Function that halves the angle of a complex point
3
$begingroup$
How would a function mapping a complex point $z=re^{itheta}$ to $re^{ifrac{theta}{2}}$ be correctly written?
functions
asked Dec 12 '18 at 20:18
MatthewMatthew
316
$endgroup$
add a comment |
3
$begingroup$
How would a function mapping a complex point $z=re^{itheta}$ to $re^{ifrac{theta}{2}}$ be correctly written?
functions
asked Dec 12 '18 at 20:18
MatthewMatthew
316
$endgroup$
add a comment |
3
3
3
$begingroup$
How would a function mapping a complex point $z=re^{itheta}$ to $re^{ifrac{theta}{2}}$ be correctly written?
functions
asked Dec 12 '18 at 20:18
MatthewMatthew
316
$endgroup$
How would a function mapping a complex point $z=re^{itheta}$ to $re^{ifrac{theta}{2}}$ be correctly written?
functions
functions
asked Dec 12 '18 at 20:18
MatthewMatthew
316
asked Dec 12 '18 at 20:18
MatthewMatthew
316
asked Dec 12 '18 at 20:18
MatthewMatthew
316
asked Dec 12 '18 at 20:18
MatthewMatthew
316
asked Dec 12 '18 at 20:18
MatthewMatthew
316
316
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
5
$begingroup$
note that $|z|=r$, so I think the answer you are looking for (and you can test it yourself) is
$$f(z)=|z|cdot left( frac{z}{|z|}right)^{frac{1}{2}}$$
answered Dec 12 '18 at 20:23
NazimJNazimJ
36317
$endgroup$
$begingroup$
Ok, so there is no easy way to get the angle of the complex point? I have seen somewhere that $Arg~z$ might be used as a notation for that, is that correct?
$endgroup$
– Matthew
Dec 12 '18 at 20:29
$begingroup$
This can be more succinctly written as $(|z|z)^{1/2}$.
$endgroup$
– eyeballfrog
Dec 12 '18 at 20:36
$begingroup$
@Matthew Yes, it's correct. To get the angle of $a+bi$ from the real axis, just draw a diagram and notice that $frac{b}{a}$ is its tangent.
$endgroup$
– timtfj
Dec 12 '18 at 20:36
$begingroup$
Yes I just wrote it out long way's so order of operations tells the story of what's happening. Also yes, $theta = tan^{-1}(frac{b}{a})$
$endgroup$
– NazimJ
Dec 12 '18 at 20:38
$begingroup$
Is it possible to define a function as $f(a+bi)=tan^{-1}(frac{b}{a})$, or is that incorrect notation?
$endgroup$
– Matthew
Dec 12 '18 at 20:42
|
show 2 more comments
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%2f3037176%2ffunction-that-halves-the-angle-of-a-complex-point%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
5
$begingroup$
note that $|z|=r$, so I think the answer you are looking for (and you can test it yourself) is
$$f(z)=|z|cdot left( frac{z}{|z|}right)^{frac{1}{2}}$$
answered Dec 12 '18 at 20:23
NazimJNazimJ
36317
$endgroup$
$begingroup$
Ok, so there is no easy way to get the angle of the complex point? I have seen somewhere that $Arg~z$ might be used as a notation for that, is that correct?
$endgroup$
– Matthew
Dec 12 '18 at 20:29
$begingroup$
This can be more succinctly written as $(|z|z)^{1/2}$.
$endgroup$
– eyeballfrog
Dec 12 '18 at 20:36
$begingroup$
@Matthew Yes, it's correct. To get the angle of $a+bi$ from the real axis, just draw a diagram and notice that $frac{b}{a}$ is its tangent.
$endgroup$
– timtfj
Dec 12 '18 at 20:36
$begingroup$
Yes I just wrote it out long way's so order of operations tells the story of what's happening. Also yes, $theta = tan^{-1}(frac{b}{a})$
$endgroup$
– NazimJ
Dec 12 '18 at 20:38
$begingroup$
Is it possible to define a function as $f(a+bi)=tan^{-1}(frac{b}{a})$, or is that incorrect notation?
$endgroup$
– Matthew
Dec 12 '18 at 20:42
|
show 2 more comments
5
$begingroup$
note that $|z|=r$, so I think the answer you are looking for (and you can test it yourself) is
$$f(z)=|z|cdot left( frac{z}{|z|}right)^{frac{1}{2}}$$
answered Dec 12 '18 at 20:23
NazimJNazimJ
36317
$endgroup$
$begingroup$
Ok, so there is no easy way to get the angle of the complex point? I have seen somewhere that $Arg~z$ might be used as a notation for that, is that correct?
$endgroup$
– Matthew
Dec 12 '18 at 20:29
$begingroup$
This can be more succinctly written as $(|z|z)^{1/2}$.
$endgroup$
– eyeballfrog
Dec 12 '18 at 20:36
$begingroup$
@Matthew Yes, it's correct. To get the angle of $a+bi$ from the real axis, just draw a diagram and notice that $frac{b}{a}$ is its tangent.
$endgroup$
– timtfj
Dec 12 '18 at 20:36
$begingroup$
Yes I just wrote it out long way's so order of operations tells the story of what's happening. Also yes, $theta = tan^{-1}(frac{b}{a})$
$endgroup$
– NazimJ
Dec 12 '18 at 20:38
$begingroup$
Is it possible to define a function as $f(a+bi)=tan^{-1}(frac{b}{a})$, or is that incorrect notation?
$endgroup$
– Matthew
Dec 12 '18 at 20:42
|
show 2 more comments
5
5
5
$begingroup$
note that $|z|=r$, so I think the answer you are looking for (and you can test it yourself) is
$$f(z)=|z|cdot left( frac{z}{|z|}right)^{frac{1}{2}}$$
answered Dec 12 '18 at 20:23
NazimJNazimJ
36317
$endgroup$
note that $|z|=r$, so I think the answer you are looking for (and you can test it yourself) is
$$f(z)=|z|cdot left( frac{z}{|z|}right)^{frac{1}{2}}$$
answered Dec 12 '18 at 20:23
NazimJNazimJ
36317
answered Dec 12 '18 at 20:23
NazimJNazimJ
36317
answered Dec 12 '18 at 20:23
NazimJNazimJ
36317
answered Dec 12 '18 at 20:23
NazimJNazimJ
36317
36317
$begingroup$
Ok, so there is no easy way to get the angle of the complex point? I have seen somewhere that $Arg~z$ might be used as a notation for that, is that correct?
$endgroup$
– Matthew
Dec 12 '18 at 20:29
$begingroup$
This can be more succinctly written as $(|z|z)^{1/2}$.
$endgroup$
– eyeballfrog
Dec 12 '18 at 20:36
$begingroup$
@Matthew Yes, it's correct. To get the angle of $a+bi$ from the real axis, just draw a diagram and notice that $frac{b}{a}$ is its tangent.
$endgroup$
– timtfj
Dec 12 '18 at 20:36
$begingroup$
Yes I just wrote it out long way's so order of operations tells the story of what's happening. Also yes, $theta = tan^{-1}(frac{b}{a})$
$endgroup$
– NazimJ
Dec 12 '18 at 20:38
$begingroup$
Is it possible to define a function as $f(a+bi)=tan^{-1}(frac{b}{a})$, or is that incorrect notation?
$endgroup$
– Matthew
Dec 12 '18 at 20:42
|
show 2 more comments
$begingroup$
Ok, so there is no easy way to get the angle of the complex point? I have seen somewhere that $Arg~z$ might be used as a notation for that, is that correct?
$endgroup$
– Matthew
Dec 12 '18 at 20:29
$begingroup$
This can be more succinctly written as $(|z|z)^{1/2}$.
$endgroup$
– eyeballfrog
Dec 12 '18 at 20:36
$begingroup$
@Matthew Yes, it's correct. To get the angle of $a+bi$ from the real axis, just draw a diagram and notice that $frac{b}{a}$ is its tangent.
$endgroup$
– timtfj
Dec 12 '18 at 20:36
$begingroup$
Yes I just wrote it out long way's so order of operations tells the story of what's happening. Also yes, $theta = tan^{-1}(frac{b}{a})$
$endgroup$
– NazimJ
Dec 12 '18 at 20:38
$begingroup$
Is it possible to define a function as $f(a+bi)=tan^{-1}(frac{b}{a})$, or is that incorrect notation?
$endgroup$
– Matthew
Dec 12 '18 at 20:42
$begingroup$
Ok, so there is no easy way to get the angle of the complex point? I have seen somewhere that $Arg~z$ might be used as a notation for that, is that correct?
$endgroup$
– Matthew
Dec 12 '18 at 20:29
$begingroup$
Ok, so there is no easy way to get the angle of the complex point? I have seen somewhere that $Arg~z$ might be used as a notation for that, is that correct?
$endgroup$
– Matthew
Dec 12 '18 at 20:29
$begingroup$
This can be more succinctly written as $(|z|z)^{1/2}$.
$endgroup$
– eyeballfrog
Dec 12 '18 at 20:36
$begingroup$
This can be more succinctly written as $(|z|z)^{1/2}$.
$endgroup$
– eyeballfrog
Dec 12 '18 at 20:36
$begingroup$
@Matthew Yes, it's correct. To get the angle of $a+bi$ from the real axis, just draw a diagram and notice that $frac{b}{a}$ is its tangent.
$endgroup$
– timtfj
Dec 12 '18 at 20:36
$begingroup$
@Matthew Yes, it's correct. To get the angle of $a+bi$ from the real axis, just draw a diagram and notice that $frac{b}{a}$ is its tangent.
$endgroup$
– timtfj
Dec 12 '18 at 20:36
$begingroup$
Yes I just wrote it out long way's so order of operations tells the story of what's happening. Also yes, $theta = tan^{-1}(frac{b}{a})$
$endgroup$
– NazimJ
Dec 12 '18 at 20:38
$begingroup$
Yes I just wrote it out long way's so order of operations tells the story of what's happening. Also yes, $theta = tan^{-1}(frac{b}{a})$
$endgroup$
– NazimJ
Dec 12 '18 at 20:38
$begingroup$
Is it possible to define a function as $f(a+bi)=tan^{-1}(frac{b}{a})$, or is that incorrect notation?
$endgroup$
– Matthew
Dec 12 '18 at 20:42
$begingroup$
Is it possible to define a function as $f(a+bi)=tan^{-1}(frac{b}{a})$, or is that incorrect notation?
$endgroup$
– Matthew
Dec 12 '18 at 20:42
|
show 2 more comments
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%2f3037176%2ffunction-that-halves-the-angle-of-a-complex-point%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
