{rc}
color{blue}{c}& color{red}{d} \
- color{red}{d} & color{blue}{c}\
end{array}
right)
$$



Addition



$$
begin{align}
%
z_{1} +
z_{2} &=
(color{blue}{a}+color{blue}{c}) + (color{red}{b} + color{red}{d} )i \[3pt]
%
&=
left(
begin{array}{rc}
color{blue}{a}+color{blue}{c}& color{red}{b}+ color{red}{d} \
-color{red}{b}- color{red}{d} & color{blue}{a}+color{blue}{c}\
end{array}
right)
end{align}
$$



Multiplication



$$
begin{align}
z_{1}
z_{2} &=
(color{blue}{ac}- color{red}{bd}) +
(color{red}{b}color{blue}{c}+color{blue}{a}color{red}{d})i\[3pt]
%
&=
%
left(
begin{array}{rc}
color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} & color{red}{b}color{blue}{c}+color{blue}{a} color{red}{d} \
- color{red}{b} color{blue}{c}-color{blue}{a} color{red}{d} & color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} \
end{array}
right)
end{align}
$$



Inversion



$$
begin{align}
%
frac{1}{z} &= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1} left( color{blue}{a} - color{red}{b} i right) \[3pt]
%
&= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1}
left(
begin{array}{cr}
color{blue}{a} & - color{red}{b} \
color{red}{b} & color{blue}{a} \
end{array}
right)
end{align}
$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology.
    $endgroup$
    – Starfall
    Apr 24 '17 at 4:04










  • $begingroup$
    @Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion.
    $endgroup$
    – dantopa
    Apr 24 '17 at 4:21














2












2








2





$begingroup$

There is a homeomorphism between the the complex numbers



$$
color{blue}{a} + color{red}{b}i
$$



and the rotation matrices



$$
left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{a} \
end{array}
right)
$$




Let the Cartesian forms be
$$
z_{1} = color{blue}{a} + color{red}{b}i, quad
z_{2} = color{blue}{c} + color{red}{d}i
$$
and the matrix forms
$$
z_{1} = left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{c} \
end{array}
right),
quad
z_{2} = left(
begin{array}{rc}
color{blue}{c}& color{red}{d} \
- color{red}{d} & color{blue}{c}\
end{array}
right)
$$



Addition



$$
begin{align}
%
z_{1} +
z_{2} &=
(color{blue}{a}+color{blue}{c}) + (color{red}{b} + color{red}{d} )i \[3pt]
%
&=
left(
begin{array}{rc}
color{blue}{a}+color{blue}{c}& color{red}{b}+ color{red}{d} \
-color{red}{b}- color{red}{d} & color{blue}{a}+color{blue}{c}\
end{array}
right)
end{align}
$$



Multiplication



$$
begin{align}
z_{1}
z_{2} &=
(color{blue}{ac}- color{red}{bd}) +
(color{red}{b}color{blue}{c}+color{blue}{a}color{red}{d})i\[3pt]
%
&=
%
left(
begin{array}{rc}
color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} & color{red}{b}color{blue}{c}+color{blue}{a} color{red}{d} \
- color{red}{b} color{blue}{c}-color{blue}{a} color{red}{d} & color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} \
end{array}
right)
end{align}
$$



Inversion



$$
begin{align}
%
frac{1}{z} &= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1} left( color{blue}{a} - color{red}{b} i right) \[3pt]
%
&= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1}
left(
begin{array}{cr}
color{blue}{a} & - color{red}{b} \
color{red}{b} & color{blue}{a} \
end{array}
right)
end{align}
$$






share|cite|improve this answer











$endgroup$



There is a homeomorphism between the the complex numbers



$$
color{blue}{a} + color{red}{b}i
$$



and the rotation matrices



$$
left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{a} \
end{array}
right)
$$




Let the Cartesian forms be
$$
z_{1} = color{blue}{a} + color{red}{b}i, quad
z_{2} = color{blue}{c} + color{red}{d}i
$$
and the matrix forms
$$
z_{1} = left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{c} \
end{array}
right),
quad
z_{2} = left(
begin{array}{rc}
color{blue}{c}& color{red}{d} \
- color{red}{d} & color{blue}{c}\
end{array}
right)
$$



Addition



$$
begin{align}
%
z_{1} +
z_{2} &=
(color{blue}{a}+color{blue}{c}) + (color{red}{b} + color{red}{d} )i \[3pt]
%
&=
left(
begin{array}{rc}
color{blue}{a}+color{blue}{c}& color{red}{b}+ color{red}{d} \
-color{red}{b}- color{red}{d} & color{blue}{a}+color{blue}{c}\
end{array}
right)
end{align}
$$



Multiplication



$$
begin{align}
z_{1}
z_{2} &=
(color{blue}{ac}- color{red}{bd}) +
(color{red}{b}color{blue}{c}+color{blue}{a}color{red}{d})i\[3pt]
%
&=
%
left(
begin{array}{rc}
color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} & color{red}{b}color{blue}{c}+color{blue}{a} color{red}{d} \
- color{red}{b} color{blue}{c}-color{blue}{a} color{red}{d} & color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} \
end{array}
right)
end{align}
$$



Inversion



$$
begin{align}
%
frac{1}{z} &= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1} left( color{blue}{a} - color{red}{b} i right) \[3pt]
%
&= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1}
left(
begin{array}{cr}
color{blue}{a} & - color{red}{b} \
color{red}{b} & color{blue}{a} \
end{array}
right)
end{align}
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 5 '18 at 16:02









amWhy

1




1










answered Apr 24 '17 at 3:39









dantopadantopa

6,46942243




6,46942243








  • 1




    $begingroup$
    This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology.
    $endgroup$
    – Starfall
    Apr 24 '17 at 4:04










  • $begingroup$
    @Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion.
    $endgroup$
    – dantopa
    Apr 24 '17 at 4:21














  • 1




    $begingroup$
    This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology.
    $endgroup$
    – Starfall
    Apr 24 '17 at 4:04










  • $begingroup$
    @Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion.
    $endgroup$
    – dantopa
    Apr 24 '17 at 4:21








1




1




$begingroup$
This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology.
$endgroup$
– Starfall
Apr 24 '17 at 4:04




$begingroup$
This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology.
$endgroup$
– Starfall
Apr 24 '17 at 4:04












$begingroup$
@Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion.
$endgroup$
– dantopa
Apr 24 '17 at 4:21




$begingroup$
@Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion.
$endgroup$
– dantopa
Apr 24 '17 at 4:21


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f498132%2fcomplex-numbers-and-their-matrix-form%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Wiesbaden

Image
Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
Read more

Marschland

Image
Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
Read more

Dieringhausen

Image
Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
Read more