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1

$begingroup$

I saw the following property in wikipedia
beginalign
(e^z)^n=e^zn tag1
endalign
where $z in mathbbC$ and $n in mathbbZ$. It's possible to change the $e$ in $(1)$ to some other number ? For example:

beginalign
(a^z)^n=a^zn
endalign

where $z in mathbbC$, $n in mathbbZ$ and $a$ is maybe a real number or even complex? Or even change the conditions on $z$ and $n$.

Another thing related to $(1)$. Gelfond's theorem says that $a^b$ is transcendental if $a$ is algebraic and $a neq 1$, $a neq 0$ and $b$ is irrational. For example $2^sqrt 2$ is transcendental. One result from this theorem is that $e^pi$ (Gelfond constant) is transcendental. Noting that:

beginalign
& e^ipi = -1 \
& (e^ipi)^-i = (-1)^-i\
& e^pi = (-1)^-i
endalign

since $(-1)$ is algebraic and $-i$ is irrational the theorem applies. But in the second line above didn't we just violate the property $(1)$ because $-i$ in not an integer?

complex-analysis number-theory transcendental-numbers

share|cite|improve this question

edited Mar 16 at 22:28

Bernard

123k741117

asked Mar 16 at 22:12

PintecoPinteco

798313

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Given two complex numbers $a,zinmathbbZ$, $a^z$ is given by $a^z=e^zlog a$, where $log$ is the multivalued complex logarithm $log a=lnleft|aright|+iarg a$. Sometimes the principal version is used with $operatornameLoga=lnleft|aright|+ioperatornameArg a$ instead. So, you are correct that $e$ can be replaced with $a$.
  $endgroup$
  – Jake
  Mar 16 at 22:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That should say $a,zinmathbbC$. I am typing up an answer now to clarify.
  $endgroup$
  – Jake
  Mar 16 at 22:30
  

  
  
  
  

add a comment | 

1

$begingroup$

I saw the following property in wikipedia
beginalign
(e^z)^n=e^zn tag1
endalign
where $z in mathbbC$ and $n in mathbbZ$. It's possible to change the $e$ in $(1)$ to some other number ? For example:

beginalign
(a^z)^n=a^zn
endalign

where $z in mathbbC$, $n in mathbbZ$ and $a$ is maybe a real number or even complex? Or even change the conditions on $z$ and $n$.

Another thing related to $(1)$. Gelfond's theorem says that $a^b$ is transcendental if $a$ is algebraic and $a neq 1$, $a neq 0$ and $b$ is irrational. For example $2^sqrt 2$ is transcendental. One result from this theorem is that $e^pi$ (Gelfond constant) is transcendental. Noting that:

beginalign
& e^ipi = -1 \
& (e^ipi)^-i = (-1)^-i\
& e^pi = (-1)^-i
endalign

since $(-1)$ is algebraic and $-i$ is irrational the theorem applies. But in the second line above didn't we just violate the property $(1)$ because $-i$ in not an integer?

complex-analysis number-theory transcendental-numbers

share|cite|improve this question

edited Mar 16 at 22:28

Bernard

123k741117

asked Mar 16 at 22:12

PintecoPinteco

798313

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Given two complex numbers $a,zinmathbbZ$, $a^z$ is given by $a^z=e^zlog a$, where $log$ is the multivalued complex logarithm $log a=lnleft|aright|+iarg a$. Sometimes the principal version is used with $operatornameLoga=lnleft|aright|+ioperatornameArg a$ instead. So, you are correct that $e$ can be replaced with $a$.
  $endgroup$
  – Jake
  Mar 16 at 22:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That should say $a,zinmathbbC$. I am typing up an answer now to clarify.
  $endgroup$
  – Jake
  Mar 16 at 22:30
  

  
  
  
  

add a comment | 

$begingroup$

I saw the following property in wikipedia
beginalign
(e^z)^n=e^zn tag1
endalign
where $z in mathbbC$ and $n in mathbbZ$. It's possible to change the $e$ in $(1)$ to some other number ? For example:

beginalign
(a^z)^n=a^zn
endalign

where $z in mathbbC$, $n in mathbbZ$ and $a$ is maybe a real number or even complex? Or even change the conditions on $z$ and $n$.

Another thing related to $(1)$. Gelfond's theorem says that $a^b$ is transcendental if $a$ is algebraic and $a neq 1$, $a neq 0$ and $b$ is irrational. For example $2^sqrt 2$ is transcendental. One result from this theorem is that $e^pi$ (Gelfond constant) is transcendental. Noting that:

beginalign
& e^ipi = -1 \
& (e^ipi)^-i = (-1)^-i\
& e^pi = (-1)^-i
endalign

since $(-1)$ is algebraic and $-i$ is irrational the theorem applies. But in the second line above didn't we just violate the property $(1)$ because $-i$ in not an integer?

complex-analysis number-theory transcendental-numbers

share|cite|improve this question

edited Mar 16 at 22:28

Bernard

123k741117

asked Mar 16 at 22:12

PintecoPinteco

798313

$endgroup$

share|cite|improve this question

edited Mar 16 at 22:28

Bernard

123k741117

asked Mar 16 at 22:12

PintecoPinteco

798313

share|cite|improve this question

edited Mar 16 at 22:28

Bernard

123k741117

asked Mar 16 at 22:12

PintecoPinteco

798313

edited Mar 16 at 22:28

Bernard

123k741117

edited Mar 16 at 22:28

Bernard

123k741117

asked Mar 16 at 22:12

PintecoPinteco

798313

asked Mar 16 at 22:12

PintecoPinteco

798313

1 Answer
1

active

oldest

votes

2

$begingroup$

Given two complex numbers $a,zinmathbbC$, we have $a^z=e^zlog a$. You are correct that for any $a,zinmathbbC$ and $ninmathbbZ$, $left(a^zright)^n=a^zn$, which can be verified by substituting $e^zlog a$. If $n$ is no longer an integer, then it is no longer true because it becomes multivalued.

Using your example, we have
beginalign*
left(-1right)^-i=e^-ilogleft(-1right)=e^-1right=e^-ileft(0+iargleft(-1right)right)=e^-i^2left(pi+2pi kright)=e^pi+2pi k,,kinmathbbZ.
endalign*

share|cite|improve this answer

answered Mar 16 at 22:35

JakeJake

540314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd add a comment that we are taking the branch of logarithm for which $log(-1) = pi+2pi k$, otherwise it doesn't make much sense since it should be a set then.
  $endgroup$
  – Jakobian
  Mar 16 at 23:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry if this a dumb question but... is true that $(a^n)^z=a^zn$ where $n$ is integer and $z$ is complex? I did the following $a^n=e^nln a implies (a^n)^z = e^nzln a = a^nz$. And $arg(z) = -iln(z/|z|)$ is true? This complex part with the arguments, principal branches, multivalues, properties of logarithms/expoents don't apply anymore etc. it's all new and confusing to me, especially because I just read stuff online, I don't have a classes involving this topic.
  $endgroup$
  – Pinteco
  Mar 16 at 23:33
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

Given two complex numbers $a,zinmathbbC$, we have $a^z=e^zlog a$. You are correct that for any $a,zinmathbbC$ and $ninmathbbZ$, $left(a^zright)^n=a^zn$, which can be verified by substituting $e^zlog a$. If $n$ is no longer an integer, then it is no longer true because it becomes multivalued.

Using your example, we have
beginalign*
left(-1right)^-i=e^-ilogleft(-1right)=e^-1right=e^-ileft(0+iargleft(-1right)right)=e^-i^2left(pi+2pi kright)=e^pi+2pi k,,kinmathbbZ.
endalign*

share|cite|improve this answer

answered Mar 16 at 22:35

JakeJake

540314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd add a comment that we are taking the branch of logarithm for which $log(-1) = pi+2pi k$, otherwise it doesn't make much sense since it should be a set then.
  $endgroup$
  – Jakobian
  Mar 16 at 23:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry if this a dumb question but... is true that $(a^n)^z=a^zn$ where $n$ is integer and $z$ is complex? I did the following $a^n=e^nln a implies (a^n)^z = e^nzln a = a^nz$. And $arg(z) = -iln(z/|z|)$ is true? This complex part with the arguments, principal branches, multivalues, properties of logarithms/expoents don't apply anymore etc. it's all new and confusing to me, especially because I just read stuff online, I don't have a classes involving this topic.
  $endgroup$
  – Pinteco
  Mar 16 at 23:33
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Given two complex numbers $a,zinmathbbC$, we have $a^z=e^zlog a$. You are correct that for any $a,zinmathbbC$ and $ninmathbbZ$, $left(a^zright)^n=a^zn$, which can be verified by substituting $e^zlog a$. If $n$ is no longer an integer, then it is no longer true because it becomes multivalued.

Using your example, we have
beginalign*
left(-1right)^-i=e^-ilogleft(-1right)=e^-1right=e^-ileft(0+iargleft(-1right)right)=e^-i^2left(pi+2pi kright)=e^pi+2pi k,,kinmathbbZ.
endalign*

share|cite|improve this answer

answered Mar 16 at 22:35

JakeJake

540314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd add a comment that we are taking the branch of logarithm for which $log(-1) = pi+2pi k$, otherwise it doesn't make much sense since it should be a set then.
  $endgroup$
  – Jakobian
  Mar 16 at 23:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry if this a dumb question but... is true that $(a^n)^z=a^zn$ where $n$ is integer and $z$ is complex? I did the following $a^n=e^nln a implies (a^n)^z = e^nzln a = a^nz$. And $arg(z) = -iln(z/|z|)$ is true? This complex part with the arguments, principal branches, multivalues, properties of logarithms/expoents don't apply anymore etc. it's all new and confusing to me, especially because I just read stuff online, I don't have a classes involving this topic.
  $endgroup$
  – Pinteco
  Mar 16 at 23:33
  
  

  
  
  
  

add a comment | 

$begingroup$

Given two complex numbers $a,zinmathbbC$, we have $a^z=e^zlog a$. You are correct that for any $a,zinmathbbC$ and $ninmathbbZ$, $left(a^zright)^n=a^zn$, which can be verified by substituting $e^zlog a$. If $n$ is no longer an integer, then it is no longer true because it becomes multivalued.

Using your example, we have
beginalign*
left(-1right)^-i=e^-ilogleft(-1right)=e^-1right=e^-ileft(0+iargleft(-1right)right)=e^-i^2left(pi+2pi kright)=e^pi+2pi k,,kinmathbbZ.
endalign*

share|cite|improve this answer

answered Mar 16 at 22:35

JakeJake

540314

$endgroup$

share|cite|improve this answer

answered Mar 16 at 22:35

JakeJake

540314

answered Mar 16 at 22:35

JakeJake

540314

answered Mar 16 at 22:35

JakeJake

540314