Absolute value of a complex expression
0
$begingroup$
I've been stuck on this problem for a while:
$$vert 1-e^{-i2pi f}+.5e^{-i2pi fcdot 2}vert^2,$$
where $i =$ the imaginary unit, $(2pi f) =$ a real value, and $(2pi fcdot 2)=$ a real value.
I just don't know how to begin.
calculus algebra-precalculus exponentiation polar-coordinates absolute-value
edited Jan 4 at 17:58
Namaste
1
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
$endgroup$
|
show 3 more comments
0
$begingroup$
I've been stuck on this problem for a while:
$$vert 1-e^{-i2pi f}+.5e^{-i2pi fcdot 2}vert^2,$$
where $i =$ the imaginary unit, $(2pi f) =$ a real value, and $(2pi fcdot 2)=$ a real value.
I just don't know how to begin.
calculus algebra-precalculus exponentiation polar-coordinates absolute-value
edited Jan 4 at 17:58
Namaste
1
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
$endgroup$
2
$begingroup$
First off, what have you tried? Second, the notation is a little unclear. Some clarification would be helpful. And third, what exactly do you want out of this? This is not a polynomial, unlike what your title suggests.
$endgroup$
– Don Thousand
Jan 4 at 17:14
$begingroup$
Hello LowEnergy, why is the expression in latex so huge? :)
$endgroup$
– Ixion
Jan 4 at 17:24
$begingroup$
i = imaginary, (2*pi*f) = real value
$endgroup$
– LowEnergy
Jan 4 at 17:26
$begingroup$
So it's the absolute value of a complex number....
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 17:28
$begingroup$
Yeah, I should've probably written complex value in the title.
$endgroup$
– LowEnergy
Jan 4 at 17:32
|
show 3 more comments
0
0
0
$begingroup$
I've been stuck on this problem for a while:
$$vert 1-e^{-i2pi f}+.5e^{-i2pi fcdot 2}vert^2,$$
where $i =$ the imaginary unit, $(2pi f) =$ a real value, and $(2pi fcdot 2)=$ a real value.
I just don't know how to begin.
calculus algebra-precalculus exponentiation polar-coordinates absolute-value
edited Jan 4 at 17:58
Namaste
1
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
$endgroup$
I've been stuck on this problem for a while:
$$vert 1-e^{-i2pi f}+.5e^{-i2pi fcdot 2}vert^2,$$
where $i =$ the imaginary unit, $(2pi f) =$ a real value, and $(2pi fcdot 2)=$ a real value.
I just don't know how to begin.
calculus algebra-precalculus exponentiation polar-coordinates absolute-value
calculus algebra-precalculus exponentiation polar-coordinates absolute-value
edited Jan 4 at 17:58
Namaste
1
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
edited Jan 4 at 17:58
Namaste
1
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
edited Jan 4 at 17:58
Namaste
1
edited Jan 4 at 17:58
Namaste
1
edited Jan 4 at 17:58
Namaste
1
1
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
11
2
$begingroup$
First off, what have you tried? Second, the notation is a little unclear. Some clarification would be helpful. And third, what exactly do you want out of this? This is not a polynomial, unlike what your title suggests.
$endgroup$
– Don Thousand
Jan 4 at 17:14
$begingroup$
Hello LowEnergy, why is the expression in latex so huge? :)
$endgroup$
– Ixion
Jan 4 at 17:24
$begingroup$
i = imaginary, (2*pi*f) = real value
$endgroup$
– LowEnergy
Jan 4 at 17:26
$begingroup$
So it's the absolute value of a complex number....
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 17:28
$begingroup$
Yeah, I should've probably written complex value in the title.
$endgroup$
– LowEnergy
Jan 4 at 17:32
|
show 3 more comments
2
$begingroup$
First off, what have you tried? Second, the notation is a little unclear. Some clarification would be helpful. And third, what exactly do you want out of this? This is not a polynomial, unlike what your title suggests.
$endgroup$
– Don Thousand
Jan 4 at 17:14
$begingroup$
Hello LowEnergy, why is the expression in latex so huge? :)
$endgroup$
– Ixion
Jan 4 at 17:24
$begingroup$
i = imaginary, (2*pi*f) = real value
$endgroup$
– LowEnergy
Jan 4 at 17:26
$begingroup$
So it's the absolute value of a complex number....
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 17:28
$begingroup$
Yeah, I should've probably written complex value in the title.
$endgroup$
– LowEnergy
Jan 4 at 17:32
2
2
$begingroup$
First off, what have you tried? Second, the notation is a little unclear. Some clarification would be helpful. And third, what exactly do you want out of this? This is not a polynomial, unlike what your title suggests.
$endgroup$
– Don Thousand
Jan 4 at 17:14
$begingroup$
First off, what have you tried? Second, the notation is a little unclear. Some clarification would be helpful. And third, what exactly do you want out of this? This is not a polynomial, unlike what your title suggests.
$endgroup$
– Don Thousand
Jan 4 at 17:14
$begingroup$
Hello LowEnergy, why is the expression in latex so huge? :)
$endgroup$
– Ixion
Jan 4 at 17:24
$begingroup$
Hello LowEnergy, why is the expression in latex so huge? :)
$endgroup$
– Ixion
Jan 4 at 17:24
$begingroup$
i = imaginary, (2*pi*f) = real value
$endgroup$
– LowEnergy
Jan 4 at 17:26
$begingroup$
i = imaginary, (2*pi*f) = real value
$endgroup$
– LowEnergy
Jan 4 at 17:26
$begingroup$
So it's the absolute value of a complex number....
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 17:28
$begingroup$
So it's the absolute value of a complex number....
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 17:28
$begingroup$
Yeah, I should've probably written complex value in the title.
$endgroup$
– LowEnergy
Jan 4 at 17:32
$begingroup$
Yeah, I should've probably written complex value in the title.
$endgroup$
– LowEnergy
Jan 4 at 17:32
|
show 3 more comments
1 Answer
1
active
oldest
votes
0
$begingroup$
Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$
answered Jan 4 at 17:46
EnnarEnnar
14.8k32445
$endgroup$
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
});
}
});
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%2f3061840%2fabsolute-value-of-a-complex-expression%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
0
$begingroup$
Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$
answered Jan 4 at 17:46
EnnarEnnar
14.8k32445
$endgroup$
add a comment |
0
$begingroup$
Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$
answered Jan 4 at 17:46
EnnarEnnar
14.8k32445
$endgroup$
add a comment |
0
0
0
$begingroup$
Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$
answered Jan 4 at 17:46
EnnarEnnar
14.8k32445
$endgroup$
Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$
answered Jan 4 at 17:46
EnnarEnnar
14.8k32445
answered Jan 4 at 17:46
EnnarEnnar
14.8k32445
answered Jan 4 at 17:46
EnnarEnnar
14.8k32445
answered Jan 4 at 17:46
EnnarEnnar
14.8k32445
14.8k32445
add a comment |
add a comment |
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%2f3061840%2fabsolute-value-of-a-complex-expression%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
2
$begingroup$
First off, what have you tried? Second, the notation is a little unclear. Some clarification would be helpful. And third, what exactly do you want out of this? This is not a polynomial, unlike what your title suggests.
$endgroup$
– Don Thousand
Jan 4 at 17:14
$begingroup$
Hello LowEnergy, why is the expression in latex so huge? :)
$endgroup$
– Ixion
Jan 4 at 17:24
$begingroup$
i = imaginary, (2*pi*f) = real value
$endgroup$
– LowEnergy
Jan 4 at 17:26
$begingroup$
So it's the absolute value of a complex number....
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 17:28
$begingroup$
Yeah, I should've probably written complex value in the title.
$endgroup$
– LowEnergy
Jan 4 at 17:32