A question on Complex numbers.












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If $$z=dfrac{sqrt{3}-i}{2}$$ then $$(z^{95}+i^{67})^{94}=z^n$$ then, $text{find the smallest positive integral value of}$ $n$ $text{where}$ $i=sqrt{-1}$




$text{My Attempt:}$ First of all I tried to convert $z$ into $text{Euler's Form}$ so, $z=e^{-i(frac{π}{6})}$
Then, I raised $z$ to the $text{95th}$ power. Then I'm getting stuck. And, not being able to proceed. Help.






complex-numbers







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  • $begingroup$
    You can use that $e^{2pi i}=1$, from that $z^{95}$ can be expressed as $e^{ix}$ for $x$ in $[0,2pi)$, what is this $x$?
    $endgroup$
    – lEm
    Jan 8 at 5:08






  • 1




    $begingroup$
    Hello: for future reference, it is always a bad choice to title your question "question on <subject area>". It would be better to copy your actual question into the title, at that rate, since it fits.
    $endgroup$
    – rschwieb
    Jan 8 at 14:48
















0












$begingroup$



If $$z=dfrac{sqrt{3}-i}{2}$$ then $$(z^{95}+i^{67})^{94}=z^n$$ then, $text{find the smallest positive integral value of}$ $n$ $text{where}$ $i=sqrt{-1}$




$text{My Attempt:}$ First of all I tried to convert $z$ into $text{Euler's Form}$ so, $z=e^{-i(frac{π}{6})}$
Then, I raised $z$ to the $text{95th}$ power. Then I'm getting stuck. And, not being able to proceed. Help.






complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can use that $e^{2pi i}=1$, from that $z^{95}$ can be expressed as $e^{ix}$ for $x$ in $[0,2pi)$, what is this $x$?
    $endgroup$
    – lEm
    Jan 8 at 5:08






  • 1




    $begingroup$
    Hello: for future reference, it is always a bad choice to title your question "question on <subject area>". It would be better to copy your actual question into the title, at that rate, since it fits.
    $endgroup$
    – rschwieb
    Jan 8 at 14:48














0












0








0





$begingroup$



If $$z=dfrac{sqrt{3}-i}{2}$$ then $$(z^{95}+i^{67})^{94}=z^n$$ then, $text{find the smallest positive integral value of}$ $n$ $text{where}$ $i=sqrt{-1}$




$text{My Attempt:}$ First of all I tried to convert $z$ into $text{Euler's Form}$ so, $z=e^{-i(frac{π}{6})}$
Then, I raised $z$ to the $text{95th}$ power. Then I'm getting stuck. And, not being able to proceed. Help.






complex-numbers







share|cite|improve this question











$endgroup$





If $$z=dfrac{sqrt{3}-i}{2}$$ then $$(z^{95}+i^{67})^{94}=z^n$$ then, $text{find the smallest positive integral value of}$ $n$ $text{where}$ $i=sqrt{-1}$




$text{My Attempt:}$ First of all I tried to convert $z$ into $text{Euler's Form}$ so, $z=e^{-i(frac{π}{6})}$
Then, I raised $z$ to the $text{95th}$ power. Then I'm getting stuck. And, not being able to proceed. Help.






complex-numbers




complex-numbers






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edited Jan 8 at 5:55







Pritam Acharya

















asked Jan 8 at 5:01









Pritam AcharyaPritam Acharya

828




828












  • $begingroup$
    You can use that $e^{2pi i}=1$, from that $z^{95}$ can be expressed as $e^{ix}$ for $x$ in $[0,2pi)$, what is this $x$?
    $endgroup$
    – lEm
    Jan 8 at 5:08






  • 1




    $begingroup$
    Hello: for future reference, it is always a bad choice to title your question "question on <subject area>". It would be better to copy your actual question into the title, at that rate, since it fits.
    $endgroup$
    – rschwieb
    Jan 8 at 14:48


















  • $begingroup$
    You can use that $e^{2pi i}=1$, from that $z^{95}$ can be expressed as $e^{ix}$ for $x$ in $[0,2pi)$, what is this $x$?
    $endgroup$
    – lEm
    Jan 8 at 5:08






  • 1




    $begingroup$
    Hello: for future reference, it is always a bad choice to title your question "question on <subject area>". It would be better to copy your actual question into the title, at that rate, since it fits.
    $endgroup$
    – rschwieb
    Jan 8 at 14:48
















$begingroup$
You can use that $e^{2pi i}=1$, from that $z^{95}$ can be expressed as $e^{ix}$ for $x$ in $[0,2pi)$, what is this $x$?
$endgroup$
– lEm
Jan 8 at 5:08




$begingroup$
You can use that $e^{2pi i}=1$, from that $z^{95}$ can be expressed as $e^{ix}$ for $x$ in $[0,2pi)$, what is this $x$?
$endgroup$
– lEm
Jan 8 at 5:08




1




1




$begingroup$
Hello: for future reference, it is always a bad choice to title your question "question on <subject area>". It would be better to copy your actual question into the title, at that rate, since it fits.
$endgroup$
– rschwieb
Jan 8 at 14:48




$begingroup$
Hello: for future reference, it is always a bad choice to title your question "question on <subject area>". It would be better to copy your actual question into the title, at that rate, since it fits.
$endgroup$
– rschwieb
Jan 8 at 14:48










2 Answers
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$begingroup$

$$z=e^{-ipi/6}implies z^{95}=e^{-95ipi/6}=e^{-16ipi+ipi/6}=e^{ipi/6}\i^{67}=i^{64}(-i)=-i$$So we have $$z^{95}+i^{67}=e^{ipi/6}+e^{-ipi/2}=frac{sqrt3+i}{2}-i=frac{sqrt{3}-i}2=z$$



$$z^{94}=z^{-2}=z^{10}$$So the smallest positive integer for which $(z^{95}+i^{67})^{94}=z^n$ is $n=10$.






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    2












    $begingroup$

    $z^6=-1$



    $z^{12m-1}=z^{-1}=e^{ipi/6}$



    $i^{4n+3}=i^3=-i$



    $z^{12m-1}+i^{4n+3}=$cis$(pi/6)-i=$cis$(-pi/6)=z$



    where cis $=cos +isin$



    So, we have $z^{n-94}=1$



    $implies12$ divides $n-94$






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      $begingroup$

      $$z=e^{-ipi/6}implies z^{95}=e^{-95ipi/6}=e^{-16ipi+ipi/6}=e^{ipi/6}\i^{67}=i^{64}(-i)=-i$$So we have $$z^{95}+i^{67}=e^{ipi/6}+e^{-ipi/2}=frac{sqrt3+i}{2}-i=frac{sqrt{3}-i}2=z$$



      $$z^{94}=z^{-2}=z^{10}$$So the smallest positive integer for which $(z^{95}+i^{67})^{94}=z^n$ is $n=10$.






      share|cite|improve this answer









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        2












        $begingroup$

        $$z=e^{-ipi/6}implies z^{95}=e^{-95ipi/6}=e^{-16ipi+ipi/6}=e^{ipi/6}\i^{67}=i^{64}(-i)=-i$$So we have $$z^{95}+i^{67}=e^{ipi/6}+e^{-ipi/2}=frac{sqrt3+i}{2}-i=frac{sqrt{3}-i}2=z$$



        $$z^{94}=z^{-2}=z^{10}$$So the smallest positive integer for which $(z^{95}+i^{67})^{94}=z^n$ is $n=10$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          $$z=e^{-ipi/6}implies z^{95}=e^{-95ipi/6}=e^{-16ipi+ipi/6}=e^{ipi/6}\i^{67}=i^{64}(-i)=-i$$So we have $$z^{95}+i^{67}=e^{ipi/6}+e^{-ipi/2}=frac{sqrt3+i}{2}-i=frac{sqrt{3}-i}2=z$$



          $$z^{94}=z^{-2}=z^{10}$$So the smallest positive integer for which $(z^{95}+i^{67})^{94}=z^n$ is $n=10$.






          share|cite|improve this answer









          $endgroup$



          $$z=e^{-ipi/6}implies z^{95}=e^{-95ipi/6}=e^{-16ipi+ipi/6}=e^{ipi/6}\i^{67}=i^{64}(-i)=-i$$So we have $$z^{95}+i^{67}=e^{ipi/6}+e^{-ipi/2}=frac{sqrt3+i}{2}-i=frac{sqrt{3}-i}2=z$$



          $$z^{94}=z^{-2}=z^{10}$$So the smallest positive integer for which $(z^{95}+i^{67})^{94}=z^n$ is $n=10$.







          share|cite|improve this answer












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          share|cite|improve this answer










          answered Jan 8 at 5:14









          John DoeJohn Doe

          12.1k11339




          12.1k11339























              2












              $begingroup$

              $z^6=-1$



              $z^{12m-1}=z^{-1}=e^{ipi/6}$



              $i^{4n+3}=i^3=-i$



              $z^{12m-1}+i^{4n+3}=$cis$(pi/6)-i=$cis$(-pi/6)=z$



              where cis $=cos +isin$



              So, we have $z^{n-94}=1$



              $implies12$ divides $n-94$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                $z^6=-1$



                $z^{12m-1}=z^{-1}=e^{ipi/6}$



                $i^{4n+3}=i^3=-i$



                $z^{12m-1}+i^{4n+3}=$cis$(pi/6)-i=$cis$(-pi/6)=z$



                where cis $=cos +isin$



                So, we have $z^{n-94}=1$



                $implies12$ divides $n-94$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  $z^6=-1$



                  $z^{12m-1}=z^{-1}=e^{ipi/6}$



                  $i^{4n+3}=i^3=-i$



                  $z^{12m-1}+i^{4n+3}=$cis$(pi/6)-i=$cis$(-pi/6)=z$



                  where cis $=cos +isin$



                  So, we have $z^{n-94}=1$



                  $implies12$ divides $n-94$






                  share|cite|improve this answer









                  $endgroup$



                  $z^6=-1$



                  $z^{12m-1}=z^{-1}=e^{ipi/6}$



                  $i^{4n+3}=i^3=-i$



                  $z^{12m-1}+i^{4n+3}=$cis$(pi/6)-i=$cis$(-pi/6)=z$



                  where cis $=cos +isin$



                  So, we have $z^{n-94}=1$



                  $implies12$ divides $n-94$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 5:12









                  lab bhattacharjeelab bhattacharjee

                  229k15159279




                  229k15159279






























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