Issue with root of unity












0














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







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Yazid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $exp(2i pi n/m)$ 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.
    – reuns
    Jan 4 at 0:05








  • 1




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










  • @reuns can you show me the proof of this ?
    – Yazid
    2 days ago
















0














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









New contributor




Yazid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • $exp(2i pi n/m)$ 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.
    – reuns
    Jan 4 at 0:05








  • 1




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










  • @reuns can you show me the proof of this ?
    – Yazid
    2 days ago














0












0








0







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









New contributor




Yazid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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









New contributor




Yazid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Yazid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 0:16









José Carlos Santos

152k22123225




152k22123225






New contributor




Yazid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Jan 3 at 23:58









Yazid

31




31




New contributor




Yazid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Yazid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Yazid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $exp(2i pi n/m)$ 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.
    – reuns
    Jan 4 at 0:05








  • 1




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










  • @reuns can you show me the proof of this ?
    – Yazid
    2 days ago


















  • $exp(2i pi n/m)$ 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.
    – reuns
    Jan 4 at 0:05








  • 1




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










  • @reuns can you show me the proof of this ?
    – Yazid
    2 days ago
















$exp(2i pi n/m)$ 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.
– reuns
Jan 4 at 0:05






$exp(2i pi n/m)$ 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.
– reuns
Jan 4 at 0:05






1




1




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




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












@reuns can you show me the proof of this ?
– Yazid
2 days ago




@reuns can you show me the proof of this ?
– Yazid
2 days ago










1 Answer
1






active

oldest

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2














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





















  • Can you explain why it doesn't make sense in general please ?
    – Yazid
    2 days ago










  • 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?
    – José Carlos Santos
    2 days ago










  • I provided an answer to that, did I not?
    – José Carlos Santos
    2 days ago











Your Answer





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1 Answer
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active

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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





















  • Can you explain why it doesn't make sense in general please ?
    – Yazid
    2 days ago










  • 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?
    – José Carlos Santos
    2 days ago










  • I provided an answer to that, did I not?
    – José Carlos Santos
    2 days ago
















2














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





















  • Can you explain why it doesn't make sense in general please ?
    – Yazid
    2 days ago










  • 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?
    – José Carlos Santos
    2 days ago










  • I provided an answer to that, did I not?
    – José Carlos Santos
    2 days ago














2












2








2






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












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 Santos

152k22123225




152k22123225












  • Can you explain why it doesn't make sense in general please ?
    – Yazid
    2 days ago










  • 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?
    – José Carlos Santos
    2 days ago










  • I provided an answer to that, did I not?
    – José Carlos Santos
    2 days ago


















  • Can you explain why it doesn't make sense in general please ?
    – Yazid
    2 days ago










  • 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?
    – José Carlos Santos
    2 days ago










  • I provided an answer to that, did I not?
    – José Carlos Santos
    2 days ago
















Can you explain why it doesn't make sense in general please ?
– Yazid
2 days ago




Can you explain why it doesn't make sense in general please ?
– Yazid
2 days ago












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?
– José Carlos Santos
2 days ago




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?
– José Carlos Santos
2 days ago












I provided an answer to that, did I not?
– José Carlos Santos
2 days ago




I provided an answer to that, did I not?
– José Carlos Santos
2 days ago










Yazid is a new contributor. Be nice, and check out our Code of Conduct.










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