Determine all solutions to $operatorname{Log}(z^2) = operatorname{Log}(z)$ (Complex Equation)
$begingroup$
I am new to complex analysis and i was trying to solve the given exercise
Determine all solutions to $operatorname{Log}(z^2) = operatorname{Log}(z)$
However, i am not sure whether my attempt is correct or complete.
My attempt:
It holds $$operatorname{Log}(z^2) = ln( vert z vert^2) + ioperatorname{Arg}(z^2) = 2ln(vert z vert) + 2ioperatorname{Arg}(z) $$
therefore
$$operatorname{Log}(z^2) = operatorname{Log}(z)$$ is true for all $z in mathbb{C}$ for which
$$2ln(vert z vert) + 2ioperatorname{Arg}(z) = ln(vert z vert) + ioperatorname{Arg}(z)$$
$$Leftrightarrow ln(vert z vert) + ioperatorname{Arg}(z) = 0$$
This is how far i got. Can i somehow solve this equation?
Thanks for any hints!
complex-analysis complex-numbers logarithms
asked Jan 8 at 5:49
ZestZest
32111
$endgroup$
add a comment |
$begingroup$
I am new to complex analysis and i was trying to solve the given exercise
Determine all solutions to $operatorname{Log}(z^2) = operatorname{Log}(z)$
However, i am not sure whether my attempt is correct or complete.
My attempt:
It holds $$operatorname{Log}(z^2) = ln( vert z vert^2) + ioperatorname{Arg}(z^2) = 2ln(vert z vert) + 2ioperatorname{Arg}(z) $$
therefore
$$operatorname{Log}(z^2) = operatorname{Log}(z)$$ is true for all $z in mathbb{C}$ for which
$$2ln(vert z vert) + 2ioperatorname{Arg}(z) = ln(vert z vert) + ioperatorname{Arg}(z)$$
$$Leftrightarrow ln(vert z vert) + ioperatorname{Arg}(z) = 0$$
This is how far i got. Can i somehow solve this equation?
Thanks for any hints!
complex-analysis complex-numbers logarithms
asked Jan 8 at 5:49
ZestZest
32111
$endgroup$
add a comment |
$begingroup$
I am new to complex analysis and i was trying to solve the given exercise
Determine all solutions to $operatorname{Log}(z^2) = operatorname{Log}(z)$
However, i am not sure whether my attempt is correct or complete.
My attempt:
It holds $$operatorname{Log}(z^2) = ln( vert z vert^2) + ioperatorname{Arg}(z^2) = 2ln(vert z vert) + 2ioperatorname{Arg}(z) $$
therefore
$$operatorname{Log}(z^2) = operatorname{Log}(z)$$ is true for all $z in mathbb{C}$ for which
$$2ln(vert z vert) + 2ioperatorname{Arg}(z) = ln(vert z vert) + ioperatorname{Arg}(z)$$
$$Leftrightarrow ln(vert z vert) + ioperatorname{Arg}(z) = 0$$
This is how far i got. Can i somehow solve this equation?
Thanks for any hints!
complex-analysis complex-numbers logarithms
asked Jan 8 at 5:49
ZestZest
32111
$endgroup$
I am new to complex analysis and i was trying to solve the given exercise
Determine all solutions to $operatorname{Log}(z^2) = operatorname{Log}(z)$
However, i am not sure whether my attempt is correct or complete.
My attempt:
It holds $$operatorname{Log}(z^2) = ln( vert z vert^2) + ioperatorname{Arg}(z^2) = 2ln(vert z vert) + 2ioperatorname{Arg}(z) $$
therefore
$$operatorname{Log}(z^2) = operatorname{Log}(z)$$ is true for all $z in mathbb{C}$ for which
$$2ln(vert z vert) + 2ioperatorname{Arg}(z) = ln(vert z vert) + ioperatorname{Arg}(z)$$
$$Leftrightarrow ln(vert z vert) + ioperatorname{Arg}(z) = 0$$
This is how far i got. Can i somehow solve this equation?
Thanks for any hints!
complex-analysis complex-numbers logarithms
complex-analysis complex-numbers logarithms
asked Jan 8 at 5:49
ZestZest
32111
asked Jan 8 at 5:49
ZestZest
32111
asked Jan 8 at 5:49
ZestZest
32111
asked Jan 8 at 5:49
ZestZest
32111
asked Jan 8 at 5:49
ZestZest
32111
32111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Just put together the two conditions you have got right there and you're done.
The $ln$ term is the real part of that last expression, and $itext{Arg}$ is the imaginary part, so they must both be $0$. $ln |z|$ is $0$ when $|z| = 1$, so you know such a $z$ must live on the unit circle. However $i text{Arg} z = 0 implies text{Arg} z = 0$, which means that $z$ is actually a positive real number.
The only $z$ satisfying both conditions is $z = 1$.
answered Jan 8 at 6:02
Alfred YergerAlfred Yerger
10.3k2148
$endgroup$
$begingroup$
thank you very much for your help!
$endgroup$
– Zest
Jan 8 at 6:08
add a comment |
$begingroup$
If $z ne 0$ and $operatorname{Log}(z^2) = operatorname{Log}(z)$, then
$z^2=e^{operatorname{Log}(z^2)}=e^{operatorname{Log}(z)}=z$, hence $z=1$.
answered Jan 8 at 6:06
FredFred
44.5k1846
$endgroup$
$begingroup$
thank you very much for your additional hint, highly appreciating it.
$endgroup$
– Zest
Jan 8 at 6:07
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%2f3065840%2fdetermine-all-solutions-to-operatornamelogz2-operatornamelogz-co%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
$begingroup$
Just put together the two conditions you have got right there and you're done.
The $ln$ term is the real part of that last expression, and $itext{Arg}$ is the imaginary part, so they must both be $0$. $ln |z|$ is $0$ when $|z| = 1$, so you know such a $z$ must live on the unit circle. However $i text{Arg} z = 0 implies text{Arg} z = 0$, which means that $z$ is actually a positive real number.
The only $z$ satisfying both conditions is $z = 1$.
answered Jan 8 at 6:02
Alfred YergerAlfred Yerger
10.3k2148
$endgroup$
$begingroup$
thank you very much for your help!
$endgroup$
– Zest
Jan 8 at 6:08
add a comment |
$begingroup$
Just put together the two conditions you have got right there and you're done.
The $ln$ term is the real part of that last expression, and $itext{Arg}$ is the imaginary part, so they must both be $0$. $ln |z|$ is $0$ when $|z| = 1$, so you know such a $z$ must live on the unit circle. However $i text{Arg} z = 0 implies text{Arg} z = 0$, which means that $z$ is actually a positive real number.
The only $z$ satisfying both conditions is $z = 1$.
answered Jan 8 at 6:02
Alfred YergerAlfred Yerger
10.3k2148
$endgroup$
$begingroup$
thank you very much for your help!
$endgroup$
– Zest
Jan 8 at 6:08
add a comment |
$begingroup$
Just put together the two conditions you have got right there and you're done.
The $ln$ term is the real part of that last expression, and $itext{Arg}$ is the imaginary part, so they must both be $0$. $ln |z|$ is $0$ when $|z| = 1$, so you know such a $z$ must live on the unit circle. However $i text{Arg} z = 0 implies text{Arg} z = 0$, which means that $z$ is actually a positive real number.
The only $z$ satisfying both conditions is $z = 1$.
answered Jan 8 at 6:02
Alfred YergerAlfred Yerger
10.3k2148
$endgroup$
Just put together the two conditions you have got right there and you're done.
The $ln$ term is the real part of that last expression, and $itext{Arg}$ is the imaginary part, so they must both be $0$. $ln |z|$ is $0$ when $|z| = 1$, so you know such a $z$ must live on the unit circle. However $i text{Arg} z = 0 implies text{Arg} z = 0$, which means that $z$ is actually a positive real number.
The only $z$ satisfying both conditions is $z = 1$.
answered Jan 8 at 6:02
Alfred YergerAlfred Yerger
10.3k2148
answered Jan 8 at 6:02
Alfred YergerAlfred Yerger
10.3k2148
answered Jan 8 at 6:02
Alfred YergerAlfred Yerger
10.3k2148
answered Jan 8 at 6:02
Alfred YergerAlfred Yerger
10.3k2148
10.3k2148
$begingroup$
thank you very much for your help!
$endgroup$
– Zest
Jan 8 at 6:08
add a comment |
$begingroup$
thank you very much for your help!
$endgroup$
– Zest
Jan 8 at 6:08
$begingroup$
thank you very much for your help!
$endgroup$
– Zest
Jan 8 at 6:08
$begingroup$
thank you very much for your help!
$endgroup$
– Zest
Jan 8 at 6:08
add a comment |
$begingroup$
If $z ne 0$ and $operatorname{Log}(z^2) = operatorname{Log}(z)$, then
$z^2=e^{operatorname{Log}(z^2)}=e^{operatorname{Log}(z)}=z$, hence $z=1$.
answered Jan 8 at 6:06
FredFred
44.5k1846
$endgroup$
$begingroup$
thank you very much for your additional hint, highly appreciating it.
$endgroup$
– Zest
Jan 8 at 6:07
add a comment |
$begingroup$
If $z ne 0$ and $operatorname{Log}(z^2) = operatorname{Log}(z)$, then
$z^2=e^{operatorname{Log}(z^2)}=e^{operatorname{Log}(z)}=z$, hence $z=1$.
answered Jan 8 at 6:06
FredFred
44.5k1846
$endgroup$
$begingroup$
thank you very much for your additional hint, highly appreciating it.
$endgroup$
– Zest
Jan 8 at 6:07
add a comment |
$begingroup$
If $z ne 0$ and $operatorname{Log}(z^2) = operatorname{Log}(z)$, then
$z^2=e^{operatorname{Log}(z^2)}=e^{operatorname{Log}(z)}=z$, hence $z=1$.
answered Jan 8 at 6:06
FredFred
44.5k1846
$endgroup$
If $z ne 0$ and $operatorname{Log}(z^2) = operatorname{Log}(z)$, then
$z^2=e^{operatorname{Log}(z^2)}=e^{operatorname{Log}(z)}=z$, hence $z=1$.
answered Jan 8 at 6:06
FredFred
44.5k1846
answered Jan 8 at 6:06
FredFred
44.5k1846
answered Jan 8 at 6:06
FredFred
44.5k1846
answered Jan 8 at 6:06
FredFred
44.5k1846
44.5k1846
$begingroup$
thank you very much for your additional hint, highly appreciating it.
$endgroup$
– Zest
Jan 8 at 6:07
add a comment |
$begingroup$
thank you very much for your additional hint, highly appreciating it.
$endgroup$
– Zest
Jan 8 at 6:07
$begingroup$
thank you very much for your additional hint, highly appreciating it.
$endgroup$
– Zest
Jan 8 at 6:07
$begingroup$
thank you very much for your additional hint, highly appreciating it.
$endgroup$
– Zest
Jan 8 at 6:07
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%2f3065840%2fdetermine-all-solutions-to-operatornamelogz2-operatornamelogz-co%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
