algebraic conjugate and sum of roots of unity

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In above lemma, why $|a'| leq 1$ still holds? I didn't see how it relates to "algebraic conjugate of a root of unity is also a root of unity", since $a$ is the sum of unity.



(definition of algebraic conjugate: http://planetmath.org/algebraicconjugates)






abstract-algebra number-theory algebraic-number-theory







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  • Each $|epsilon_i|$. Their average is inside their convex hull, their convex hull is inside the unit disc.
    – Pp..
    Jan 16 '15 at 22:48










  • @Pp.. How is that related to $a'$?
    – CYC
    Jan 16 '15 at 22:50







  • 1




    When you conjugate $a$ you get an average of some $epsilon_i$. You know, the $epsilon_i$ just get moved around. $sigma(fracsum epsilon_in)=fracsumsigma(epsilon_i)n$
    – Pp..
    Jan 16 '15 at 22:57















up vote
1
down vote

favorite












enter image description here



In above lemma, why $|a'| leq 1$ still holds? I didn't see how it relates to "algebraic conjugate of a root of unity is also a root of unity", since $a$ is the sum of unity.



(definition of algebraic conjugate: http://planetmath.org/algebraicconjugates)






abstract-algebra number-theory algebraic-number-theory







share|cite|improve this question























  • Each $|epsilon_i|$. Their average is inside their convex hull, their convex hull is inside the unit disc.
    – Pp..
    Jan 16 '15 at 22:48










  • @Pp.. How is that related to $a'$?
    – CYC
    Jan 16 '15 at 22:50







  • 1




    When you conjugate $a$ you get an average of some $epsilon_i$. You know, the $epsilon_i$ just get moved around. $sigma(fracsum epsilon_in)=fracsumsigma(epsilon_i)n$
    – Pp..
    Jan 16 '15 at 22:57













up vote
1
down vote

favorite









up vote
1
down vote

favorite











enter image description here



In above lemma, why $|a'| leq 1$ still holds? I didn't see how it relates to "algebraic conjugate of a root of unity is also a root of unity", since $a$ is the sum of unity.



(definition of algebraic conjugate: http://planetmath.org/algebraicconjugates)






abstract-algebra number-theory algebraic-number-theory







share|cite|improve this question















enter image description here



In above lemma, why $|a'| leq 1$ still holds? I didn't see how it relates to "algebraic conjugate of a root of unity is also a root of unity", since $a$ is the sum of unity.



(definition of algebraic conjugate: http://planetmath.org/algebraicconjugates)






abstract-algebra number-theory algebraic-number-theory




abstract-algebra number-theory algebraic-number-theory






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edited Jan 16 '15 at 22:56

























asked Jan 16 '15 at 22:43









CYC

940711




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  • Each $|epsilon_i|$. Their average is inside their convex hull, their convex hull is inside the unit disc.
    – Pp..
    Jan 16 '15 at 22:48










  • @Pp.. How is that related to $a'$?
    – CYC
    Jan 16 '15 at 22:50







  • 1




    When you conjugate $a$ you get an average of some $epsilon_i$. You know, the $epsilon_i$ just get moved around. $sigma(fracsum epsilon_in)=fracsumsigma(epsilon_i)n$
    – Pp..
    Jan 16 '15 at 22:57

















  • Each $|epsilon_i|$. Their average is inside their convex hull, their convex hull is inside the unit disc.
    – Pp..
    Jan 16 '15 at 22:48










  • @Pp.. How is that related to $a'$?
    – CYC
    Jan 16 '15 at 22:50







  • 1




    When you conjugate $a$ you get an average of some $epsilon_i$. You know, the $epsilon_i$ just get moved around. $sigma(fracsum epsilon_in)=fracsumsigma(epsilon_i)n$
    – Pp..
    Jan 16 '15 at 22:57
















Each $|epsilon_i|$. Their average is inside their convex hull, their convex hull is inside the unit disc.
– Pp..
Jan 16 '15 at 22:48




Each $|epsilon_i|$. Their average is inside their convex hull, their convex hull is inside the unit disc.
– Pp..
Jan 16 '15 at 22:48












@Pp.. How is that related to $a'$?
– CYC
Jan 16 '15 at 22:50





@Pp.. How is that related to $a'$?
– CYC
Jan 16 '15 at 22:50





1




1




When you conjugate $a$ you get an average of some $epsilon_i$. You know, the $epsilon_i$ just get moved around. $sigma(fracsum epsilon_in)=fracsumsigma(epsilon_i)n$
– Pp..
Jan 16 '15 at 22:57





When you conjugate $a$ you get an average of some $epsilon_i$. You know, the $epsilon_i$ just get moved around. $sigma(fracsum epsilon_in)=fracsumsigma(epsilon_i)n$
– Pp..
Jan 16 '15 at 22:57











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Just use the triangle inequality.



(Each conjugate of each $varepsilon_i$ is a root of unity, and so has absolute value $1$.)






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    We have $sigma (sum epsilon_i/n) = sumsigma(epsilon_i)/n$, since each $|sigma(epsilon_i)|=1$, $|a'|=|sigma(a)|=1$.






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      2 Answers
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      up vote
      0
      down vote













      Just use the triangle inequality.



      (Each conjugate of each $varepsilon_i$ is a root of unity, and so has absolute value $1$.)






      share|cite|improve this answer
























        up vote
        0
        down vote













        Just use the triangle inequality.



        (Each conjugate of each $varepsilon_i$ is a root of unity, and so has absolute value $1$.)






        share|cite|improve this answer






















          up vote
          0
          down vote










          up vote
          0
          down vote









          Just use the triangle inequality.



          (Each conjugate of each $varepsilon_i$ is a root of unity, and so has absolute value $1$.)






          share|cite|improve this answer












          Just use the triangle inequality.



          (Each conjugate of each $varepsilon_i$ is a root of unity, and so has absolute value $1$.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 17 '15 at 2:23









          tracing

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              We have $sigma (sum epsilon_i/n) = sumsigma(epsilon_i)/n$, since each $|sigma(epsilon_i)|=1$, $|a'|=|sigma(a)|=1$.






              share|cite|improve this answer
























                up vote
                0
                down vote



                accepted










                We have $sigma (sum epsilon_i/n) = sumsigma(epsilon_i)/n$, since each $|sigma(epsilon_i)|=1$, $|a'|=|sigma(a)|=1$.






                share|cite|improve this answer






















                  up vote
                  0
                  down vote



                  accepted







                  up vote
                  0
                  down vote



                  accepted






                  We have $sigma (sum epsilon_i/n) = sumsigma(epsilon_i)/n$, since each $|sigma(epsilon_i)|=1$, $|a'|=|sigma(a)|=1$.






                  share|cite|improve this answer












                  We have $sigma (sum epsilon_i/n) = sumsigma(epsilon_i)/n$, since each $|sigma(epsilon_i)|=1$, $|a'|=|sigma(a)|=1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 9 at 11:47









                  CYC

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