Order of a Complex Number?












2














Bit stuck on this question: The aim of this exercise is to determine all elements of finite order in $mathbb{C}^{∗}$, the multiplicative group of non-zero complex numbers.



What is the order of $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})$?



I know $z$ can be of the form $z=re^{itheta}$ but don't know what to do to determine the 'order' or what I can do to the formula to solve this.



Any help would be great, thanks.






abstract-algebra group-theory







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  • I haven't heard of complex numbers being referred to as having any such property. Do you have any more context? What did previous parts of the question say for instance?
    – John Doe
    1 hour ago












  • This is the only previous part of the question, after it just states De Moivres Theorem. The aim of this exercise is to determine all elements of finite order in $C^{∗}$, the multiplicative group of non-zero complex numbers.@John Doe
    – Reety
    1 hour ago








  • 1




    Ahh ok - now that you mention group theory, this may make more sense. The order of a group element is the lowest (integer) power to which that element has to be raised for you to reach the identity. In your group, this is any $ainBbb C$ such that $a^n=1$ for $ninBbb N$. Then the order of your element $a$ will be $n$.
    – John Doe
    57 mins ago
















2














Bit stuck on this question: The aim of this exercise is to determine all elements of finite order in $mathbb{C}^{∗}$, the multiplicative group of non-zero complex numbers.



What is the order of $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})$?



I know $z$ can be of the form $z=re^{itheta}$ but don't know what to do to determine the 'order' or what I can do to the formula to solve this.



Any help would be great, thanks.






abstract-algebra group-theory







share|cite|improve this question
























  • I haven't heard of complex numbers being referred to as having any such property. Do you have any more context? What did previous parts of the question say for instance?
    – John Doe
    1 hour ago












  • This is the only previous part of the question, after it just states De Moivres Theorem. The aim of this exercise is to determine all elements of finite order in $C^{∗}$, the multiplicative group of non-zero complex numbers.@John Doe
    – Reety
    1 hour ago








  • 1




    Ahh ok - now that you mention group theory, this may make more sense. The order of a group element is the lowest (integer) power to which that element has to be raised for you to reach the identity. In your group, this is any $ainBbb C$ such that $a^n=1$ for $ninBbb N$. Then the order of your element $a$ will be $n$.
    – John Doe
    57 mins ago














2












2








2


1





Bit stuck on this question: The aim of this exercise is to determine all elements of finite order in $mathbb{C}^{∗}$, the multiplicative group of non-zero complex numbers.



What is the order of $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})$?



I know $z$ can be of the form $z=re^{itheta}$ but don't know what to do to determine the 'order' or what I can do to the formula to solve this.



Any help would be great, thanks.






abstract-algebra group-theory







share|cite|improve this question















Bit stuck on this question: The aim of this exercise is to determine all elements of finite order in $mathbb{C}^{∗}$, the multiplicative group of non-zero complex numbers.



What is the order of $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})$?



I know $z$ can be of the form $z=re^{itheta}$ but don't know what to do to determine the 'order' or what I can do to the formula to solve this.



Any help would be great, thanks.






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 10 mins ago









Gaby Alfonso

674315




674315










asked 1 hour ago









Reety

12910




12910












  • I haven't heard of complex numbers being referred to as having any such property. Do you have any more context? What did previous parts of the question say for instance?
    – John Doe
    1 hour ago












  • This is the only previous part of the question, after it just states De Moivres Theorem. The aim of this exercise is to determine all elements of finite order in $C^{∗}$, the multiplicative group of non-zero complex numbers.@John Doe
    – Reety
    1 hour ago








  • 1




    Ahh ok - now that you mention group theory, this may make more sense. The order of a group element is the lowest (integer) power to which that element has to be raised for you to reach the identity. In your group, this is any $ainBbb C$ such that $a^n=1$ for $ninBbb N$. Then the order of your element $a$ will be $n$.
    – John Doe
    57 mins ago


















  • I haven't heard of complex numbers being referred to as having any such property. Do you have any more context? What did previous parts of the question say for instance?
    – John Doe
    1 hour ago












  • This is the only previous part of the question, after it just states De Moivres Theorem. The aim of this exercise is to determine all elements of finite order in $C^{∗}$, the multiplicative group of non-zero complex numbers.@John Doe
    – Reety
    1 hour ago








  • 1




    Ahh ok - now that you mention group theory, this may make more sense. The order of a group element is the lowest (integer) power to which that element has to be raised for you to reach the identity. In your group, this is any $ainBbb C$ such that $a^n=1$ for $ninBbb N$. Then the order of your element $a$ will be $n$.
    – John Doe
    57 mins ago
















I haven't heard of complex numbers being referred to as having any such property. Do you have any more context? What did previous parts of the question say for instance?
– John Doe
1 hour ago






I haven't heard of complex numbers being referred to as having any such property. Do you have any more context? What did previous parts of the question say for instance?
– John Doe
1 hour ago














This is the only previous part of the question, after it just states De Moivres Theorem. The aim of this exercise is to determine all elements of finite order in $C^{∗}$, the multiplicative group of non-zero complex numbers.@John Doe
– Reety
1 hour ago






This is the only previous part of the question, after it just states De Moivres Theorem. The aim of this exercise is to determine all elements of finite order in $C^{∗}$, the multiplicative group of non-zero complex numbers.@John Doe
– Reety
1 hour ago






1




1




Ahh ok - now that you mention group theory, this may make more sense. The order of a group element is the lowest (integer) power to which that element has to be raised for you to reach the identity. In your group, this is any $ainBbb C$ such that $a^n=1$ for $ninBbb N$. Then the order of your element $a$ will be $n$.
– John Doe
57 mins ago




Ahh ok - now that you mention group theory, this may make more sense. The order of a group element is the lowest (integer) power to which that element has to be raised for you to reach the identity. In your group, this is any $ainBbb C$ such that $a^n=1$ for $ninBbb N$. Then the order of your element $a$ will be $n$.
– John Doe
57 mins ago










3 Answers
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Hint: $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{i2pi/3}$. You need to find the smallest $nin Bbb N $ such that $z^n=1$,i.e, $e^{in2pi/3}=1$. Note that $e^{i2pi}=1$. Can you find such an $n $?






share|cite|improve this answer



















  • 1




    Thank you very much, its when n=3 so if I'm correct this must mean it is of order 3?
    – Reety
    52 mins ago










  • Yes. ${}{} {}{}{}$
    – Thomas Shelby
    46 mins ago



















1














If the order of z is the smallest positive integer n such that $z^n=1$, then this is fairly simple.



z has a finite order iff |z| = 1, and arg(z)=$pi cdot q$ where q is a rational number. To find the order, just find the smallest positive integer n such that $q cdot n = 2k$ for some integer k. This is because multiplying the complex numbers, their arguments add together. Hopefully this makes sense and I interpreted the question right.






share|cite|edit








New contributor




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


























    0














    Find the smallest natural $n$ so that $z^n = 1$.



    $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{ifrac {2pi}{3}}$



    So $z^n = e^{ifrac {2npi} 3}$



    Now $e^{i 2kpi} = 1$ for all integer $k$ and $e^{itheta} = 1$ only if $theta = 2kpi$ for some integer $k$.



    So $z^n = e^{ifrac {2npi} 3} = 1 iff frac {2npi}3 = 2pi*k$ for some integer $k$.



    So we need to find then smallest natural $n$ so and $frac {2npi}3 = 2pi *k$ for some integer value of $k$.



    That's.... not a hard question.






    share|cite|improve this answer





















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      3 Answers
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      active

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      3 Answers
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      1














      Hint: $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{i2pi/3}$. You need to find the smallest $nin Bbb N $ such that $z^n=1$,i.e, $e^{in2pi/3}=1$. Note that $e^{i2pi}=1$. Can you find such an $n $?






      share|cite|improve this answer



















      • 1




        Thank you very much, its when n=3 so if I'm correct this must mean it is of order 3?
        – Reety
        52 mins ago










      • Yes. ${}{} {}{}{}$
        – Thomas Shelby
        46 mins ago
















      1














      Hint: $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{i2pi/3}$. You need to find the smallest $nin Bbb N $ such that $z^n=1$,i.e, $e^{in2pi/3}=1$. Note that $e^{i2pi}=1$. Can you find such an $n $?






      share|cite|improve this answer



















      • 1




        Thank you very much, its when n=3 so if I'm correct this must mean it is of order 3?
        – Reety
        52 mins ago










      • Yes. ${}{} {}{}{}$
        – Thomas Shelby
        46 mins ago














      1












      1








      1






      Hint: $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{i2pi/3}$. You need to find the smallest $nin Bbb N $ such that $z^n=1$,i.e, $e^{in2pi/3}=1$. Note that $e^{i2pi}=1$. Can you find such an $n $?






      share|cite|improve this answer














      Hint: $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{i2pi/3}$. You need to find the smallest $nin Bbb N $ such that $z^n=1$,i.e, $e^{in2pi/3}=1$. Note that $e^{i2pi}=1$. Can you find such an $n $?







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 45 mins ago

























      answered 54 mins ago









      Thomas Shelby

      1,383216




      1,383216








      • 1




        Thank you very much, its when n=3 so if I'm correct this must mean it is of order 3?
        – Reety
        52 mins ago










      • Yes. ${}{} {}{}{}$
        – Thomas Shelby
        46 mins ago














      • 1




        Thank you very much, its when n=3 so if I'm correct this must mean it is of order 3?
        – Reety
        52 mins ago










      • Yes. ${}{} {}{}{}$
        – Thomas Shelby
        46 mins ago








      1




      1




      Thank you very much, its when n=3 so if I'm correct this must mean it is of order 3?
      – Reety
      52 mins ago




      Thank you very much, its when n=3 so if I'm correct this must mean it is of order 3?
      – Reety
      52 mins ago












      Yes. ${}{} {}{}{}$
      – Thomas Shelby
      46 mins ago




      Yes. ${}{} {}{}{}$
      – Thomas Shelby
      46 mins ago











      1














      If the order of z is the smallest positive integer n such that $z^n=1$, then this is fairly simple.



      z has a finite order iff |z| = 1, and arg(z)=$pi cdot q$ where q is a rational number. To find the order, just find the smallest positive integer n such that $q cdot n = 2k$ for some integer k. This is because multiplying the complex numbers, their arguments add together. Hopefully this makes sense and I interpreted the question right.






      share|cite|edit








      New contributor




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























        1














        If the order of z is the smallest positive integer n such that $z^n=1$, then this is fairly simple.



        z has a finite order iff |z| = 1, and arg(z)=$pi cdot q$ where q is a rational number. To find the order, just find the smallest positive integer n such that $q cdot n = 2k$ for some integer k. This is because multiplying the complex numbers, their arguments add together. Hopefully this makes sense and I interpreted the question right.






        share|cite|edit








        New contributor




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





















          1












          1








          1






          If the order of z is the smallest positive integer n such that $z^n=1$, then this is fairly simple.



          z has a finite order iff |z| = 1, and arg(z)=$pi cdot q$ where q is a rational number. To find the order, just find the smallest positive integer n such that $q cdot n = 2k$ for some integer k. This is because multiplying the complex numbers, their arguments add together. Hopefully this makes sense and I interpreted the question right.






          share|cite|edit








          New contributor




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









          If the order of z is the smallest positive integer n such that $z^n=1$, then this is fairly simple.



          z has a finite order iff |z| = 1, and arg(z)=$pi cdot q$ where q is a rational number. To find the order, just find the smallest positive integer n such that $q cdot n = 2k$ for some integer k. This is because multiplying the complex numbers, their arguments add together. Hopefully this makes sense and I interpreted the question right.







          share|cite|edit








          New contributor




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









          share|cite|edit



          share|cite|edit






          New contributor




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









          answered 47 mins ago









          Zachary Hunter

          111




          111




          New contributor




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





          New contributor





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






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























              0














              Find the smallest natural $n$ so that $z^n = 1$.



              $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{ifrac {2pi}{3}}$



              So $z^n = e^{ifrac {2npi} 3}$



              Now $e^{i 2kpi} = 1$ for all integer $k$ and $e^{itheta} = 1$ only if $theta = 2kpi$ for some integer $k$.



              So $z^n = e^{ifrac {2npi} 3} = 1 iff frac {2npi}3 = 2pi*k$ for some integer $k$.



              So we need to find then smallest natural $n$ so and $frac {2npi}3 = 2pi *k$ for some integer value of $k$.



              That's.... not a hard question.






              share|cite|improve this answer


























                0














                Find the smallest natural $n$ so that $z^n = 1$.



                $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{ifrac {2pi}{3}}$



                So $z^n = e^{ifrac {2npi} 3}$



                Now $e^{i 2kpi} = 1$ for all integer $k$ and $e^{itheta} = 1$ only if $theta = 2kpi$ for some integer $k$.



                So $z^n = e^{ifrac {2npi} 3} = 1 iff frac {2npi}3 = 2pi*k$ for some integer $k$.



                So we need to find then smallest natural $n$ so and $frac {2npi}3 = 2pi *k$ for some integer value of $k$.



                That's.... not a hard question.






                share|cite|improve this answer
























                  0












                  0








                  0






                  Find the smallest natural $n$ so that $z^n = 1$.



                  $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{ifrac {2pi}{3}}$



                  So $z^n = e^{ifrac {2npi} 3}$



                  Now $e^{i 2kpi} = 1$ for all integer $k$ and $e^{itheta} = 1$ only if $theta = 2kpi$ for some integer $k$.



                  So $z^n = e^{ifrac {2npi} 3} = 1 iff frac {2npi}3 = 2pi*k$ for some integer $k$.



                  So we need to find then smallest natural $n$ so and $frac {2npi}3 = 2pi *k$ for some integer value of $k$.



                  That's.... not a hard question.






                  share|cite|improve this answer












                  Find the smallest natural $n$ so that $z^n = 1$.



                  $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{ifrac {2pi}{3}}$



                  So $z^n = e^{ifrac {2npi} 3}$



                  Now $e^{i 2kpi} = 1$ for all integer $k$ and $e^{itheta} = 1$ only if $theta = 2kpi$ for some integer $k$.



                  So $z^n = e^{ifrac {2npi} 3} = 1 iff frac {2npi}3 = 2pi*k$ for some integer $k$.



                  So we need to find then smallest natural $n$ so and $frac {2npi}3 = 2pi *k$ for some integer value of $k$.



                  That's.... not a hard question.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 38 mins ago









                  fleablood

                  68.1k22684




                  68.1k22684






























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                      広島県道265号伴広島線

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                      一般県道 広島県道265号伴広島線 路線延長 km 起点 広島県広島市安佐南区伴東7丁目 終点 広島県広島市中区榎町 接続する 主な道路 （記法） 広島県道38号広島豊平線 広島県道71号広島湯来線 山陽自動車道五日市IC 国道183号（ 国道261号重複） ■テンプレート（■ノート ■使い方） ■PJ道路 広島県道265号伴広島線 （ひろしまけんどう265ごう ともひろしません）は広島県広島市安佐南区から広島県広島市中区を結ぶ一般県道である。 目次 1 路線データ 2 通過する自治体 3 見どころ 4 接続道路 4.1 本線 4.2 旧道 5 別名 6 関連項目 路線データ 本線 起点：広島市安佐南区伴東7丁目・大原下橋南詰交差点（広島県道38号広島豊平線（本線・旧道）交点） 終点：広島市中区榎町・十日市交差点（国道183号（国道261号重用）交点） 総延長： 旧道 起点：広島市安佐南区伴東4丁目・新細坂交差点（広島県道38号広島豊平線交点） 終点：広島市安佐南区伴中央4丁目・寺組橋(南)交差点（広島県道265号伴広島線（本線）交点） 総延長： 通過する自治体 広島県 広島市（安佐南区、佐伯区、西区、中区） 見どころ 広島高速交通広島新交通1号線（アストラムライン） 広島広域公園 広島広域公園陸上競技場（広島ビッグアーチ） 己斐峠 己斐橋（太田川放水路） 天満橋（天満川） 接続道路 本線 広島県道38号広島豊平線（本線・旧道）（広島市安佐南区伴東7丁目・大原下橋南詰交差点〔起点〕） 広島県道71号広島湯来線（本線）（広島市安佐南区伴中央2丁目・伴交番入口交差点 - 広島市安佐南区伴中央4丁目・寺組橋(南)交差点間で重用） 広島県道71号広島湯来線（旧道）（広島県道265号伴広島線（旧道）重用）（広島市安佐南区伴中央4丁目・寺組橋(南)交差点） 広島県道71号広島湯来線（広島市安佐南区伴中央4丁目・寺組橋(南)交差点 - 広島市佐伯区五日市町石内・五月が丘(南)交差点間で重用）
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                      Setup Asymptote in Texstudio

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                      0 Good time of the day! I am trying to compile the following file: documentclass[12pt]{article} usepackage[inline]{asymptote} begin{document} Here is a venn diagram produced with Asymptote, drawn to width 4cm: defA{A} defB{V{B}} begin{figure} begin{center} begin{asy} size(4cm,0); pen colour1=red; pen colour2=green; pair z0=(0,0); pair z1=(-1,0); pair z2=(1,0); real r=1.5; path c1=circle(z1,r); path c2=circle(z2,r); fill(c1,colour1); fill(c2,colour2); picture intersection=new picture; fill(intersection,c1,colour1+colour2); clip(intersection,c2); add(intersection); draw(c1); draw(c2); //draw("$A$",box,z1)
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