How do the roots of unity relate to De Moivre's Theorem?
How do the roots of unity relate to De Moivre's Theorem? Others always pair the two up, but I do not understand why.
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How do the roots of unity relate to De Moivre's Theorem? Others always pair the two up, but I do not understand why.
algebra-precalculus
edited yesterday
Blue
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asked yesterday
M. C.
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M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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How do the roots of unity relate to De Moivre's Theorem? Others always pair the two up, but I do not understand why.
algebra-precalculus
edited yesterday
Blue
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asked yesterday
M. C.
213
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M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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How do the roots of unity relate to De Moivre's Theorem? Others always pair the two up, but I do not understand why.
algebra-precalculus
algebra-precalculus
edited yesterday
Blue
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M. C.
213
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Blue
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Blue
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Blue
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M. C.
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asked yesterday
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213
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If we let the $n$th root of unity be $$z=cos(theta)+icdot sin(theta)$$
then by De Moivre's Theorem$$z^n=(cos(theta)+icdot sin(theta))^n$$$$=cos(ncdot theta)+icdot sin(ncdot theta)$$where $ninmathbb{N}$. But:$$z^n=1=cos(2picdot m)+icdot sin(pi+2picdot m)$$ Where $minmathbb{Z}$. So, equating real parts,$$cos(ncdot theta)=cos(2picdot m)$$$$thereforetheta=frac{2picdot m}{n}$$
i.e. the $n$th roots of unity are $$z=cos(frac{2picdot m}{n})+icdot sin(frac{2picdot m}{n})$$Where $0le m le n-1$ as all other values of $m$ will produce the same values of $z$.
answered yesterday
Peter Foreman
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In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root".
– DanielWainfleet
yesterday
add a comment |
The roots of unity are all in the form $cos(x)+isin(x)$. Nth roots of unity, raised to the n power, will be 1. Thus, De Moivre's Theorem, which states that $cis(x)^n=cis(nx)$ also tells us that if $cis(x)$ is a nth root of unity, then $cis(x)^n=cis(nx)=1$. This process can also be reversed to find nth roots of unity, as, substituting in a n value, we then have a trig equation we can solve to find the values of x, and the nth roots of unity, $cis(x)$.
answered yesterday
H Huang
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2 Answers
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2 Answers
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If we let the $n$th root of unity be $$z=cos(theta)+icdot sin(theta)$$
then by De Moivre's Theorem$$z^n=(cos(theta)+icdot sin(theta))^n$$$$=cos(ncdot theta)+icdot sin(ncdot theta)$$where $ninmathbb{N}$. But:$$z^n=1=cos(2picdot m)+icdot sin(pi+2picdot m)$$ Where $minmathbb{Z}$. So, equating real parts,$$cos(ncdot theta)=cos(2picdot m)$$$$thereforetheta=frac{2picdot m}{n}$$
i.e. the $n$th roots of unity are $$z=cos(frac{2picdot m}{n})+icdot sin(frac{2picdot m}{n})$$Where $0le m le n-1$ as all other values of $m$ will produce the same values of $z$.
answered yesterday
Peter Foreman
2056
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In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root".
– DanielWainfleet
yesterday
add a comment |
If we let the $n$th root of unity be $$z=cos(theta)+icdot sin(theta)$$
then by De Moivre's Theorem$$z^n=(cos(theta)+icdot sin(theta))^n$$$$=cos(ncdot theta)+icdot sin(ncdot theta)$$where $ninmathbb{N}$. But:$$z^n=1=cos(2picdot m)+icdot sin(pi+2picdot m)$$ Where $minmathbb{Z}$. So, equating real parts,$$cos(ncdot theta)=cos(2picdot m)$$$$thereforetheta=frac{2picdot m}{n}$$
i.e. the $n$th roots of unity are $$z=cos(frac{2picdot m}{n})+icdot sin(frac{2picdot m}{n})$$Where $0le m le n-1$ as all other values of $m$ will produce the same values of $z$.
answered yesterday
Peter Foreman
2056
New contributor
Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root".
– DanielWainfleet
yesterday
add a comment |
If we let the $n$th root of unity be $$z=cos(theta)+icdot sin(theta)$$
then by De Moivre's Theorem$$z^n=(cos(theta)+icdot sin(theta))^n$$$$=cos(ncdot theta)+icdot sin(ncdot theta)$$where $ninmathbb{N}$. But:$$z^n=1=cos(2picdot m)+icdot sin(pi+2picdot m)$$ Where $minmathbb{Z}$. So, equating real parts,$$cos(ncdot theta)=cos(2picdot m)$$$$thereforetheta=frac{2picdot m}{n}$$
i.e. the $n$th roots of unity are $$z=cos(frac{2picdot m}{n})+icdot sin(frac{2picdot m}{n})$$Where $0le m le n-1$ as all other values of $m$ will produce the same values of $z$.
answered yesterday
Peter Foreman
2056
New contributor
Peter Foreman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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If we let the $n$th root of unity be $$z=cos(theta)+icdot sin(theta)$$
then by De Moivre's Theorem$$z^n=(cos(theta)+icdot sin(theta))^n$$$$=cos(ncdot theta)+icdot sin(ncdot theta)$$where $ninmathbb{N}$. But:$$z^n=1=cos(2picdot m)+icdot sin(pi+2picdot m)$$ Where $minmathbb{Z}$. So, equating real parts,$$cos(ncdot theta)=cos(2picdot m)$$$$thereforetheta=frac{2picdot m}{n}$$
i.e. the $n$th roots of unity are $$z=cos(frac{2picdot m}{n})+icdot sin(frac{2picdot m}{n})$$Where $0le m le n-1$ as all other values of $m$ will produce the same values of $z$.
answered yesterday
Peter Foreman
2056
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edited yesterday
answered yesterday
Peter Foreman
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answered yesterday
Peter Foreman
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answered yesterday
Peter Foreman
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In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root".
– DanielWainfleet
yesterday
add a comment |
In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root".
– DanielWainfleet
yesterday
In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root".
– DanielWainfleet
yesterday
In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root".
– DanielWainfleet
yesterday
add a comment |
The roots of unity are all in the form $cos(x)+isin(x)$. Nth roots of unity, raised to the n power, will be 1. Thus, De Moivre's Theorem, which states that $cis(x)^n=cis(nx)$ also tells us that if $cis(x)$ is a nth root of unity, then $cis(x)^n=cis(nx)=1$. This process can also be reversed to find nth roots of unity, as, substituting in a n value, we then have a trig equation we can solve to find the values of x, and the nth roots of unity, $cis(x)$.
answered yesterday
H Huang
437
add a comment |
The roots of unity are all in the form $cos(x)+isin(x)$. Nth roots of unity, raised to the n power, will be 1. Thus, De Moivre's Theorem, which states that $cis(x)^n=cis(nx)$ also tells us that if $cis(x)$ is a nth root of unity, then $cis(x)^n=cis(nx)=1$. This process can also be reversed to find nth roots of unity, as, substituting in a n value, we then have a trig equation we can solve to find the values of x, and the nth roots of unity, $cis(x)$.
answered yesterday
H Huang
437
add a comment |
The roots of unity are all in the form $cos(x)+isin(x)$. Nth roots of unity, raised to the n power, will be 1. Thus, De Moivre's Theorem, which states that $cis(x)^n=cis(nx)$ also tells us that if $cis(x)$ is a nth root of unity, then $cis(x)^n=cis(nx)=1$. This process can also be reversed to find nth roots of unity, as, substituting in a n value, we then have a trig equation we can solve to find the values of x, and the nth roots of unity, $cis(x)$.
answered yesterday
H Huang
437
The roots of unity are all in the form $cos(x)+isin(x)$. Nth roots of unity, raised to the n power, will be 1. Thus, De Moivre's Theorem, which states that $cis(x)^n=cis(nx)$ also tells us that if $cis(x)$ is a nth root of unity, then $cis(x)^n=cis(nx)=1$. This process can also be reversed to find nth roots of unity, as, substituting in a n value, we then have a trig equation we can solve to find the values of x, and the nth roots of unity, $cis(x)$.
answered yesterday
H Huang
437
answered yesterday
H Huang
437
answered yesterday
H Huang
437
answered yesterday
H Huang
437
437
add a comment |
add a comment |
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