Find all the solutions for $e^z=e^{iz}$












0












$begingroup$


so iam trying to solve this



$e^z=e^{iz} $



so, since z = x+iy



$ e^{x+iy} = e^{-y+ix} $



so should i take the Log for the two sides ? or what should i do ?!!



I really want the help!



Thanks






calculus complex-analysis complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    where z is in the form x+iy , where x,y belong to R (real values).
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 18:56










  • $begingroup$
    this is work if z ∈ $Real$ ?
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:13


















0












$begingroup$


so iam trying to solve this



$e^z=e^{iz} $



so, since z = x+iy



$ e^{x+iy} = e^{-y+ix} $



so should i take the Log for the two sides ? or what should i do ?!!



I really want the help!



Thanks






calculus complex-analysis complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    where z is in the form x+iy , where x,y belong to R (real values).
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 18:56










  • $begingroup$
    this is work if z ∈ $Real$ ?
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:13
















0












0








0





$begingroup$


so iam trying to solve this



$e^z=e^{iz} $



so, since z = x+iy



$ e^{x+iy} = e^{-y+ix} $



so should i take the Log for the two sides ? or what should i do ?!!



I really want the help!



Thanks






calculus complex-analysis complex-numbers







share|cite|improve this question











$endgroup$




so iam trying to solve this



$e^z=e^{iz} $



so, since z = x+iy



$ e^{x+iy} = e^{-y+ix} $



so should i take the Log for the two sides ? or what should i do ?!!



I really want the help!



Thanks






calculus complex-analysis complex-numbers




calculus complex-analysis complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 18:48







Smb Youz

















asked Nov 28 '18 at 18:45









Smb YouzSmb Youz

165




165












  • $begingroup$
    where z is in the form x+iy , where x,y belong to R (real values).
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 18:56










  • $begingroup$
    this is work if z ∈ $Real$ ?
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:13




















  • $begingroup$
    where z is in the form x+iy , where x,y belong to R (real values).
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 18:56










  • $begingroup$
    this is work if z ∈ $Real$ ?
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:13


















$begingroup$
where z is in the form x+iy , where x,y belong to R (real values).
$endgroup$
– Smb Youz
Nov 28 '18 at 18:56




$begingroup$
where z is in the form x+iy , where x,y belong to R (real values).
$endgroup$
– Smb Youz
Nov 28 '18 at 18:56












$begingroup$
this is work if z ∈ $Real$ ?
$endgroup$
– Smb Youz
Nov 28 '18 at 19:13






$begingroup$
this is work if z ∈ $Real$ ?
$endgroup$
– Smb Youz
Nov 28 '18 at 19:13












3 Answers
3






active

oldest

votes


















1












$begingroup$

Hint:



$$e^{z(1-i)}=1=e^{2mpi i}$$ where $m$ is any integer






share|cite|improve this answer









$endgroup$













  • $begingroup$
    amazing , so that's is the all solution in order to $e^z=e^{iz} $
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:00












  • $begingroup$
    so i should continue as follow (x+iy)(1-i)=2m$pi$i ??
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:08



















1












$begingroup$

Notice that $$|e^z|=|e^{iz}|$$therefore $$e^x=e^{-y}$$which leads to $$x=-y$$therefore $$e^z=e^{iz}=e^{x+ix}=e^{x-ix}$$which reduce to $$e^{ix}=e^{-ix}$$hence the solutions are$$x=-y=kpi$$and $$z=kpi (1-i)quad,quad kin Bbb Z$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    that's why I am confused lol !!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 18:50










  • $begingroup$
    Why then? If two complex numbers are equal, so are their absolute values while the angles do not need to be so.
    $endgroup$
    – Mostafa Ayaz
    Nov 28 '18 at 18:51












  • $begingroup$
    yes you are right , but what is the solutions ? right !
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:10










  • $begingroup$
    that's better sir , thanks
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:19












  • $begingroup$
    You're welcome!!
    $endgroup$
    – Mostafa Ayaz
    Nov 28 '18 at 20:07



















0












$begingroup$

If $z,winmathbb C$, then $e^z=e^w$ if and only if there is an integer $k$ such that $z=w+2kpi i$. So, when do we have $z-iz=2kpi i$ for some integer $k$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    z,w∈R numbers !!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:09










  • $begingroup$
    Why should I suppose that $z,winmathbb R$?
    $endgroup$
    – José Carlos Santos
    Nov 28 '18 at 19:11










  • $begingroup$
    The question we are trying to solve giving me this hint !!!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:11












  • $begingroup$
    The question that you are trying to solve has no $w$.
    $endgroup$
    – José Carlos Santos
    Nov 28 '18 at 19:12










  • $begingroup$
    i got it sir , thanks alot
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:21











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Hint:



$$e^{z(1-i)}=1=e^{2mpi i}$$ where $m$ is any integer






share|cite|improve this answer









$endgroup$













  • $begingroup$
    amazing , so that's is the all solution in order to $e^z=e^{iz} $
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:00












  • $begingroup$
    so i should continue as follow (x+iy)(1-i)=2m$pi$i ??
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:08
















1












$begingroup$

Hint:



$$e^{z(1-i)}=1=e^{2mpi i}$$ where $m$ is any integer






share|cite|improve this answer









$endgroup$













  • $begingroup$
    amazing , so that's is the all solution in order to $e^z=e^{iz} $
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:00












  • $begingroup$
    so i should continue as follow (x+iy)(1-i)=2m$pi$i ??
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:08














1












1








1





$begingroup$

Hint:



$$e^{z(1-i)}=1=e^{2mpi i}$$ where $m$ is any integer






share|cite|improve this answer









$endgroup$



Hint:



$$e^{z(1-i)}=1=e^{2mpi i}$$ where $m$ is any integer







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 28 '18 at 18:47









lab bhattacharjeelab bhattacharjee

224k15156274




224k15156274












  • $begingroup$
    amazing , so that's is the all solution in order to $e^z=e^{iz} $
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:00












  • $begingroup$
    so i should continue as follow (x+iy)(1-i)=2m$pi$i ??
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:08


















  • $begingroup$
    amazing , so that's is the all solution in order to $e^z=e^{iz} $
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:00












  • $begingroup$
    so i should continue as follow (x+iy)(1-i)=2m$pi$i ??
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:08
















$begingroup$
amazing , so that's is the all solution in order to $e^z=e^{iz} $
$endgroup$
– Smb Youz
Nov 28 '18 at 19:00






$begingroup$
amazing , so that's is the all solution in order to $e^z=e^{iz} $
$endgroup$
– Smb Youz
Nov 28 '18 at 19:00














$begingroup$
so i should continue as follow (x+iy)(1-i)=2m$pi$i ??
$endgroup$
– Smb Youz
Nov 28 '18 at 19:08




$begingroup$
so i should continue as follow (x+iy)(1-i)=2m$pi$i ??
$endgroup$
– Smb Youz
Nov 28 '18 at 19:08











1












$begingroup$

Notice that $$|e^z|=|e^{iz}|$$therefore $$e^x=e^{-y}$$which leads to $$x=-y$$therefore $$e^z=e^{iz}=e^{x+ix}=e^{x-ix}$$which reduce to $$e^{ix}=e^{-ix}$$hence the solutions are$$x=-y=kpi$$and $$z=kpi (1-i)quad,quad kin Bbb Z$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    that's why I am confused lol !!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 18:50










  • $begingroup$
    Why then? If two complex numbers are equal, so are their absolute values while the angles do not need to be so.
    $endgroup$
    – Mostafa Ayaz
    Nov 28 '18 at 18:51












  • $begingroup$
    yes you are right , but what is the solutions ? right !
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:10










  • $begingroup$
    that's better sir , thanks
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:19












  • $begingroup$
    You're welcome!!
    $endgroup$
    – Mostafa Ayaz
    Nov 28 '18 at 20:07
















1












$begingroup$

Notice that $$|e^z|=|e^{iz}|$$therefore $$e^x=e^{-y}$$which leads to $$x=-y$$therefore $$e^z=e^{iz}=e^{x+ix}=e^{x-ix}$$which reduce to $$e^{ix}=e^{-ix}$$hence the solutions are$$x=-y=kpi$$and $$z=kpi (1-i)quad,quad kin Bbb Z$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    that's why I am confused lol !!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 18:50










  • $begingroup$
    Why then? If two complex numbers are equal, so are their absolute values while the angles do not need to be so.
    $endgroup$
    – Mostafa Ayaz
    Nov 28 '18 at 18:51












  • $begingroup$
    yes you are right , but what is the solutions ? right !
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:10










  • $begingroup$
    that's better sir , thanks
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:19












  • $begingroup$
    You're welcome!!
    $endgroup$
    – Mostafa Ayaz
    Nov 28 '18 at 20:07














1












1








1





$begingroup$

Notice that $$|e^z|=|e^{iz}|$$therefore $$e^x=e^{-y}$$which leads to $$x=-y$$therefore $$e^z=e^{iz}=e^{x+ix}=e^{x-ix}$$which reduce to $$e^{ix}=e^{-ix}$$hence the solutions are$$x=-y=kpi$$and $$z=kpi (1-i)quad,quad kin Bbb Z$$






share|cite|improve this answer











$endgroup$



Notice that $$|e^z|=|e^{iz}|$$therefore $$e^x=e^{-y}$$which leads to $$x=-y$$therefore $$e^z=e^{iz}=e^{x+ix}=e^{x-ix}$$which reduce to $$e^{ix}=e^{-ix}$$hence the solutions are$$x=-y=kpi$$and $$z=kpi (1-i)quad,quad kin Bbb Z$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 '18 at 19:15

























answered Nov 28 '18 at 18:48









Mostafa AyazMostafa Ayaz

14.8k3938




14.8k3938












  • $begingroup$
    that's why I am confused lol !!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 18:50










  • $begingroup$
    Why then? If two complex numbers are equal, so are their absolute values while the angles do not need to be so.
    $endgroup$
    – Mostafa Ayaz
    Nov 28 '18 at 18:51












  • $begingroup$
    yes you are right , but what is the solutions ? right !
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:10










  • $begingroup$
    that's better sir , thanks
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:19












  • $begingroup$
    You're welcome!!
    $endgroup$
    – Mostafa Ayaz
    Nov 28 '18 at 20:07


















  • $begingroup$
    that's why I am confused lol !!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 18:50










  • $begingroup$
    Why then? If two complex numbers are equal, so are their absolute values while the angles do not need to be so.
    $endgroup$
    – Mostafa Ayaz
    Nov 28 '18 at 18:51












  • $begingroup$
    yes you are right , but what is the solutions ? right !
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:10










  • $begingroup$
    that's better sir , thanks
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:19












  • $begingroup$
    You're welcome!!
    $endgroup$
    – Mostafa Ayaz
    Nov 28 '18 at 20:07
















$begingroup$
that's why I am confused lol !!
$endgroup$
– Smb Youz
Nov 28 '18 at 18:50




$begingroup$
that's why I am confused lol !!
$endgroup$
– Smb Youz
Nov 28 '18 at 18:50












$begingroup$
Why then? If two complex numbers are equal, so are their absolute values while the angles do not need to be so.
$endgroup$
– Mostafa Ayaz
Nov 28 '18 at 18:51






$begingroup$
Why then? If two complex numbers are equal, so are their absolute values while the angles do not need to be so.
$endgroup$
– Mostafa Ayaz
Nov 28 '18 at 18:51














$begingroup$
yes you are right , but what is the solutions ? right !
$endgroup$
– Smb Youz
Nov 28 '18 at 19:10




$begingroup$
yes you are right , but what is the solutions ? right !
$endgroup$
– Smb Youz
Nov 28 '18 at 19:10












$begingroup$
that's better sir , thanks
$endgroup$
– Smb Youz
Nov 28 '18 at 19:19






$begingroup$
that's better sir , thanks
$endgroup$
– Smb Youz
Nov 28 '18 at 19:19














$begingroup$
You're welcome!!
$endgroup$
– Mostafa Ayaz
Nov 28 '18 at 20:07




$begingroup$
You're welcome!!
$endgroup$
– Mostafa Ayaz
Nov 28 '18 at 20:07











0












$begingroup$

If $z,winmathbb C$, then $e^z=e^w$ if and only if there is an integer $k$ such that $z=w+2kpi i$. So, when do we have $z-iz=2kpi i$ for some integer $k$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    z,w∈R numbers !!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:09










  • $begingroup$
    Why should I suppose that $z,winmathbb R$?
    $endgroup$
    – José Carlos Santos
    Nov 28 '18 at 19:11










  • $begingroup$
    The question we are trying to solve giving me this hint !!!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:11












  • $begingroup$
    The question that you are trying to solve has no $w$.
    $endgroup$
    – José Carlos Santos
    Nov 28 '18 at 19:12










  • $begingroup$
    i got it sir , thanks alot
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:21
















0












$begingroup$

If $z,winmathbb C$, then $e^z=e^w$ if and only if there is an integer $k$ such that $z=w+2kpi i$. So, when do we have $z-iz=2kpi i$ for some integer $k$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    z,w∈R numbers !!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:09










  • $begingroup$
    Why should I suppose that $z,winmathbb R$?
    $endgroup$
    – José Carlos Santos
    Nov 28 '18 at 19:11










  • $begingroup$
    The question we are trying to solve giving me this hint !!!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:11












  • $begingroup$
    The question that you are trying to solve has no $w$.
    $endgroup$
    – José Carlos Santos
    Nov 28 '18 at 19:12










  • $begingroup$
    i got it sir , thanks alot
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:21














0












0








0





$begingroup$

If $z,winmathbb C$, then $e^z=e^w$ if and only if there is an integer $k$ such that $z=w+2kpi i$. So, when do we have $z-iz=2kpi i$ for some integer $k$?






share|cite|improve this answer









$endgroup$



If $z,winmathbb C$, then $e^z=e^w$ if and only if there is an integer $k$ such that $z=w+2kpi i$. So, when do we have $z-iz=2kpi i$ for some integer $k$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 28 '18 at 18:48









José Carlos SantosJosé Carlos Santos

153k22123225




153k22123225












  • $begingroup$
    z,w∈R numbers !!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:09










  • $begingroup$
    Why should I suppose that $z,winmathbb R$?
    $endgroup$
    – José Carlos Santos
    Nov 28 '18 at 19:11










  • $begingroup$
    The question we are trying to solve giving me this hint !!!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:11












  • $begingroup$
    The question that you are trying to solve has no $w$.
    $endgroup$
    – José Carlos Santos
    Nov 28 '18 at 19:12










  • $begingroup$
    i got it sir , thanks alot
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:21


















  • $begingroup$
    z,w∈R numbers !!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:09










  • $begingroup$
    Why should I suppose that $z,winmathbb R$?
    $endgroup$
    – José Carlos Santos
    Nov 28 '18 at 19:11










  • $begingroup$
    The question we are trying to solve giving me this hint !!!
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:11












  • $begingroup$
    The question that you are trying to solve has no $w$.
    $endgroup$
    – José Carlos Santos
    Nov 28 '18 at 19:12










  • $begingroup$
    i got it sir , thanks alot
    $endgroup$
    – Smb Youz
    Nov 28 '18 at 19:21
















$begingroup$
z,w∈R numbers !!
$endgroup$
– Smb Youz
Nov 28 '18 at 19:09




$begingroup$
z,w∈R numbers !!
$endgroup$
– Smb Youz
Nov 28 '18 at 19:09












$begingroup$
Why should I suppose that $z,winmathbb R$?
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:11




$begingroup$
Why should I suppose that $z,winmathbb R$?
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:11












$begingroup$
The question we are trying to solve giving me this hint !!!
$endgroup$
– Smb Youz
Nov 28 '18 at 19:11






$begingroup$
The question we are trying to solve giving me this hint !!!
$endgroup$
– Smb Youz
Nov 28 '18 at 19:11














$begingroup$
The question that you are trying to solve has no $w$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:12




$begingroup$
The question that you are trying to solve has no $w$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:12












$begingroup$
i got it sir , thanks alot
$endgroup$
– Smb Youz
Nov 28 '18 at 19:21




$begingroup$
i got it sir , thanks alot
$endgroup$
– Smb Youz
Nov 28 '18 at 19:21


















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Plaza Victoria

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Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
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In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
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How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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