$z^c = e^c log z$ 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 $(z,c)$?
complex-analysis complex-numbers definition
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up vote
2
down vote
favorite
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 $(z,c)$?
complex-analysis complex-numbers definition
1
Isn't $z^c:=mathrme^log z^c$ circular?
â user1337
Aug 11 at 20:14
1
What did you try? With the usual principal value, $$ log((-i)^2)=i pi ne -i pi = 2log(-i) $$
â 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 $..
â Edi
Aug 11 at 20:16
1
it means defining $z^c$ in terms of $z^c$.
â 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..
â Edi
Aug 11 at 20:23
 |Â
show 3 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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 $(z,c)$?
complex-analysis complex-numbers definition
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 $(z,c)$?
complex-analysis complex-numbers definition
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?
â user1337
Aug 11 at 20:14
1
What did you try? With the usual principal value, $$ log((-i)^2)=i pi ne -i pi = 2log(-i) $$
â 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 $..
â Edi
Aug 11 at 20:16
1
it means defining $z^c$ in terms of $z^c$.
â 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..
â Edi
Aug 11 at 20:23
 |Â
show 3 more comments
1
Isn't $z^c:=mathrme^log z^c$ circular?
â user1337
Aug 11 at 20:14
1
What did you try? With the usual principal value, $$ log((-i)^2)=i pi ne -i pi = 2log(-i) $$
â 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 $..
â Edi
Aug 11 at 20:16
1
it means defining $z^c$ in terms of $z^c$.
â 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..
â Edi
Aug 11 at 20:23
1
1
Isn't $z^c:=mathrme^log z^c$ circular?
â user1337
Aug 11 at 20:14
Isn't $z^c:=mathrme^log z^c$ circular?
â user1337
Aug 11 at 20:14
1
1
What did you try? With the usual principal value, $$ log((-i)^2)=i pi ne -i pi = 2log(-i) $$
â GEdgar
Aug 11 at 20:15
What did you try? With the usual principal value, $$ log((-i)^2)=i pi ne -i pi = 2log(-i) $$
â 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 $..
â 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 $..
â Edi
Aug 11 at 20:16
1
1
it means defining $z^c$ in terms of $z^c$.
â user1337
Aug 11 at 20:17
it means defining $z^c$ in terms of $z^c$.
â 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..
â 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..
â Edi
Aug 11 at 20:23
 |Â
show 3 more comments
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1
Isn't $z^c:=mathrme^log z^c$ circular?
â user1337
Aug 11 at 20:14
1
What did you try? With the usual principal value, $$ log((-i)^2)=i pi ne -i pi = 2log(-i) $$
â 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 $..
â Edi
Aug 11 at 20:16
1
it means defining $z^c$ in terms of $z^c$.
â 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..
â Edi
Aug 11 at 20:23