Order of a Complex Number?












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.










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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
    1 hour ago
















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.










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
    1 hour ago














1












1








1


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.










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.







complex-numbers






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edited 16 mins ago









Did

246k23220454




246k23220454










asked 1 hour ago









Reety

12711




12711












  • 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
    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








  • 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
    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






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
1 hour 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
1 hour ago










4 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
    1 hour ago










  • Yes. ${}{} {}{}{}$
    – Thomas Shelby
    1 hour ago



















1














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





























    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|improve this answer










    New contributor




    Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      0














      Let us define $e^{iA}= cos A +isin A$ when $Ain Bbb R$ without any thought to the motivation for it. By DeMoivre's Theorem we have $e^{inA}=(e^{iA})^n$ for $nin Bbb Z^+ $ and $Ain Bbb R.$



      Let $0ne zin Bbb C.$ Then $z=re^{it}$ where $|z|=rin Bbb R^+$ and $t$ is some member of $Bbb R.$



      (1). Suppose $z^n=1$ for some $nin Bbb Z^+.$ Then $1=|z^n|=|r^n(e^{it})^n|=|r^ne^{nit}|=r^n,$ so $r=1.$ So $1=z^n=e^{nit},$ which requires $ntin {2pi m:min Bbb Z}.$ So $t/2pi in Bbb Q.$ So $zin {e^{2pi iq}:qin Bbb Q}.$



      (2). Suppose $zin {e^{2pi iq}:qin Bbb Q}.$ Then $z=e^{2pi i m/n}$ for some $nin Bbb Z^+$ and some $min Bbb Z,$ so $z^n=e^{2pi i m}=1.$



      (3). From (1) and (2), $z$ has a finite order iff $zin {e^{2pi i q}: qin Bbb Q}=$ $={cos 2pi q +isin 2pi q: qin Bbb Qcap [0,1)}.$



      Note. For any real $u,v$ with $u^2+v^2=1$ there exists $tinBbb R$ such that $(cos t, sin t)=(u,v).$ So if $0ne zin Bbb C,$ let $z=x+iy$ with $x,y in Bbb R.$ Since $x,y$ are not both $0,$ we have $|z|=sqrt {x^2+y^2},>0.$ Now let $(u,v)=(x/|z|, y/|z|).$ Since $u^2+v^2=1,$ there exists $tin Bbb R$ with $(u,v)=(cos t,sin t). $ So $z=|z|(u+iv)=|z|(cos t +isin t)=|z|e^{it}.$






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        4 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
          1 hour ago










        • Yes. ${}{} {}{}{}$
          – Thomas Shelby
          1 hour ago
















        2














        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
          1 hour ago










        • Yes. ${}{} {}{}{}$
          – Thomas Shelby
          1 hour ago














        2












        2








        2






        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 1 hour ago

























        answered 1 hour ago









        Thomas Shelby

        1,393216




        1,393216








        • 1




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










        • Yes. ${}{} {}{}{}$
          – Thomas Shelby
          1 hour ago














        • 1




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










        • Yes. ${}{} {}{}{}$
          – Thomas Shelby
          1 hour 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
        1 hour ago




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












        Yes. ${}{} {}{}{}$
        – Thomas Shelby
        1 hour ago




        Yes. ${}{} {}{}{}$
        – Thomas Shelby
        1 hour ago











        1














        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


























          1














          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
























            1












            1








            1






            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 1 hour ago









            fleablood

            68.1k22684




            68.1k22684























                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|improve this answer










                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|improve this answer










                  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|improve this answer










                    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|improve this answer










                    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|improve this answer



                    share|cite|improve this answer








                    edited 25 mins ago









                    Gaby Alfonso

                    676315




                    676315






                    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 1 hour 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














                        Let us define $e^{iA}= cos A +isin A$ when $Ain Bbb R$ without any thought to the motivation for it. By DeMoivre's Theorem we have $e^{inA}=(e^{iA})^n$ for $nin Bbb Z^+ $ and $Ain Bbb R.$



                        Let $0ne zin Bbb C.$ Then $z=re^{it}$ where $|z|=rin Bbb R^+$ and $t$ is some member of $Bbb R.$



                        (1). Suppose $z^n=1$ for some $nin Bbb Z^+.$ Then $1=|z^n|=|r^n(e^{it})^n|=|r^ne^{nit}|=r^n,$ so $r=1.$ So $1=z^n=e^{nit},$ which requires $ntin {2pi m:min Bbb Z}.$ So $t/2pi in Bbb Q.$ So $zin {e^{2pi iq}:qin Bbb Q}.$



                        (2). Suppose $zin {e^{2pi iq}:qin Bbb Q}.$ Then $z=e^{2pi i m/n}$ for some $nin Bbb Z^+$ and some $min Bbb Z,$ so $z^n=e^{2pi i m}=1.$



                        (3). From (1) and (2), $z$ has a finite order iff $zin {e^{2pi i q}: qin Bbb Q}=$ $={cos 2pi q +isin 2pi q: qin Bbb Qcap [0,1)}.$



                        Note. For any real $u,v$ with $u^2+v^2=1$ there exists $tinBbb R$ such that $(cos t, sin t)=(u,v).$ So if $0ne zin Bbb C,$ let $z=x+iy$ with $x,y in Bbb R.$ Since $x,y$ are not both $0,$ we have $|z|=sqrt {x^2+y^2},>0.$ Now let $(u,v)=(x/|z|, y/|z|).$ Since $u^2+v^2=1,$ there exists $tin Bbb R$ with $(u,v)=(cos t,sin t). $ So $z=|z|(u+iv)=|z|(cos t +isin t)=|z|e^{it}.$






                        share|cite|improve this answer




























                          0














                          Let us define $e^{iA}= cos A +isin A$ when $Ain Bbb R$ without any thought to the motivation for it. By DeMoivre's Theorem we have $e^{inA}=(e^{iA})^n$ for $nin Bbb Z^+ $ and $Ain Bbb R.$



                          Let $0ne zin Bbb C.$ Then $z=re^{it}$ where $|z|=rin Bbb R^+$ and $t$ is some member of $Bbb R.$



                          (1). Suppose $z^n=1$ for some $nin Bbb Z^+.$ Then $1=|z^n|=|r^n(e^{it})^n|=|r^ne^{nit}|=r^n,$ so $r=1.$ So $1=z^n=e^{nit},$ which requires $ntin {2pi m:min Bbb Z}.$ So $t/2pi in Bbb Q.$ So $zin {e^{2pi iq}:qin Bbb Q}.$



                          (2). Suppose $zin {e^{2pi iq}:qin Bbb Q}.$ Then $z=e^{2pi i m/n}$ for some $nin Bbb Z^+$ and some $min Bbb Z,$ so $z^n=e^{2pi i m}=1.$



                          (3). From (1) and (2), $z$ has a finite order iff $zin {e^{2pi i q}: qin Bbb Q}=$ $={cos 2pi q +isin 2pi q: qin Bbb Qcap [0,1)}.$



                          Note. For any real $u,v$ with $u^2+v^2=1$ there exists $tinBbb R$ such that $(cos t, sin t)=(u,v).$ So if $0ne zin Bbb C,$ let $z=x+iy$ with $x,y in Bbb R.$ Since $x,y$ are not both $0,$ we have $|z|=sqrt {x^2+y^2},>0.$ Now let $(u,v)=(x/|z|, y/|z|).$ Since $u^2+v^2=1,$ there exists $tin Bbb R$ with $(u,v)=(cos t,sin t). $ So $z=|z|(u+iv)=|z|(cos t +isin t)=|z|e^{it}.$






                          share|cite|improve this answer


























                            0












                            0








                            0






                            Let us define $e^{iA}= cos A +isin A$ when $Ain Bbb R$ without any thought to the motivation for it. By DeMoivre's Theorem we have $e^{inA}=(e^{iA})^n$ for $nin Bbb Z^+ $ and $Ain Bbb R.$



                            Let $0ne zin Bbb C.$ Then $z=re^{it}$ where $|z|=rin Bbb R^+$ and $t$ is some member of $Bbb R.$



                            (1). Suppose $z^n=1$ for some $nin Bbb Z^+.$ Then $1=|z^n|=|r^n(e^{it})^n|=|r^ne^{nit}|=r^n,$ so $r=1.$ So $1=z^n=e^{nit},$ which requires $ntin {2pi m:min Bbb Z}.$ So $t/2pi in Bbb Q.$ So $zin {e^{2pi iq}:qin Bbb Q}.$



                            (2). Suppose $zin {e^{2pi iq}:qin Bbb Q}.$ Then $z=e^{2pi i m/n}$ for some $nin Bbb Z^+$ and some $min Bbb Z,$ so $z^n=e^{2pi i m}=1.$



                            (3). From (1) and (2), $z$ has a finite order iff $zin {e^{2pi i q}: qin Bbb Q}=$ $={cos 2pi q +isin 2pi q: qin Bbb Qcap [0,1)}.$



                            Note. For any real $u,v$ with $u^2+v^2=1$ there exists $tinBbb R$ such that $(cos t, sin t)=(u,v).$ So if $0ne zin Bbb C,$ let $z=x+iy$ with $x,y in Bbb R.$ Since $x,y$ are not both $0,$ we have $|z|=sqrt {x^2+y^2},>0.$ Now let $(u,v)=(x/|z|, y/|z|).$ Since $u^2+v^2=1,$ there exists $tin Bbb R$ with $(u,v)=(cos t,sin t). $ So $z=|z|(u+iv)=|z|(cos t +isin t)=|z|e^{it}.$






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                            Let us define $e^{iA}= cos A +isin A$ when $Ain Bbb R$ without any thought to the motivation for it. By DeMoivre's Theorem we have $e^{inA}=(e^{iA})^n$ for $nin Bbb Z^+ $ and $Ain Bbb R.$



                            Let $0ne zin Bbb C.$ Then $z=re^{it}$ where $|z|=rin Bbb R^+$ and $t$ is some member of $Bbb R.$



                            (1). Suppose $z^n=1$ for some $nin Bbb Z^+.$ Then $1=|z^n|=|r^n(e^{it})^n|=|r^ne^{nit}|=r^n,$ so $r=1.$ So $1=z^n=e^{nit},$ which requires $ntin {2pi m:min Bbb Z}.$ So $t/2pi in Bbb Q.$ So $zin {e^{2pi iq}:qin Bbb Q}.$



                            (2). Suppose $zin {e^{2pi iq}:qin Bbb Q}.$ Then $z=e^{2pi i m/n}$ for some $nin Bbb Z^+$ and some $min Bbb Z,$ so $z^n=e^{2pi i m}=1.$



                            (3). From (1) and (2), $z$ has a finite order iff $zin {e^{2pi i q}: qin Bbb Q}=$ $={cos 2pi q +isin 2pi q: qin Bbb Qcap [0,1)}.$



                            Note. For any real $u,v$ with $u^2+v^2=1$ there exists $tinBbb R$ such that $(cos t, sin t)=(u,v).$ So if $0ne zin Bbb C,$ let $z=x+iy$ with $x,y in Bbb R.$ Since $x,y$ are not both $0,$ we have $|z|=sqrt {x^2+y^2},>0.$ Now let $(u,v)=(x/|z|, y/|z|).$ Since $u^2+v^2=1,$ there exists $tin Bbb R$ with $(u,v)=(cos t,sin t). $ So $z=|z|(u+iv)=|z|(cos t +isin t)=|z|e^{it}.$







                            share|cite|improve this answer














                            share|cite|improve this answer



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                            edited 19 mins ago

























                            answered 24 mins ago









                            DanielWainfleet

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