Roots Across the Complex Numbers
$begingroup$
Why is it the case that an even root (square root, quartic, etc) can be positive or negative across the complex numbers, but is limited to postive in the reals? Is there a good mathematical reason for this, or is it simply notation?
radicals
asked Dec 28 '18 at 17:07
John GaltJohn Galt
185
$endgroup$
add a comment |
$begingroup$
Why is it the case that an even root (square root, quartic, etc) can be positive or negative across the complex numbers, but is limited to postive in the reals? Is there a good mathematical reason for this, or is it simply notation?
radicals
asked Dec 28 '18 at 17:07
John GaltJohn Galt
185
$endgroup$
$begingroup$
We have defined $sqrt a$ to be the principal root. Even in the complex numbers there is a principal root.
$endgroup$
– Doug M
Dec 28 '18 at 20:29
add a comment |
$begingroup$
Why is it the case that an even root (square root, quartic, etc) can be positive or negative across the complex numbers, but is limited to postive in the reals? Is there a good mathematical reason for this, or is it simply notation?
radicals
asked Dec 28 '18 at 17:07
John GaltJohn Galt
185
$endgroup$
Why is it the case that an even root (square root, quartic, etc) can be positive or negative across the complex numbers, but is limited to postive in the reals? Is there a good mathematical reason for this, or is it simply notation?
radicals
radicals
asked Dec 28 '18 at 17:07
John GaltJohn Galt
185
asked Dec 28 '18 at 17:07
John GaltJohn Galt
185
edited Dec 28 '18 at 17:09
John Galt
asked Dec 28 '18 at 17:07
John GaltJohn Galt
185
asked Dec 28 '18 at 17:07
John GaltJohn Galt
185
asked Dec 28 '18 at 17:07
John GaltJohn Galt
185
185
$begingroup$
We have defined $sqrt a$ to be the principal root. Even in the complex numbers there is a principal root.
$endgroup$
– Doug M
Dec 28 '18 at 20:29
add a comment |
$begingroup$
We have defined $sqrt a$ to be the principal root. Even in the complex numbers there is a principal root.
$endgroup$
– Doug M
Dec 28 '18 at 20:29
$begingroup$
We have defined $sqrt a$ to be the principal root. Even in the complex numbers there is a principal root.
$endgroup$
– Doug M
Dec 28 '18 at 20:29
$begingroup$
We have defined $sqrt a$ to be the principal root. Even in the complex numbers there is a principal root.
$endgroup$
– Doug M
Dec 28 '18 at 20:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Essentially, the reason is that complex multiplication, and by extension exponentiation, are far more complicated than in the reals. There is a rotational aspect, and for an $n$th root, there are $n$ answers. It would be silly to declare that only one of these is the "actual" answer, whereas in the reals it makes much more sense. Glad to see that you have joined the site. Please continue asking questions.
answered Dec 28 '18 at 17:12
William GrannisWilliam Grannis
1,008521
$endgroup$
add a comment |
$begingroup$
Adding to William Grannis' answer about it not making sense to choose "the actual root": here's another reason why.
For positive $x$, we can easily distinguish between $sqrt x$ and $-sqrt x$, and define $sqrt x$ to be the positive one.
Now think about
$$i^2=-1$$
This has two solutions. We "choose" one of them to be $i$, and call the other one $-i$, but there's absolutely no way to say which one we "chose". $i$ and $-i$ have completely identical properties, other than one being the negative of the other.
That is: if we said "Let's define $j=-i$, and rewrite all our mathematics in terms of $j$", nothing would change except for the name of the imaginary unit.
I've always found this a slightly alarming feature of the complex number system.
But alarming or not: it makes no sense to prefer a "positive imaginary" value to a "negative imaginary" one, since ultimately it's arbitrary which is which.
answered Dec 28 '18 at 20:22
timtfjtimtfj
2,468420
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055089%2froots-across-the-complex-numbers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Essentially, the reason is that complex multiplication, and by extension exponentiation, are far more complicated than in the reals. There is a rotational aspect, and for an $n$th root, there are $n$ answers. It would be silly to declare that only one of these is the "actual" answer, whereas in the reals it makes much more sense. Glad to see that you have joined the site. Please continue asking questions.
answered Dec 28 '18 at 17:12
William GrannisWilliam Grannis
1,008521
$endgroup$
add a comment |
$begingroup$
Essentially, the reason is that complex multiplication, and by extension exponentiation, are far more complicated than in the reals. There is a rotational aspect, and for an $n$th root, there are $n$ answers. It would be silly to declare that only one of these is the "actual" answer, whereas in the reals it makes much more sense. Glad to see that you have joined the site. Please continue asking questions.
answered Dec 28 '18 at 17:12
William GrannisWilliam Grannis
1,008521
$endgroup$
add a comment |
$begingroup$
Essentially, the reason is that complex multiplication, and by extension exponentiation, are far more complicated than in the reals. There is a rotational aspect, and for an $n$th root, there are $n$ answers. It would be silly to declare that only one of these is the "actual" answer, whereas in the reals it makes much more sense. Glad to see that you have joined the site. Please continue asking questions.
answered Dec 28 '18 at 17:12
William GrannisWilliam Grannis
1,008521
$endgroup$
Essentially, the reason is that complex multiplication, and by extension exponentiation, are far more complicated than in the reals. There is a rotational aspect, and for an $n$th root, there are $n$ answers. It would be silly to declare that only one of these is the "actual" answer, whereas in the reals it makes much more sense. Glad to see that you have joined the site. Please continue asking questions.
answered Dec 28 '18 at 17:12
William GrannisWilliam Grannis
1,008521
answered Dec 28 '18 at 17:12
William GrannisWilliam Grannis
1,008521
answered Dec 28 '18 at 17:12
William GrannisWilliam Grannis
1,008521
answered Dec 28 '18 at 17:12
William GrannisWilliam Grannis
1,008521
1,008521
add a comment |
add a comment |
$begingroup$
Adding to William Grannis' answer about it not making sense to choose "the actual root": here's another reason why.
For positive $x$, we can easily distinguish between $sqrt x$ and $-sqrt x$, and define $sqrt x$ to be the positive one.
Now think about
$$i^2=-1$$
This has two solutions. We "choose" one of them to be $i$, and call the other one $-i$, but there's absolutely no way to say which one we "chose". $i$ and $-i$ have completely identical properties, other than one being the negative of the other.
That is: if we said "Let's define $j=-i$, and rewrite all our mathematics in terms of $j$", nothing would change except for the name of the imaginary unit.
I've always found this a slightly alarming feature of the complex number system.
But alarming or not: it makes no sense to prefer a "positive imaginary" value to a "negative imaginary" one, since ultimately it's arbitrary which is which.
answered Dec 28 '18 at 20:22
timtfjtimtfj
2,468420
$endgroup$
add a comment |
$begingroup$
Adding to William Grannis' answer about it not making sense to choose "the actual root": here's another reason why.
For positive $x$, we can easily distinguish between $sqrt x$ and $-sqrt x$, and define $sqrt x$ to be the positive one.
Now think about
$$i^2=-1$$
This has two solutions. We "choose" one of them to be $i$, and call the other one $-i$, but there's absolutely no way to say which one we "chose". $i$ and $-i$ have completely identical properties, other than one being the negative of the other.
That is: if we said "Let's define $j=-i$, and rewrite all our mathematics in terms of $j$", nothing would change except for the name of the imaginary unit.
I've always found this a slightly alarming feature of the complex number system.
But alarming or not: it makes no sense to prefer a "positive imaginary" value to a "negative imaginary" one, since ultimately it's arbitrary which is which.
answered Dec 28 '18 at 20:22
timtfjtimtfj
2,468420
$endgroup$
add a comment |
$begingroup$
Adding to William Grannis' answer about it not making sense to choose "the actual root": here's another reason why.
For positive $x$, we can easily distinguish between $sqrt x$ and $-sqrt x$, and define $sqrt x$ to be the positive one.
Now think about
$$i^2=-1$$
This has two solutions. We "choose" one of them to be $i$, and call the other one $-i$, but there's absolutely no way to say which one we "chose". $i$ and $-i$ have completely identical properties, other than one being the negative of the other.
That is: if we said "Let's define $j=-i$, and rewrite all our mathematics in terms of $j$", nothing would change except for the name of the imaginary unit.
I've always found this a slightly alarming feature of the complex number system.
But alarming or not: it makes no sense to prefer a "positive imaginary" value to a "negative imaginary" one, since ultimately it's arbitrary which is which.
answered Dec 28 '18 at 20:22
timtfjtimtfj
2,468420
$endgroup$
Adding to William Grannis' answer about it not making sense to choose "the actual root": here's another reason why.
For positive $x$, we can easily distinguish between $sqrt x$ and $-sqrt x$, and define $sqrt x$ to be the positive one.
Now think about
$$i^2=-1$$
This has two solutions. We "choose" one of them to be $i$, and call the other one $-i$, but there's absolutely no way to say which one we "chose". $i$ and $-i$ have completely identical properties, other than one being the negative of the other.
That is: if we said "Let's define $j=-i$, and rewrite all our mathematics in terms of $j$", nothing would change except for the name of the imaginary unit.
I've always found this a slightly alarming feature of the complex number system.
But alarming or not: it makes no sense to prefer a "positive imaginary" value to a "negative imaginary" one, since ultimately it's arbitrary which is which.
answered Dec 28 '18 at 20:22
timtfjtimtfj
2,468420
edited Dec 28 '18 at 20:32
answered Dec 28 '18 at 20:22
timtfjtimtfj
2,468420
answered Dec 28 '18 at 20:22
timtfjtimtfj
2,468420
answered Dec 28 '18 at 20:22
timtfjtimtfj
2,468420
2,468420
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055089%2froots-across-the-complex-numbers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
We have defined $sqrt a$ to be the principal root. Even in the complex numbers there is a principal root.
$endgroup$
– Doug M
Dec 28 '18 at 20:29