$z^c = e^c log z$ vs. $z^c = e^log z^c$ and the domain of equality

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Definition of $x^r$ for $xin mathbbR$ and $rin mathbbQ^c$ makes sense if we define it as a limit of $x^r_n$ for a sequence $r_n subset mathbbQ$ for $lim_infty r_n=r$. And the reasons are that $x^r$ is a continuous function of $r$ and both $mathbbQ$ and mathbbQ^c are dense in $mathbbR$.



Churchill's book defines $z^c = e^c log z$ because




[it] provides a consistent definition of $z^c$ in the sense that it is
already known to be valid when $c =n$ for $n in mathbbN$ and $c =
dfrac1n$ for $n in mathbbN$.




Also the book defines $log z$ based on $z = e^log z$. So, my Question 1. is that isn't it more reasonable to to define $z^c = e^log z^c$ since



1- "Defining" a number (or a set of numbers) makes sense only if it indicates the value of it and we can't define whatever we want;



2- Extending from $c =n$ for $n in mathbbN$ and $c = dfrac1n$ for $n in mathbbN$ to any $c in mathbbC$ doesn't seem logical because $mathbbC$ is much "larger" set [I mean it may not be possible uniquely extrapolate to $mathbbC$ from $mathbbN cup frac1n$] and it may result in different extensions.



3- Though may(?) $e^log z^c=e^c log z$ but $log z^c ne c log z$ in general even for $c in mathbbN$ ?



And, Question 2. Is it possible $e^c log z ne e^log z^c$ for some pair $(z,c)$?







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  • 1




    Isn't $z^c:=mathrme^log z^c$ circular?
    – user1337
    Aug 11 at 20:14







  • 1




    What did you try? With the usual principal value, $$ log((-i)^2)=i pi ne -i pi = 2log(-i) $$
    – GEdgar
    Aug 11 at 20:15










  • @user1337, what does it mean "circular"? I suggest $z^c:=mathrme^log z^c$ because $z:=mathrme^log z $..
    – Edi
    Aug 11 at 20:16






  • 1




    it means defining $z^c$ in terms of $z^c$.
    – user1337
    Aug 11 at 20:17










  • @user1337, Aha I got why the motivation for that definition, but still I can't find out answers for the questions I mentioned..
    – Edi
    Aug 11 at 20:23














up vote
2
down vote

favorite
1












Definition of $x^r$ for $xin mathbbR$ and $rin mathbbQ^c$ makes sense if we define it as a limit of $x^r_n$ for a sequence $r_n subset mathbbQ$ for $lim_infty r_n=r$. And the reasons are that $x^r$ is a continuous function of $r$ and both $mathbbQ$ and mathbbQ^c are dense in $mathbbR$.



Churchill's book defines $z^c = e^c log z$ because




[it] provides a consistent definition of $z^c$ in the sense that it is
already known to be valid when $c =n$ for $n in mathbbN$ and $c =
dfrac1n$ for $n in mathbbN$.




Also the book defines $log z$ based on $z = e^log z$. So, my Question 1. is that isn't it more reasonable to to define $z^c = e^log z^c$ since



1- "Defining" a number (or a set of numbers) makes sense only if it indicates the value of it and we can't define whatever we want;



2- Extending from $c =n$ for $n in mathbbN$ and $c = dfrac1n$ for $n in mathbbN$ to any $c in mathbbC$ doesn't seem logical because $mathbbC$ is much "larger" set [I mean it may not be possible uniquely extrapolate to $mathbbC$ from $mathbbN cup frac1n$] and it may result in different extensions.



3- Though may(?) $e^log z^c=e^c log z$ but $log z^c ne c log z$ in general even for $c in mathbbN$ ?



And, Question 2. Is it possible $e^c log z ne e^log z^c$ for some pair $(z,c)$?







share|cite|improve this question


















  • 1




    Isn't $z^c:=mathrme^log z^c$ circular?
    – user1337
    Aug 11 at 20:14







  • 1




    What did you try? With the usual principal value, $$ log((-i)^2)=i pi ne -i pi = 2log(-i) $$
    – GEdgar
    Aug 11 at 20:15










  • @user1337, what does it mean "circular"? I suggest $z^c:=mathrme^log z^c$ because $z:=mathrme^log z $..
    – Edi
    Aug 11 at 20:16






  • 1




    it means defining $z^c$ in terms of $z^c$.
    – user1337
    Aug 11 at 20:17










  • @user1337, Aha I got why the motivation for that definition, but still I can't find out answers for the questions I mentioned..
    – Edi
    Aug 11 at 20:23












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





Definition of $x^r$ for $xin mathbbR$ and $rin mathbbQ^c$ makes sense if we define it as a limit of $x^r_n$ for a sequence $r_n subset mathbbQ$ for $lim_infty r_n=r$. And the reasons are that $x^r$ is a continuous function of $r$ and both $mathbbQ$ and mathbbQ^c are dense in $mathbbR$.



Churchill's book defines $z^c = e^c log z$ because




[it] provides a consistent definition of $z^c$ in the sense that it is
already known to be valid when $c =n$ for $n in mathbbN$ and $c =
dfrac1n$ for $n in mathbbN$.




Also the book defines $log z$ based on $z = e^log z$. So, my Question 1. is that isn't it more reasonable to to define $z^c = e^log z^c$ since



1- "Defining" a number (or a set of numbers) makes sense only if it indicates the value of it and we can't define whatever we want;



2- Extending from $c =n$ for $n in mathbbN$ and $c = dfrac1n$ for $n in mathbbN$ to any $c in mathbbC$ doesn't seem logical because $mathbbC$ is much "larger" set [I mean it may not be possible uniquely extrapolate to $mathbbC$ from $mathbbN cup frac1n$] and it may result in different extensions.



3- Though may(?) $e^log z^c=e^c log z$ but $log z^c ne c log z$ in general even for $c in mathbbN$ ?



And, Question 2. Is it possible $e^c log z ne e^log z^c$ for some pair $(z,c)$?







share|cite|improve this question














Definition of $x^r$ for $xin mathbbR$ and $rin mathbbQ^c$ makes sense if we define it as a limit of $x^r_n$ for a sequence $r_n subset mathbbQ$ for $lim_infty r_n=r$. And the reasons are that $x^r$ is a continuous function of $r$ and both $mathbbQ$ and mathbbQ^c are dense in $mathbbR$.



Churchill's book defines $z^c = e^c log z$ because




[it] provides a consistent definition of $z^c$ in the sense that it is
already known to be valid when $c =n$ for $n in mathbbN$ and $c =
dfrac1n$ for $n in mathbbN$.




Also the book defines $log z$ based on $z = e^log z$. So, my Question 1. is that isn't it more reasonable to to define $z^c = e^log z^c$ since



1- "Defining" a number (or a set of numbers) makes sense only if it indicates the value of it and we can't define whatever we want;



2- Extending from $c =n$ for $n in mathbbN$ and $c = dfrac1n$ for $n in mathbbN$ to any $c in mathbbC$ doesn't seem logical because $mathbbC$ is much "larger" set [I mean it may not be possible uniquely extrapolate to $mathbbC$ from $mathbbN cup frac1n$] and it may result in different extensions.



3- Though may(?) $e^log z^c=e^c log z$ but $log z^c ne c log z$ in general even for $c in mathbbN$ ?



And, Question 2. Is it possible $e^c log z ne e^log z^c$ for some pair $(z,c)$?









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 12 at 1:41

























asked Aug 11 at 19:40









Edi

1,111829




1,111829







  • 1




    Isn't $z^c:=mathrme^log z^c$ circular?
    – user1337
    Aug 11 at 20:14







  • 1




    What did you try? With the usual principal value, $$ log((-i)^2)=i pi ne -i pi = 2log(-i) $$
    – GEdgar
    Aug 11 at 20:15










  • @user1337, what does it mean "circular"? I suggest $z^c:=mathrme^log z^c$ because $z:=mathrme^log z $..
    – Edi
    Aug 11 at 20:16






  • 1




    it means defining $z^c$ in terms of $z^c$.
    – user1337
    Aug 11 at 20:17










  • @user1337, Aha I got why the motivation for that definition, but still I can't find out answers for the questions I mentioned..
    – Edi
    Aug 11 at 20:23












  • 1




    Isn't $z^c:=mathrme^log z^c$ circular?
    – user1337
    Aug 11 at 20:14







  • 1




    What did you try? With the usual principal value, $$ log((-i)^2)=i pi ne -i pi = 2log(-i) $$
    – GEdgar
    Aug 11 at 20:15










  • @user1337, what does it mean "circular"? I suggest $z^c:=mathrme^log z^c$ because $z:=mathrme^log z $..
    – Edi
    Aug 11 at 20:16






  • 1




    it means defining $z^c$ in terms of $z^c$.
    – user1337
    Aug 11 at 20:17










  • @user1337, Aha I got why the motivation for that definition, but still I can't find out answers for the questions I mentioned..
    – Edi
    Aug 11 at 20:23







1




1




Isn't $z^c:=mathrme^log z^c$ circular?
– user1337
Aug 11 at 20:14





Isn't $z^c:=mathrme^log z^c$ circular?
– user1337
Aug 11 at 20:14





1




1




What did you try? With the usual principal value, $$ log((-i)^2)=i pi ne -i pi = 2log(-i) $$
– GEdgar
Aug 11 at 20:15




What did you try? With the usual principal value, $$ log((-i)^2)=i pi ne -i pi = 2log(-i) $$
– GEdgar
Aug 11 at 20:15












@user1337, what does it mean "circular"? I suggest $z^c:=mathrme^log z^c$ because $z:=mathrme^log z $..
– Edi
Aug 11 at 20:16




@user1337, what does it mean "circular"? I suggest $z^c:=mathrme^log z^c$ because $z:=mathrme^log z $..
– Edi
Aug 11 at 20:16




1




1




it means defining $z^c$ in terms of $z^c$.
– user1337
Aug 11 at 20:17




it means defining $z^c$ in terms of $z^c$.
– user1337
Aug 11 at 20:17












@user1337, Aha I got why the motivation for that definition, but still I can't find out answers for the questions I mentioned..
– Edi
Aug 11 at 20:23




@user1337, Aha I got why the motivation for that definition, but still I can't find out answers for the questions I mentioned..
– Edi
Aug 11 at 20:23















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