Applying a general result to a specific case: if $e^a = 1$, then $a = 2inpi$












0












$begingroup$


In a (Partial Differential Equations / Laplace Equation) , I try to solving a problem of Laplace eq. by using separation of variables method.



I usually using the rule : if $e^{2 sqrt{k} b} = 1$, then I have: $2sqrt{k} b = 2nipi$.



Now in my problem I have : $e^{2 sqrt{k}pi} = 1$
Can I use the same rule which lead to cancel the $pi$ ?






complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    I've edited your question to make the formatting look fancy: you might want to click "edit" on it to see how it works.
    $endgroup$
    – user3482749
    Jan 11 at 19:36










  • $begingroup$
    Do you mean $$e^{2sqrt{k}b}=1$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 11 at 19:36










  • $begingroup$
    thanx and yes I mean I have $e^{2 sqrt{k} b} = 1$
    $endgroup$
    – Ahmed Naji
    Jan 12 at 12:34
















0












$begingroup$


In a (Partial Differential Equations / Laplace Equation) , I try to solving a problem of Laplace eq. by using separation of variables method.



I usually using the rule : if $e^{2 sqrt{k} b} = 1$, then I have: $2sqrt{k} b = 2nipi$.



Now in my problem I have : $e^{2 sqrt{k}pi} = 1$
Can I use the same rule which lead to cancel the $pi$ ?






complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    I've edited your question to make the formatting look fancy: you might want to click "edit" on it to see how it works.
    $endgroup$
    – user3482749
    Jan 11 at 19:36










  • $begingroup$
    Do you mean $$e^{2sqrt{k}b}=1$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 11 at 19:36










  • $begingroup$
    thanx and yes I mean I have $e^{2 sqrt{k} b} = 1$
    $endgroup$
    – Ahmed Naji
    Jan 12 at 12:34














0












0








0





$begingroup$


In a (Partial Differential Equations / Laplace Equation) , I try to solving a problem of Laplace eq. by using separation of variables method.



I usually using the rule : if $e^{2 sqrt{k} b} = 1$, then I have: $2sqrt{k} b = 2nipi$.



Now in my problem I have : $e^{2 sqrt{k}pi} = 1$
Can I use the same rule which lead to cancel the $pi$ ?






complex-numbers







share|cite|improve this question











$endgroup$




In a (Partial Differential Equations / Laplace Equation) , I try to solving a problem of Laplace eq. by using separation of variables method.



I usually using the rule : if $e^{2 sqrt{k} b} = 1$, then I have: $2sqrt{k} b = 2nipi$.



Now in my problem I have : $e^{2 sqrt{k}pi} = 1$
Can I use the same rule which lead to cancel the $pi$ ?






complex-numbers




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 12:06









MaoWao

3,953618




3,953618










asked Jan 11 at 19:32









Ahmed NajiAhmed Naji

11




11












  • $begingroup$
    I've edited your question to make the formatting look fancy: you might want to click "edit" on it to see how it works.
    $endgroup$
    – user3482749
    Jan 11 at 19:36










  • $begingroup$
    Do you mean $$e^{2sqrt{k}b}=1$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 11 at 19:36










  • $begingroup$
    thanx and yes I mean I have $e^{2 sqrt{k} b} = 1$
    $endgroup$
    – Ahmed Naji
    Jan 12 at 12:34


















  • $begingroup$
    I've edited your question to make the formatting look fancy: you might want to click "edit" on it to see how it works.
    $endgroup$
    – user3482749
    Jan 11 at 19:36










  • $begingroup$
    Do you mean $$e^{2sqrt{k}b}=1$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 11 at 19:36










  • $begingroup$
    thanx and yes I mean I have $e^{2 sqrt{k} b} = 1$
    $endgroup$
    – Ahmed Naji
    Jan 12 at 12:34
















$begingroup$
I've edited your question to make the formatting look fancy: you might want to click "edit" on it to see how it works.
$endgroup$
– user3482749
Jan 11 at 19:36




$begingroup$
I've edited your question to make the formatting look fancy: you might want to click "edit" on it to see how it works.
$endgroup$
– user3482749
Jan 11 at 19:36












$begingroup$
Do you mean $$e^{2sqrt{k}b}=1$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 19:36




$begingroup$
Do you mean $$e^{2sqrt{k}b}=1$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 19:36












$begingroup$
thanx and yes I mean I have $e^{2 sqrt{k} b} = 1$
$endgroup$
– Ahmed Naji
Jan 12 at 12:34




$begingroup$
thanx and yes I mean I have $e^{2 sqrt{k} b} = 1$
$endgroup$
– Ahmed Naji
Jan 12 at 12:34










2 Answers
2






active

oldest

votes


















1












$begingroup$

Yes, indeed: the result that you quote (after correcting the missing/surplus $2$) is true for all complex values of $b$, and so, if $e^{2sqrt{k}pi} = 1$, then you have $2sqrt{k}pi = 2nipi$, and hence $sqrt{k} = ni$, for some integer $n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Here's how I'm interpretting the question: the asker is aware of the general fact that if $e^a = 1$, then $a = 2nipi$, for some integer $n$. The asker wishes to know if they can apply this fact to the case where $a = 2sqrt{k}pi$, and has made some typos in writing their formula (probably due to the almost-unreadable format in which they were originally written, prior to my correcting them to LaTeX).
    $endgroup$
    – user3482749
    Jan 11 at 19:39










  • $begingroup$
    @DietrichBurde The question does, however. Are you really surprised that the question has a poor title?
    $endgroup$
    – user3482749
    Jan 11 at 19:42










  • $begingroup$
    @DietrichBurde The title is essentially gibberish. There's nothing to understand. In fact, I'll go and fix it now.
    $endgroup$
    – user3482749
    Jan 11 at 19:44










  • $begingroup$
    Thanx I got it and thanx again for fixixng the title.
    $endgroup$
    – Ahmed Naji
    Jan 12 at 12:38



















0












$begingroup$

You're using Euler's formula for complex numbers



$$e^{itheta}=cos(theta)+isin(theta)$$



We know that $cos(2npi)=1, sin(2npi)=0$, for $nin Bbb Z$ so:



$$nin Bbb Z implies e^{2nipi}=1$$



You have, therefore, in the first part of your question:



$$2bsqrt k =2nipito bsqrt k=nipito k=-(frac{npi}{b})^2$$



This CAN be applied to part b as well:



$$2pisqrt k=2nipitosqrt k=nito k=-n^2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanx for your detail I got it.
    $endgroup$
    – Ahmed Naji
    Jan 12 at 12:39












Your Answer








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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070258%2fapplying-a-general-result-to-a-specific-case-if-ea-1-then-a-2in-pi%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









1












$begingroup$

Yes, indeed: the result that you quote (after correcting the missing/surplus $2$) is true for all complex values of $b$, and so, if $e^{2sqrt{k}pi} = 1$, then you have $2sqrt{k}pi = 2nipi$, and hence $sqrt{k} = ni$, for some integer $n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Here's how I'm interpretting the question: the asker is aware of the general fact that if $e^a = 1$, then $a = 2nipi$, for some integer $n$. The asker wishes to know if they can apply this fact to the case where $a = 2sqrt{k}pi$, and has made some typos in writing their formula (probably due to the almost-unreadable format in which they were originally written, prior to my correcting them to LaTeX).
    $endgroup$
    – user3482749
    Jan 11 at 19:39










  • $begingroup$
    @DietrichBurde The question does, however. Are you really surprised that the question has a poor title?
    $endgroup$
    – user3482749
    Jan 11 at 19:42










  • $begingroup$
    @DietrichBurde The title is essentially gibberish. There's nothing to understand. In fact, I'll go and fix it now.
    $endgroup$
    – user3482749
    Jan 11 at 19:44










  • $begingroup$
    Thanx I got it and thanx again for fixixng the title.
    $endgroup$
    – Ahmed Naji
    Jan 12 at 12:38
















1












$begingroup$

Yes, indeed: the result that you quote (after correcting the missing/surplus $2$) is true for all complex values of $b$, and so, if $e^{2sqrt{k}pi} = 1$, then you have $2sqrt{k}pi = 2nipi$, and hence $sqrt{k} = ni$, for some integer $n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Here's how I'm interpretting the question: the asker is aware of the general fact that if $e^a = 1$, then $a = 2nipi$, for some integer $n$. The asker wishes to know if they can apply this fact to the case where $a = 2sqrt{k}pi$, and has made some typos in writing their formula (probably due to the almost-unreadable format in which they were originally written, prior to my correcting them to LaTeX).
    $endgroup$
    – user3482749
    Jan 11 at 19:39










  • $begingroup$
    @DietrichBurde The question does, however. Are you really surprised that the question has a poor title?
    $endgroup$
    – user3482749
    Jan 11 at 19:42










  • $begingroup$
    @DietrichBurde The title is essentially gibberish. There's nothing to understand. In fact, I'll go and fix it now.
    $endgroup$
    – user3482749
    Jan 11 at 19:44










  • $begingroup$
    Thanx I got it and thanx again for fixixng the title.
    $endgroup$
    – Ahmed Naji
    Jan 12 at 12:38














1












1








1





$begingroup$

Yes, indeed: the result that you quote (after correcting the missing/surplus $2$) is true for all complex values of $b$, and so, if $e^{2sqrt{k}pi} = 1$, then you have $2sqrt{k}pi = 2nipi$, and hence $sqrt{k} = ni$, for some integer $n$.






share|cite|improve this answer











$endgroup$



Yes, indeed: the result that you quote (after correcting the missing/surplus $2$) is true for all complex values of $b$, and so, if $e^{2sqrt{k}pi} = 1$, then you have $2sqrt{k}pi = 2nipi$, and hence $sqrt{k} = ni$, for some integer $n$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 11 at 19:38

























answered Jan 11 at 19:36









user3482749user3482749

4,3291119




4,3291119












  • $begingroup$
    Here's how I'm interpretting the question: the asker is aware of the general fact that if $e^a = 1$, then $a = 2nipi$, for some integer $n$. The asker wishes to know if they can apply this fact to the case where $a = 2sqrt{k}pi$, and has made some typos in writing their formula (probably due to the almost-unreadable format in which they were originally written, prior to my correcting them to LaTeX).
    $endgroup$
    – user3482749
    Jan 11 at 19:39










  • $begingroup$
    @DietrichBurde The question does, however. Are you really surprised that the question has a poor title?
    $endgroup$
    – user3482749
    Jan 11 at 19:42










  • $begingroup$
    @DietrichBurde The title is essentially gibberish. There's nothing to understand. In fact, I'll go and fix it now.
    $endgroup$
    – user3482749
    Jan 11 at 19:44










  • $begingroup$
    Thanx I got it and thanx again for fixixng the title.
    $endgroup$
    – Ahmed Naji
    Jan 12 at 12:38


















  • $begingroup$
    Here's how I'm interpretting the question: the asker is aware of the general fact that if $e^a = 1$, then $a = 2nipi$, for some integer $n$. The asker wishes to know if they can apply this fact to the case where $a = 2sqrt{k}pi$, and has made some typos in writing their formula (probably due to the almost-unreadable format in which they were originally written, prior to my correcting them to LaTeX).
    $endgroup$
    – user3482749
    Jan 11 at 19:39










  • $begingroup$
    @DietrichBurde The question does, however. Are you really surprised that the question has a poor title?
    $endgroup$
    – user3482749
    Jan 11 at 19:42










  • $begingroup$
    @DietrichBurde The title is essentially gibberish. There's nothing to understand. In fact, I'll go and fix it now.
    $endgroup$
    – user3482749
    Jan 11 at 19:44










  • $begingroup$
    Thanx I got it and thanx again for fixixng the title.
    $endgroup$
    – Ahmed Naji
    Jan 12 at 12:38
















$begingroup$
Here's how I'm interpretting the question: the asker is aware of the general fact that if $e^a = 1$, then $a = 2nipi$, for some integer $n$. The asker wishes to know if they can apply this fact to the case where $a = 2sqrt{k}pi$, and has made some typos in writing their formula (probably due to the almost-unreadable format in which they were originally written, prior to my correcting them to LaTeX).
$endgroup$
– user3482749
Jan 11 at 19:39




$begingroup$
Here's how I'm interpretting the question: the asker is aware of the general fact that if $e^a = 1$, then $a = 2nipi$, for some integer $n$. The asker wishes to know if they can apply this fact to the case where $a = 2sqrt{k}pi$, and has made some typos in writing their formula (probably due to the almost-unreadable format in which they were originally written, prior to my correcting them to LaTeX).
$endgroup$
– user3482749
Jan 11 at 19:39












$begingroup$
@DietrichBurde The question does, however. Are you really surprised that the question has a poor title?
$endgroup$
– user3482749
Jan 11 at 19:42




$begingroup$
@DietrichBurde The question does, however. Are you really surprised that the question has a poor title?
$endgroup$
– user3482749
Jan 11 at 19:42












$begingroup$
@DietrichBurde The title is essentially gibberish. There's nothing to understand. In fact, I'll go and fix it now.
$endgroup$
– user3482749
Jan 11 at 19:44




$begingroup$
@DietrichBurde The title is essentially gibberish. There's nothing to understand. In fact, I'll go and fix it now.
$endgroup$
– user3482749
Jan 11 at 19:44












$begingroup$
Thanx I got it and thanx again for fixixng the title.
$endgroup$
– Ahmed Naji
Jan 12 at 12:38




$begingroup$
Thanx I got it and thanx again for fixixng the title.
$endgroup$
– Ahmed Naji
Jan 12 at 12:38











0












$begingroup$

You're using Euler's formula for complex numbers



$$e^{itheta}=cos(theta)+isin(theta)$$



We know that $cos(2npi)=1, sin(2npi)=0$, for $nin Bbb Z$ so:



$$nin Bbb Z implies e^{2nipi}=1$$



You have, therefore, in the first part of your question:



$$2bsqrt k =2nipito bsqrt k=nipito k=-(frac{npi}{b})^2$$



This CAN be applied to part b as well:



$$2pisqrt k=2nipitosqrt k=nito k=-n^2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanx for your detail I got it.
    $endgroup$
    – Ahmed Naji
    Jan 12 at 12:39
















0












$begingroup$

You're using Euler's formula for complex numbers



$$e^{itheta}=cos(theta)+isin(theta)$$



We know that $cos(2npi)=1, sin(2npi)=0$, for $nin Bbb Z$ so:



$$nin Bbb Z implies e^{2nipi}=1$$



You have, therefore, in the first part of your question:



$$2bsqrt k =2nipito bsqrt k=nipito k=-(frac{npi}{b})^2$$



This CAN be applied to part b as well:



$$2pisqrt k=2nipitosqrt k=nito k=-n^2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanx for your detail I got it.
    $endgroup$
    – Ahmed Naji
    Jan 12 at 12:39














0












0








0





$begingroup$

You're using Euler's formula for complex numbers



$$e^{itheta}=cos(theta)+isin(theta)$$



We know that $cos(2npi)=1, sin(2npi)=0$, for $nin Bbb Z$ so:



$$nin Bbb Z implies e^{2nipi}=1$$



You have, therefore, in the first part of your question:



$$2bsqrt k =2nipito bsqrt k=nipito k=-(frac{npi}{b})^2$$



This CAN be applied to part b as well:



$$2pisqrt k=2nipitosqrt k=nito k=-n^2$$






share|cite|improve this answer









$endgroup$



You're using Euler's formula for complex numbers



$$e^{itheta}=cos(theta)+isin(theta)$$



We know that $cos(2npi)=1, sin(2npi)=0$, for $nin Bbb Z$ so:



$$nin Bbb Z implies e^{2nipi}=1$$



You have, therefore, in the first part of your question:



$$2bsqrt k =2nipito bsqrt k=nipito k=-(frac{npi}{b})^2$$



This CAN be applied to part b as well:



$$2pisqrt k=2nipitosqrt k=nito k=-n^2$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 19:52









Rhys HughesRhys Hughes

7,0501630




7,0501630












  • $begingroup$
    Thanx for your detail I got it.
    $endgroup$
    – Ahmed Naji
    Jan 12 at 12:39


















  • $begingroup$
    Thanx for your detail I got it.
    $endgroup$
    – Ahmed Naji
    Jan 12 at 12:39
















$begingroup$
Thanx for your detail I got it.
$endgroup$
– Ahmed Naji
Jan 12 at 12:39




$begingroup$
Thanx for your detail I got it.
$endgroup$
– Ahmed Naji
Jan 12 at 12:39


















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














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070258%2fapplying-a-general-result-to-a-specific-case-if-ea-1-then-a-2in-pi%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more