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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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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