Mixing
Issue with root of unity
$begingroup$
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
edited Jan 4 at 0:16
José Carlos Santos
155k22124227
asked Jan 3 at 23:58
YazidYazid
31
$endgroup$
add a comment |
$begingroup$
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
edited Jan 4 at 0:16
José Carlos Santos
155k22124227
asked Jan 3 at 23:58
YazidYazid
31
$endgroup$
$begingroup$
$exp(2i pi n/m)$ 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
add a comment |
$begingroup$
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
edited Jan 4 at 0:16
José Carlos Santos
155k22124227
asked Jan 3 at 23:58
YazidYazid
31
$endgroup$
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
edited Jan 4 at 0:16
José Carlos Santos
155k22124227
asked Jan 3 at 23:58
YazidYazid
31
edited Jan 4 at 0:16
José Carlos Santos
155k22124227
asked Jan 3 at 23:58
YazidYazid
31
edited Jan 4 at 0:16
José Carlos Santos
155k22124227
edited Jan 4 at 0:16
José Carlos Santos
155k22124227
edited Jan 4 at 0:16
José Carlos Santos
155k22124227
155k22124227
asked Jan 3 at 23:58
YazidYazid
31
asked Jan 3 at 23:58
YazidYazid
31
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
add a comment |
$begingroup$
$exp(2i pi n/m)$ 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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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}
answered Jan 4 at 0:13
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
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}
answered Jan 4 at 0:13
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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}
answered Jan 4 at 0:13
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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}
answered Jan 4 at 0:13
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
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}
answered Jan 4 at 0:13
José Carlos SantosJosé Carlos Santos
155k22124227
answered Jan 4 at 0:13
José Carlos SantosJosé Carlos Santos
155k22124227
answered Jan 4 at 0:13
José Carlos SantosJosé Carlos Santos
155k22124227
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
$exp(2i pi n/m)$ 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