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Show the nth root of unity are the vertice of regular polygon and a formula for the perimeter…
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1) Show the nth root of unity are the vertice of regular polygon
2) Find the formula for the perimeter of that polygon called "ln" and prove
$lim_nrightarrow infty l_n=2pi$
My attempt
Let $z=1$such that $zin mathbbC$. We need the nth root of the unity.
Let $win mathbbC$. such that $w^frac1n=z$
Then
$w_k=cos(fractheta+2kpin)+isin(fractheta+2kpin),,,,(1)$ for $k=0,1,...,n-1$
As $z=1$ then $theta=0$. Replacing in $(1)$ we have:
$w_k=cos(frac2kpin)+isin(frac2kpin)=e^ifrac2kpin,,,,(2)$
Here, i'm stuck.
I make a graphich representation for $k=4$ and is a polygin of four vertices. But for $n$ i'm stuck.
For the 2) question i don't have idea. Can someone help me?
complex-analysis
asked 4 hours ago
Bvss12
1,700516
add a comment |Â
up vote
3
down vote
favorite
1) Show the nth root of unity are the vertice of regular polygon
2) Find the formula for the perimeter of that polygon called "ln" and prove
$lim_nrightarrow infty l_n=2pi$
My attempt
Let $z=1$such that $zin mathbbC$. We need the nth root of the unity.
Let $win mathbbC$. such that $w^frac1n=z$
Then
$w_k=cos(fractheta+2kpin)+isin(fractheta+2kpin),,,,(1)$ for $k=0,1,...,n-1$
As $z=1$ then $theta=0$. Replacing in $(1)$ we have:
$w_k=cos(frac2kpin)+isin(frac2kpin)=e^ifrac2kpin,,,,(2)$
Here, i'm stuck.
I make a graphich representation for $k=4$ and is a polygin of four vertices. But for $n$ i'm stuck.
For the 2) question i don't have idea. Can someone help me?
complex-analysis
asked 4 hours ago
Bvss12
1,700516
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
1) Show the nth root of unity are the vertice of regular polygon
2) Find the formula for the perimeter of that polygon called "ln" and prove
$lim_nrightarrow infty l_n=2pi$
My attempt
Let $z=1$such that $zin mathbbC$. We need the nth root of the unity.
Let $win mathbbC$. such that $w^frac1n=z$
Then
$w_k=cos(fractheta+2kpin)+isin(fractheta+2kpin),,,,(1)$ for $k=0,1,...,n-1$
As $z=1$ then $theta=0$. Replacing in $(1)$ we have:
$w_k=cos(frac2kpin)+isin(frac2kpin)=e^ifrac2kpin,,,,(2)$
Here, i'm stuck.
I make a graphich representation for $k=4$ and is a polygin of four vertices. But for $n$ i'm stuck.
For the 2) question i don't have idea. Can someone help me?
complex-analysis
asked 4 hours ago
Bvss12
1,700516
1) Show the nth root of unity are the vertice of regular polygon
2) Find the formula for the perimeter of that polygon called "ln" and prove
$lim_nrightarrow infty l_n=2pi$
My attempt
Let $z=1$such that $zin mathbbC$. We need the nth root of the unity.
Let $win mathbbC$. such that $w^frac1n=z$
Then
$w_k=cos(fractheta+2kpin)+isin(fractheta+2kpin),,,,(1)$ for $k=0,1,...,n-1$
As $z=1$ then $theta=0$. Replacing in $(1)$ we have:
$w_k=cos(frac2kpin)+isin(frac2kpin)=e^ifrac2kpin,,,,(2)$
Here, i'm stuck.
I make a graphich representation for $k=4$ and is a polygin of four vertices. But for $n$ i'm stuck.
For the 2) question i don't have idea. Can someone help me?
complex-analysis
complex-analysis
asked 4 hours ago
Bvss12
1,700516
asked 4 hours ago
Bvss12
1,700516
asked 4 hours ago
Bvss12
1,700516
asked 4 hours ago
Bvss12
1,700516
asked 4 hours ago
Bvss12
1,700516
1,700516
add a comment |Â
add a comment |Â
2 Answers
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You want to show that the angle between $e^2pi k i /n$ and $e^2pi (k+1)i/n$ is constant.
Note that if you divide these two complex numbers you get the resulting angle of rotation between the two. $$ frac e^2pi (k+1)i/ne^2pi ki /n= e^2pi i/n$$ which is the same for all $k$, that is they are vertices of a regular polygon, considering that they all have unit length.
For the side-length of the polygon you need to find the norm of the difference of two consecutive roots, for example $$|1-e^2pi i/n|$$
Multiply the result by n and let n goes to $infty$ to get your $2pi$
answered 3 hours ago
Mohammad Riazi-Kermani
37.3k41957
Excellent. Thanks for your answer! Very clear
– Bvss12
3 hours ago
@Bvss12 Thanks for your attention and understanding. I am glad you liked the answer.
– Mohammad Riazi-Kermani
3 hours ago
add a comment |Â
up vote
2
down vote
In De Moivre's formula, you see that the argument of two consecutive roots differ of $2pi/n$. Since the module of every root is equal to $1$, the points are distributed lie on the circumference $|z|=1$ and the angular distance between two consecutive points is $2pi/n$. Therefore the points are vertex of a regular $n$-agon. By the chord theorem, the length of the $n$-agon is $2sin(pi/n)$. Therefore $$ l_n=2nsin(pi/n)=2picdotfracsin(pi/n)pi/noversetnto inftyto 2pi. $$
answered 4 hours ago
Maurizio Moreschi
1546
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You want to show that the angle between $e^2pi k i /n$ and $e^2pi (k+1)i/n$ is constant.
Note that if you divide these two complex numbers you get the resulting angle of rotation between the two. $$ frac e^2pi (k+1)i/ne^2pi ki /n= e^2pi i/n$$ which is the same for all $k$, that is they are vertices of a regular polygon, considering that they all have unit length.
For the side-length of the polygon you need to find the norm of the difference of two consecutive roots, for example $$|1-e^2pi i/n|$$
Multiply the result by n and let n goes to $infty$ to get your $2pi$
answered 3 hours ago
Mohammad Riazi-Kermani
37.3k41957
Excellent. Thanks for your answer! Very clear
– Bvss12
3 hours ago
@Bvss12 Thanks for your attention and understanding. I am glad you liked the answer.
– Mohammad Riazi-Kermani
3 hours ago
add a comment |Â
up vote
3
down vote
accepted
You want to show that the angle between $e^2pi k i /n$ and $e^2pi (k+1)i/n$ is constant.
Note that if you divide these two complex numbers you get the resulting angle of rotation between the two. $$ frac e^2pi (k+1)i/ne^2pi ki /n= e^2pi i/n$$ which is the same for all $k$, that is they are vertices of a regular polygon, considering that they all have unit length.
For the side-length of the polygon you need to find the norm of the difference of two consecutive roots, for example $$|1-e^2pi i/n|$$
Multiply the result by n and let n goes to $infty$ to get your $2pi$
answered 3 hours ago
Mohammad Riazi-Kermani
37.3k41957
Excellent. Thanks for your answer! Very clear
– Bvss12
3 hours ago
@Bvss12 Thanks for your attention and understanding. I am glad you liked the answer.
– Mohammad Riazi-Kermani
3 hours ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You want to show that the angle between $e^2pi k i /n$ and $e^2pi (k+1)i/n$ is constant.
Note that if you divide these two complex numbers you get the resulting angle of rotation between the two. $$ frac e^2pi (k+1)i/ne^2pi ki /n= e^2pi i/n$$ which is the same for all $k$, that is they are vertices of a regular polygon, considering that they all have unit length.
For the side-length of the polygon you need to find the norm of the difference of two consecutive roots, for example $$|1-e^2pi i/n|$$
Multiply the result by n and let n goes to $infty$ to get your $2pi$
answered 3 hours ago
Mohammad Riazi-Kermani
37.3k41957
You want to show that the angle between $e^2pi k i /n$ and $e^2pi (k+1)i/n$ is constant.
Note that if you divide these two complex numbers you get the resulting angle of rotation between the two. $$ frac e^2pi (k+1)i/ne^2pi ki /n= e^2pi i/n$$ which is the same for all $k$, that is they are vertices of a regular polygon, considering that they all have unit length.
For the side-length of the polygon you need to find the norm of the difference of two consecutive roots, for example $$|1-e^2pi i/n|$$
Multiply the result by n and let n goes to $infty$ to get your $2pi$
answered 3 hours ago
Mohammad Riazi-Kermani
37.3k41957
answered 3 hours ago
Mohammad Riazi-Kermani
37.3k41957
answered 3 hours ago
Mohammad Riazi-Kermani
37.3k41957
answered 3 hours ago
Mohammad Riazi-Kermani
37.3k41957
37.3k41957
Excellent. Thanks for your answer! Very clear
– Bvss12
3 hours ago
@Bvss12 Thanks for your attention and understanding. I am glad you liked the answer.
– Mohammad Riazi-Kermani
3 hours ago
add a comment |Â
Excellent. Thanks for your answer! Very clear
– Bvss12
3 hours ago
@Bvss12 Thanks for your attention and understanding. I am glad you liked the answer.
– Mohammad Riazi-Kermani
3 hours ago
Excellent. Thanks for your answer! Very clear
– Bvss12
3 hours ago
Excellent. Thanks for your answer! Very clear
– Bvss12
3 hours ago
@Bvss12 Thanks for your attention and understanding. I am glad you liked the answer.
– Mohammad Riazi-Kermani
3 hours ago
@Bvss12 Thanks for your attention and understanding. I am glad you liked the answer.
– Mohammad Riazi-Kermani
3 hours ago
add a comment |Â
up vote
2
down vote
In De Moivre's formula, you see that the argument of two consecutive roots differ of $2pi/n$. Since the module of every root is equal to $1$, the points are distributed lie on the circumference $|z|=1$ and the angular distance between two consecutive points is $2pi/n$. Therefore the points are vertex of a regular $n$-agon. By the chord theorem, the length of the $n$-agon is $2sin(pi/n)$. Therefore $$ l_n=2nsin(pi/n)=2picdotfracsin(pi/n)pi/noversetnto inftyto 2pi. $$
answered 4 hours ago
Maurizio Moreschi
1546
add a comment |Â
up vote
2
down vote
In De Moivre's formula, you see that the argument of two consecutive roots differ of $2pi/n$. Since the module of every root is equal to $1$, the points are distributed lie on the circumference $|z|=1$ and the angular distance between two consecutive points is $2pi/n$. Therefore the points are vertex of a regular $n$-agon. By the chord theorem, the length of the $n$-agon is $2sin(pi/n)$. Therefore $$ l_n=2nsin(pi/n)=2picdotfracsin(pi/n)pi/noversetnto inftyto 2pi. $$
answered 4 hours ago
Maurizio Moreschi
1546
add a comment |Â
up vote
2
down vote
up vote
2
down vote
In De Moivre's formula, you see that the argument of two consecutive roots differ of $2pi/n$. Since the module of every root is equal to $1$, the points are distributed lie on the circumference $|z|=1$ and the angular distance between two consecutive points is $2pi/n$. Therefore the points are vertex of a regular $n$-agon. By the chord theorem, the length of the $n$-agon is $2sin(pi/n)$. Therefore $$ l_n=2nsin(pi/n)=2picdotfracsin(pi/n)pi/noversetnto inftyto 2pi. $$
answered 4 hours ago
Maurizio Moreschi
1546
In De Moivre's formula, you see that the argument of two consecutive roots differ of $2pi/n$. Since the module of every root is equal to $1$, the points are distributed lie on the circumference $|z|=1$ and the angular distance between two consecutive points is $2pi/n$. Therefore the points are vertex of a regular $n$-agon. By the chord theorem, the length of the $n$-agon is $2sin(pi/n)$. Therefore $$ l_n=2nsin(pi/n)=2picdotfracsin(pi/n)pi/noversetnto inftyto 2pi. $$
answered 4 hours ago
Maurizio Moreschi
1546
answered 4 hours ago
Maurizio Moreschi
1546
answered 4 hours ago
Maurizio Moreschi
1546
answered 4 hours ago
Maurizio Moreschi
1546
1546
add a comment |Â
add a comment |Â
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