Issue with root of unity












0












$begingroup$


I everyone !



I have to show that with $r=frac{p}{q}$ a rationnal number, an $k$ a natural number :



$exp(itimespitimes 2^k times r)$ is a $2q$-th root of unity.



What I did :



$exp(itimespitimes 2^k times r)=(e^{itimespi})^{2^k times r}$
$=((e^{itimespi})^{2^k})^r$



I used the fact that $e^{itimespi}=-1$ and $2^k$ is even :



$((e^{itimespi})^{2^k})^r=((-1)^{2^k})^r$
$=1^r=1$



I know there is a mistake somewhere because for $r=frac{1}{2}$ and $k=1$,



$exp(itimespitimes 2^k times r)=-1$



Can you help me to fix my issue please ?






complex-analysis complex-numbers exponential-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    $exp(2i pi n/m)$ is a primitive $m$-th root of unity iff $gcd(m,n)=1$. In general it is a $km$-th root of unity and a primitive $frac{m}{gcd(m,n)}$-th root of unity.
    $endgroup$
    – reuns
    Jan 4 at 0:05








  • 1




    $begingroup$
    What mistake? -1 is a square root of unity.
    $endgroup$
    – fleablood
    Jan 4 at 0:14










  • $begingroup$
    @reuns can you show me the proof of this ?
    $endgroup$
    – Yazid
    Jan 4 at 14:54
















0












$begingroup$


I everyone !



I have to show that with $r=frac{p}{q}$ a rationnal number, an $k$ a natural number :



$exp(itimespitimes 2^k times r)$ is a $2q$-th root of unity.



What I did :



$exp(itimespitimes 2^k times r)=(e^{itimespi})^{2^k times r}$
$=((e^{itimespi})^{2^k})^r$



I used the fact that $e^{itimespi}=-1$ and $2^k$ is even :



$((e^{itimespi})^{2^k})^r=((-1)^{2^k})^r$
$=1^r=1$



I know there is a mistake somewhere because for $r=frac{1}{2}$ and $k=1$,



$exp(itimespitimes 2^k times r)=-1$



Can you help me to fix my issue please ?






complex-analysis complex-numbers exponential-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    $exp(2i pi n/m)$ is a primitive $m$-th root of unity iff $gcd(m,n)=1$. In general it is a $km$-th root of unity and a primitive $frac{m}{gcd(m,n)}$-th root of unity.
    $endgroup$
    – reuns
    Jan 4 at 0:05








  • 1




    $begingroup$
    What mistake? -1 is a square root of unity.
    $endgroup$
    – fleablood
    Jan 4 at 0:14










  • $begingroup$
    @reuns can you show me the proof of this ?
    $endgroup$
    – Yazid
    Jan 4 at 14:54














0












0








0





$begingroup$


I everyone !



I have to show that with $r=frac{p}{q}$ a rationnal number, an $k$ a natural number :



$exp(itimespitimes 2^k times r)$ is a $2q$-th root of unity.



What I did :



$exp(itimespitimes 2^k times r)=(e^{itimespi})^{2^k times r}$
$=((e^{itimespi})^{2^k})^r$



I used the fact that $e^{itimespi}=-1$ and $2^k$ is even :



$((e^{itimespi})^{2^k})^r=((-1)^{2^k})^r$
$=1^r=1$



I know there is a mistake somewhere because for $r=frac{1}{2}$ and $k=1$,



$exp(itimespitimes 2^k times r)=-1$



Can you help me to fix my issue please ?






complex-analysis complex-numbers exponential-function







share|cite|improve this question











$endgroup$




I everyone !



I have to show that with $r=frac{p}{q}$ a rationnal number, an $k$ a natural number :



$exp(itimespitimes 2^k times r)$ is a $2q$-th root of unity.



What I did :



$exp(itimespitimes 2^k times r)=(e^{itimespi})^{2^k times r}$
$=((e^{itimespi})^{2^k})^r$



I used the fact that $e^{itimespi}=-1$ and $2^k$ is even :



$((e^{itimespi})^{2^k})^r=((-1)^{2^k})^r$
$=1^r=1$



I know there is a mistake somewhere because for $r=frac{1}{2}$ and $k=1$,



$exp(itimespitimes 2^k times r)=-1$



Can you help me to fix my issue please ?






complex-analysis complex-numbers exponential-function




complex-analysis complex-numbers exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 0:16









José Carlos Santos

155k22124227




155k22124227










asked Jan 3 at 23:58









YazidYazid

31




31












  • $begingroup$
    $exp(2i pi n/m)$ is a primitive $m$-th root of unity iff $gcd(m,n)=1$. In general it is a $km$-th root of unity and a primitive $frac{m}{gcd(m,n)}$-th root of unity.
    $endgroup$
    – reuns
    Jan 4 at 0:05








  • 1




    $begingroup$
    What mistake? -1 is a square root of unity.
    $endgroup$
    – fleablood
    Jan 4 at 0:14










  • $begingroup$
    @reuns can you show me the proof of this ?
    $endgroup$
    – Yazid
    Jan 4 at 14:54


















  • $begingroup$
    $exp(2i pi n/m)$ is a primitive $m$-th root of unity iff $gcd(m,n)=1$. In general it is a $km$-th root of unity and a primitive $frac{m}{gcd(m,n)}$-th root of unity.
    $endgroup$
    – reuns
    Jan 4 at 0:05








  • 1




    $begingroup$
    What mistake? -1 is a square root of unity.
    $endgroup$
    – fleablood
    Jan 4 at 0:14










  • $begingroup$
    @reuns can you show me the proof of this ?
    $endgroup$
    – Yazid
    Jan 4 at 14:54
















$begingroup$
$exp(2i pi n/m)$ is a primitive $m$-th root of unity iff $gcd(m,n)=1$. In general it is a $km$-th root of unity and a primitive $frac{m}{gcd(m,n)}$-th root of unity.
$endgroup$
– reuns
Jan 4 at 0:05






$begingroup$
$exp(2i pi n/m)$ is a primitive $m$-th root of unity iff $gcd(m,n)=1$. In general it is a $km$-th root of unity and a primitive $frac{m}{gcd(m,n)}$-th root of unity.
$endgroup$
– reuns
Jan 4 at 0:05






1




1




$begingroup$
What mistake? -1 is a square root of unity.
$endgroup$
– fleablood
Jan 4 at 0:14




$begingroup$
What mistake? -1 is a square root of unity.
$endgroup$
– fleablood
Jan 4 at 0:14












$begingroup$
@reuns can you show me the proof of this ?
$endgroup$
– Yazid
Jan 4 at 14:54




$begingroup$
@reuns can you show me the proof of this ?
$endgroup$
– Yazid
Jan 4 at 14:54










1 Answer
1






active

oldest

votes


















2












$begingroup$

Your mistake lies in assuming that $e^{ab}=(e^a)^b$. This is worst than wrong: it doesn't make sense in general (but it holds if $binmathbb Z$).



You can prove what you wish to prove this way:begin{align}expleft(ipi2^k rright)^{2q}&=expleft(ipi 2^kfrac pqtimes2qright)\&=expleft(ipi 2^{k+1}pright)\&=1.end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you explain why it doesn't make sense in general please ?
    $endgroup$
    – Yazid
    Jan 4 at 12:52










  • $begingroup$
    Of course. The number $e^a$ can be any number $winmathbb{C}setminus{0}$. But then, what is $w^b$? How do you define it?
    $endgroup$
    – José Carlos Santos
    Jan 4 at 13:03










  • $begingroup$
    I provided an answer to that, did I not?
    $endgroup$
    – José Carlos Santos
    Jan 4 at 14:59











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1 Answer
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1 Answer
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active

oldest

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oldest

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2












$begingroup$

Your mistake lies in assuming that $e^{ab}=(e^a)^b$. This is worst than wrong: it doesn't make sense in general (but it holds if $binmathbb Z$).



You can prove what you wish to prove this way:begin{align}expleft(ipi2^k rright)^{2q}&=expleft(ipi 2^kfrac pqtimes2qright)\&=expleft(ipi 2^{k+1}pright)\&=1.end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you explain why it doesn't make sense in general please ?
    $endgroup$
    – Yazid
    Jan 4 at 12:52










  • $begingroup$
    Of course. The number $e^a$ can be any number $winmathbb{C}setminus{0}$. But then, what is $w^b$? How do you define it?
    $endgroup$
    – José Carlos Santos
    Jan 4 at 13:03










  • $begingroup$
    I provided an answer to that, did I not?
    $endgroup$
    – José Carlos Santos
    Jan 4 at 14:59
















2












$begingroup$

Your mistake lies in assuming that $e^{ab}=(e^a)^b$. This is worst than wrong: it doesn't make sense in general (but it holds if $binmathbb Z$).



You can prove what you wish to prove this way:begin{align}expleft(ipi2^k rright)^{2q}&=expleft(ipi 2^kfrac pqtimes2qright)\&=expleft(ipi 2^{k+1}pright)\&=1.end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you explain why it doesn't make sense in general please ?
    $endgroup$
    – Yazid
    Jan 4 at 12:52










  • $begingroup$
    Of course. The number $e^a$ can be any number $winmathbb{C}setminus{0}$. But then, what is $w^b$? How do you define it?
    $endgroup$
    – José Carlos Santos
    Jan 4 at 13:03










  • $begingroup$
    I provided an answer to that, did I not?
    $endgroup$
    – José Carlos Santos
    Jan 4 at 14:59














2












2








2





$begingroup$

Your mistake lies in assuming that $e^{ab}=(e^a)^b$. This is worst than wrong: it doesn't make sense in general (but it holds if $binmathbb Z$).



You can prove what you wish to prove this way:begin{align}expleft(ipi2^k rright)^{2q}&=expleft(ipi 2^kfrac pqtimes2qright)\&=expleft(ipi 2^{k+1}pright)\&=1.end{align}






share|cite|improve this answer









$endgroup$



Your mistake lies in assuming that $e^{ab}=(e^a)^b$. This is worst than wrong: it doesn't make sense in general (but it holds if $binmathbb Z$).



You can prove what you wish to prove this way:begin{align}expleft(ipi2^k rright)^{2q}&=expleft(ipi 2^kfrac pqtimes2qright)\&=expleft(ipi 2^{k+1}pright)\&=1.end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 0:13









José Carlos SantosJosé Carlos Santos

155k22124227




155k22124227












  • $begingroup$
    Can you explain why it doesn't make sense in general please ?
    $endgroup$
    – Yazid
    Jan 4 at 12:52










  • $begingroup$
    Of course. The number $e^a$ can be any number $winmathbb{C}setminus{0}$. But then, what is $w^b$? How do you define it?
    $endgroup$
    – José Carlos Santos
    Jan 4 at 13:03










  • $begingroup$
    I provided an answer to that, did I not?
    $endgroup$
    – José Carlos Santos
    Jan 4 at 14:59


















  • $begingroup$
    Can you explain why it doesn't make sense in general please ?
    $endgroup$
    – Yazid
    Jan 4 at 12:52










  • $begingroup$
    Of course. The number $e^a$ can be any number $winmathbb{C}setminus{0}$. But then, what is $w^b$? How do you define it?
    $endgroup$
    – José Carlos Santos
    Jan 4 at 13:03










  • $begingroup$
    I provided an answer to that, did I not?
    $endgroup$
    – José Carlos Santos
    Jan 4 at 14:59
















$begingroup$
Can you explain why it doesn't make sense in general please ?
$endgroup$
– Yazid
Jan 4 at 12:52




$begingroup$
Can you explain why it doesn't make sense in general please ?
$endgroup$
– Yazid
Jan 4 at 12:52












$begingroup$
Of course. The number $e^a$ can be any number $winmathbb{C}setminus{0}$. But then, what is $w^b$? How do you define it?
$endgroup$
– José Carlos Santos
Jan 4 at 13:03




$begingroup$
Of course. The number $e^a$ can be any number $winmathbb{C}setminus{0}$. But then, what is $w^b$? How do you define it?
$endgroup$
– José Carlos Santos
Jan 4 at 13:03












$begingroup$
I provided an answer to that, did I not?
$endgroup$
– José Carlos Santos
Jan 4 at 14:59




$begingroup$
I provided an answer to that, did I not?
$endgroup$
– José Carlos Santos
Jan 4 at 14:59


















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