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












0














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
























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










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


















0














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
























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










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
















0












0








0







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















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 at 18:48

























asked Nov 28 at 18:45









Smb Youz

144




144












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










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




















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










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


















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




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












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






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












3 Answers
3






active

oldest

votes


















1














Hint:



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






share|cite|improve this answer





















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












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



















1














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























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










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












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










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












  • You're welcome!!
    – Mostafa Ayaz
    Nov 28 at 20:07



















0














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





















  • z,w∈R numbers !!
    – Smb Youz
    Nov 28 at 19:09










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










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












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










  • i got it sir , thanks alot
    – Smb Youz
    Nov 28 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














Hint:



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






share|cite|improve this answer





















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












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
















1














Hint:



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






share|cite|improve this answer





















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












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














1












1








1






Hint:



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






share|cite|improve this answer












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 at 18:47









lab bhattacharjee

222k15155273




222k15155273












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












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


















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












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
















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






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














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




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











1














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























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










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












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










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












  • You're welcome!!
    – Mostafa Ayaz
    Nov 28 at 20:07
















1














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























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










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












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










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












  • You're welcome!!
    – Mostafa Ayaz
    Nov 28 at 20:07














1












1








1






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














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 at 19:15

























answered Nov 28 at 18:48









Mostafa Ayaz

13.7k3836




13.7k3836












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










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












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










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












  • You're welcome!!
    – Mostafa Ayaz
    Nov 28 at 20:07


















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










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












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










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












  • You're welcome!!
    – Mostafa Ayaz
    Nov 28 at 20:07
















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




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












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






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














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




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












that's better sir , thanks
– Smb Youz
Nov 28 at 19:19






that's better sir , thanks
– Smb Youz
Nov 28 at 19:19














You're welcome!!
– Mostafa Ayaz
Nov 28 at 20:07




You're welcome!!
– Mostafa Ayaz
Nov 28 at 20:07











0














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





















  • z,w∈R numbers !!
    – Smb Youz
    Nov 28 at 19:09










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










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












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










  • i got it sir , thanks alot
    – Smb Youz
    Nov 28 at 19:21
















0














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





















  • z,w∈R numbers !!
    – Smb Youz
    Nov 28 at 19:09










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










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












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










  • i got it sir , thanks alot
    – Smb Youz
    Nov 28 at 19:21














0












0








0






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












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 at 18:48









José Carlos Santos

148k22117218




148k22117218












  • z,w∈R numbers !!
    – Smb Youz
    Nov 28 at 19:09










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










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












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










  • i got it sir , thanks alot
    – Smb Youz
    Nov 28 at 19:21


















  • z,w∈R numbers !!
    – Smb Youz
    Nov 28 at 19:09










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










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












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










  • i got it sir , thanks alot
    – Smb Youz
    Nov 28 at 19:21
















z,w∈R numbers !!
– Smb Youz
Nov 28 at 19:09




z,w∈R numbers !!
– Smb Youz
Nov 28 at 19:09












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




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












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






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














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




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












i got it sir , thanks alot
– Smb Youz
Nov 28 at 19:21




i got it sir , thanks alot
– Smb Youz
Nov 28 at 19:21


















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