Mixing
Why $|sqrt{a}-i|=sqrt{a+1}$ for every nonnegative real number $a$?
0
$begingroup$
Why $|sqrt{a}-i|=sqrt{a+1}$ for every nonnegative real number $a$?
I know that $i=sqrt{-1}$, and in my textbook I get $$|sqrt{x^2+(y-1)^2}-i|=sqrt{(x^2+(y-1)^2)^2+(-1)^2}=sqrt{x^2+(y-1)^2+1}$$
I can't find how the first two passages are connected.
Edit: corrected equation.
complex-numbers
asked Jan 10 at 16:45
KevinKevin
13311
$endgroup$
|
show 2 more comments
0
$begingroup$
Why $|sqrt{a}-i|=sqrt{a+1}$ for every nonnegative real number $a$?
I know that $i=sqrt{-1}$, and in my textbook I get $$|sqrt{x^2+(y-1)^2}-i|=sqrt{(x^2+(y-1)^2)^2+(-1)^2}=sqrt{x^2+(y-1)^2+1}$$
I can't find how the first two passages are connected.
Edit: corrected equation.
complex-numbers
asked Jan 10 at 16:45
KevinKevin
13311
$endgroup$
$begingroup$
I don't get it. Let $x=2$ and $y=1$ to see that this is false.
$endgroup$
– Randall
Jan 10 at 16:47
1
$begingroup$
They are not... Take $a=1$. I highly doubt $1-i=sqrt 2$...
$endgroup$
– Martigan
Jan 10 at 16:48
1
$begingroup$
For every nonnegative real number $a$, $$|sqrt a-i|^2=(sqrt a-i)overline{(sqrt a+i)}=(sqrt a-i)(sqrt a+i)$$ hence $$|sqrt a-i|^2=a-i^2=a+1$$ which is equivalent to the desired identity $$|sqrt a-i|=sqrt{a+1}$$
$endgroup$
– Did
Jan 10 at 16:50
$begingroup$
I'm sorry, I corrected the equation.
$endgroup$
– Kevin
Jan 10 at 16:52
$begingroup$
Thank you @Did!
$endgroup$
– Kevin
Jan 10 at 17:01
|
show 2 more comments
0
0
0
$begingroup$
Why $|sqrt{a}-i|=sqrt{a+1}$ for every nonnegative real number $a$?
I know that $i=sqrt{-1}$, and in my textbook I get $$|sqrt{x^2+(y-1)^2}-i|=sqrt{(x^2+(y-1)^2)^2+(-1)^2}=sqrt{x^2+(y-1)^2+1}$$
I can't find how the first two passages are connected.
Edit: corrected equation.
complex-numbers
asked Jan 10 at 16:45
KevinKevin
13311
$endgroup$
Why $|sqrt{a}-i|=sqrt{a+1}$ for every nonnegative real number $a$?
I know that $i=sqrt{-1}$, and in my textbook I get $$|sqrt{x^2+(y-1)^2}-i|=sqrt{(x^2+(y-1)^2)^2+(-1)^2}=sqrt{x^2+(y-1)^2+1}$$
I can't find how the first two passages are connected.
Edit: corrected equation.
complex-numbers
complex-numbers
asked Jan 10 at 16:45
KevinKevin
13311
asked Jan 10 at 16:45
KevinKevin
13311
edited Jan 10 at 16:51
Kevin
asked Jan 10 at 16:45
KevinKevin
13311
asked Jan 10 at 16:45
KevinKevin
13311
asked Jan 10 at 16:45
KevinKevin
13311
13311
$begingroup$
I don't get it. Let $x=2$ and $y=1$ to see that this is false.
$endgroup$
– Randall
Jan 10 at 16:47
1
$begingroup$
They are not... Take $a=1$. I highly doubt $1-i=sqrt 2$...
$endgroup$
– Martigan
Jan 10 at 16:48
1
$begingroup$
For every nonnegative real number $a$, $$|sqrt a-i|^2=(sqrt a-i)overline{(sqrt a+i)}=(sqrt a-i)(sqrt a+i)$$ hence $$|sqrt a-i|^2=a-i^2=a+1$$ which is equivalent to the desired identity $$|sqrt a-i|=sqrt{a+1}$$
$endgroup$
– Did
Jan 10 at 16:50
$begingroup$
I'm sorry, I corrected the equation.
$endgroup$
– Kevin
Jan 10 at 16:52
$begingroup$
Thank you @Did!
$endgroup$
– Kevin
Jan 10 at 17:01
|
show 2 more comments
$begingroup$
I don't get it. Let $x=2$ and $y=1$ to see that this is false.
$endgroup$
– Randall
Jan 10 at 16:47
1
$begingroup$
They are not... Take $a=1$. I highly doubt $1-i=sqrt 2$...
$endgroup$
– Martigan
Jan 10 at 16:48
1
$begingroup$
For every nonnegative real number $a$, $$|sqrt a-i|^2=(sqrt a-i)overline{(sqrt a+i)}=(sqrt a-i)(sqrt a+i)$$ hence $$|sqrt a-i|^2=a-i^2=a+1$$ which is equivalent to the desired identity $$|sqrt a-i|=sqrt{a+1}$$
$endgroup$
– Did
Jan 10 at 16:50
$begingroup$
I'm sorry, I corrected the equation.
$endgroup$
– Kevin
Jan 10 at 16:52
$begingroup$
Thank you @Did!
$endgroup$
– Kevin
Jan 10 at 17:01
$begingroup$
I don't get it. Let $x=2$ and $y=1$ to see that this is false.
$endgroup$
– Randall
Jan 10 at 16:47
$begingroup$
I don't get it. Let $x=2$ and $y=1$ to see that this is false.
$endgroup$
– Randall
Jan 10 at 16:47
1
1
$begingroup$
They are not... Take $a=1$. I highly doubt $1-i=sqrt 2$...
$endgroup$
– Martigan
Jan 10 at 16:48
$begingroup$
They are not... Take $a=1$. I highly doubt $1-i=sqrt 2$...
$endgroup$
– Martigan
Jan 10 at 16:48
1
1
$begingroup$
For every nonnegative real number $a$, $$|sqrt a-i|^2=(sqrt a-i)overline{(sqrt a+i)}=(sqrt a-i)(sqrt a+i)$$ hence $$|sqrt a-i|^2=a-i^2=a+1$$ which is equivalent to the desired identity $$|sqrt a-i|=sqrt{a+1}$$
$endgroup$
– Did
Jan 10 at 16:50
$begingroup$
For every nonnegative real number $a$, $$|sqrt a-i|^2=(sqrt a-i)overline{(sqrt a+i)}=(sqrt a-i)(sqrt a+i)$$ hence $$|sqrt a-i|^2=a-i^2=a+1$$ which is equivalent to the desired identity $$|sqrt a-i|=sqrt{a+1}$$
$endgroup$
– Did
Jan 10 at 16:50
$begingroup$
I'm sorry, I corrected the equation.
$endgroup$
– Kevin
Jan 10 at 16:52
$begingroup$
I'm sorry, I corrected the equation.
$endgroup$
– Kevin
Jan 10 at 16:52
$begingroup$
Thank you @Did!
$endgroup$
– Kevin
Jan 10 at 17:01
$begingroup$
Thank you @Did!
$endgroup$
– Kevin
Jan 10 at 17:01
|
show 2 more comments
0
active
oldest
votes
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
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%2f3068881%2fwhy-sqrta-i-sqrta1-for-every-nonnegative-real-number-a%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
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%2f3068881%2fwhy-sqrta-i-sqrta1-for-every-nonnegative-real-number-a%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
$begingroup$
I don't get it. Let $x=2$ and $y=1$ to see that this is false.
$endgroup$
– Randall
Jan 10 at 16:47
1
$begingroup$
They are not... Take $a=1$. I highly doubt $1-i=sqrt 2$...
$endgroup$
– Martigan
Jan 10 at 16:48
1
$begingroup$
For every nonnegative real number $a$, $$|sqrt a-i|^2=(sqrt a-i)overline{(sqrt a+i)}=(sqrt a-i)(sqrt a+i)$$ hence $$|sqrt a-i|^2=a-i^2=a+1$$ which is equivalent to the desired identity $$|sqrt a-i|=sqrt{a+1}$$
$endgroup$
– Did
Jan 10 at 16:50
$begingroup$
I'm sorry, I corrected the equation.
$endgroup$
– Kevin
Jan 10 at 16:52
$begingroup$
Thank you @Did!
$endgroup$
– Kevin
Jan 10 at 17:01