Let $ninmathbb{N}$ and $zinmathbb{C}$ with $|z|=1$ and $z^{2n}neq-1$. Prove that...

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Let $ninmathbb{N}$ and $zinmathbb{C}$ with $|z|=1$ and $z^{2n}neq-1$. Prove that $frac{z^n}{1+z^{2n}}inmathbb{R}$. [duplicate]



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Showing that $frac{z^n}{1+z^{2n}}$ is real for $|z|=1$.Proving a Complex Number is RealProve that if H ∪ K is a subgroup of G…Prove $forall n inmathbb{N}, (n|105 wedge n|70) implies 5|n$Prove that the following sequence converges to zero.How to prove $log(exp(z))neq z$ if $zin mathbb{C}$Prove that $exp(r)≥sum_{k=0}^nleft(frac{r^k}{k!} right)$ for any $n∈mathbb{N}$ and $r≥0$.Let $ A ={ninmathbb{Z}: 2 | n}$ and $B={ninmathbb{Z}: 4 | n}$. Prove that $nin (A - B)$ if and only if $n=2k$ for some odd integer k.Prove that if ${7k:kin{mathbb Z}}subsetneq{nm:min{mathbb Z}}$, then $n=1.$Prove that for all sets $A$, $B$, and $C$, if $Acap{B}=emptyset$ and $Acap{C}=emptyset$, then $Acap({B}cup {C})=emptyset$.Prove that $4mid{n}$ if and only if the integer formed by the final two digits of $n$ is divisible by 4.












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This question already has an answer here:




  • Showing that $frac{z^n}{1+z^{2n}}$ is real for $|z|=1$.

    4 answers




Let $ninmathbb{N}$ and $zinmathbb{C}$ with $|z|=1$ and $z^{2n}neq-1$. Prove that $frac{z^n}{1+z^{2n}}inmathbb{R}$.



So I know that $z^{2n}neq{-1}$ implies that the denominator cannot be zero.



Not sure how to prove this. Any and all help is appreciated, thanks!






complex-numbers proof-writing







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marked as duplicate by rtybase, egreg, Lord Shark the Unknown, Cameron Buie, Community♦ Mar 24 at 20:58


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    The detailed answer is here
    $endgroup$
    – ersh
    Mar 24 at 20:47
















0












$begingroup$



This question already has an answer here:




  • Showing that $frac{z^n}{1+z^{2n}}$ is real for $|z|=1$.

    4 answers




Let $ninmathbb{N}$ and $zinmathbb{C}$ with $|z|=1$ and $z^{2n}neq-1$. Prove that $frac{z^n}{1+z^{2n}}inmathbb{R}$.



So I know that $z^{2n}neq{-1}$ implies that the denominator cannot be zero.



Not sure how to prove this. Any and all help is appreciated, thanks!






complex-numbers proof-writing







share|cite|improve this question











$endgroup$



marked as duplicate by rtybase, egreg, Lord Shark the Unknown, Cameron Buie, Community♦ Mar 24 at 20:58


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    The detailed answer is here
    $endgroup$
    – ersh
    Mar 24 at 20:47














0












0








0





$begingroup$



This question already has an answer here:




  • Showing that $frac{z^n}{1+z^{2n}}$ is real for $|z|=1$.

    4 answers




Let $ninmathbb{N}$ and $zinmathbb{C}$ with $|z|=1$ and $z^{2n}neq-1$. Prove that $frac{z^n}{1+z^{2n}}inmathbb{R}$.



So I know that $z^{2n}neq{-1}$ implies that the denominator cannot be zero.



Not sure how to prove this. Any and all help is appreciated, thanks!






complex-numbers proof-writing







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Showing that $frac{z^n}{1+z^{2n}}$ is real for $|z|=1$.

    4 answers




Let $ninmathbb{N}$ and $zinmathbb{C}$ with $|z|=1$ and $z^{2n}neq-1$. Prove that $frac{z^n}{1+z^{2n}}inmathbb{R}$.



So I know that $z^{2n}neq{-1}$ implies that the denominator cannot be zero.



Not sure how to prove this. Any and all help is appreciated, thanks!





This question already has an answer here:




  • Showing that $frac{z^n}{1+z^{2n}}$ is real for $|z|=1$.

    4 answers







complex-numbers proof-writing




complex-numbers proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 24 at 20:42









Cameron Buie

87k773161




87k773161










asked Mar 24 at 20:36









SaniaSania

406




406




marked as duplicate by rtybase, egreg, Lord Shark the Unknown, Cameron Buie, Community♦ Mar 24 at 20:58


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by rtybase, egreg, Lord Shark the Unknown, Cameron Buie, Community♦ Mar 24 at 20:58


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    The detailed answer is here
    $endgroup$
    – ersh
    Mar 24 at 20:47














  • 1




    $begingroup$
    The detailed answer is here
    $endgroup$
    – ersh
    Mar 24 at 20:47








1




1




$begingroup$
The detailed answer is here
$endgroup$
– ersh
Mar 24 at 20:47




$begingroup$
The detailed answer is here
$endgroup$
– ersh
Mar 24 at 20:47










2 Answers
2






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It might be easier to show that $$frac{1+z^{2n}}{z^n}$$ is real. As a hint, show first that $frac1z=overline z$ when $|z|=1,$ and that in general, $overline{wcdot z}=overline{w}cdotoverline{z}.$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    If $|z|=1$, there exists $theta in mathbb{R}$ such that $z=e^{itheta}$. So
    $$frac{z^n}{1+z^{2n}}=frac{e^{intheta}}{1+e^{2intheta}} = frac{e^{intheta}}{e^{intheta}left( e^{-intheta} + e^{intheta}right)} = frac{1}{ e^{-intheta} + e^{intheta}} = frac{1}{2cos(ntheta)} in mathbb{R}$$






    share|cite|improve this answer









    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      It might be easier to show that $$frac{1+z^{2n}}{z^n}$$ is real. As a hint, show first that $frac1z=overline z$ when $|z|=1,$ and that in general, $overline{wcdot z}=overline{w}cdotoverline{z}.$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        It might be easier to show that $$frac{1+z^{2n}}{z^n}$$ is real. As a hint, show first that $frac1z=overline z$ when $|z|=1,$ and that in general, $overline{wcdot z}=overline{w}cdotoverline{z}.$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          It might be easier to show that $$frac{1+z^{2n}}{z^n}$$ is real. As a hint, show first that $frac1z=overline z$ when $|z|=1,$ and that in general, $overline{wcdot z}=overline{w}cdotoverline{z}.$






          share|cite|improve this answer











          $endgroup$



          It might be easier to show that $$frac{1+z^{2n}}{z^n}$$ is real. As a hint, show first that $frac1z=overline z$ when $|z|=1,$ and that in general, $overline{wcdot z}=overline{w}cdotoverline{z}.$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 24 at 20:46

























          answered Mar 24 at 20:40









          Cameron BuieCameron Buie

          87k773161




          87k773161























              1












              $begingroup$

              If $|z|=1$, there exists $theta in mathbb{R}$ such that $z=e^{itheta}$. So
              $$frac{z^n}{1+z^{2n}}=frac{e^{intheta}}{1+e^{2intheta}} = frac{e^{intheta}}{e^{intheta}left( e^{-intheta} + e^{intheta}right)} = frac{1}{ e^{-intheta} + e^{intheta}} = frac{1}{2cos(ntheta)} in mathbb{R}$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                If $|z|=1$, there exists $theta in mathbb{R}$ such that $z=e^{itheta}$. So
                $$frac{z^n}{1+z^{2n}}=frac{e^{intheta}}{1+e^{2intheta}} = frac{e^{intheta}}{e^{intheta}left( e^{-intheta} + e^{intheta}right)} = frac{1}{ e^{-intheta} + e^{intheta}} = frac{1}{2cos(ntheta)} in mathbb{R}$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  If $|z|=1$, there exists $theta in mathbb{R}$ such that $z=e^{itheta}$. So
                  $$frac{z^n}{1+z^{2n}}=frac{e^{intheta}}{1+e^{2intheta}} = frac{e^{intheta}}{e^{intheta}left( e^{-intheta} + e^{intheta}right)} = frac{1}{ e^{-intheta} + e^{intheta}} = frac{1}{2cos(ntheta)} in mathbb{R}$$






                  share|cite|improve this answer









                  $endgroup$



                  If $|z|=1$, there exists $theta in mathbb{R}$ such that $z=e^{itheta}$. So
                  $$frac{z^n}{1+z^{2n}}=frac{e^{intheta}}{1+e^{2intheta}} = frac{e^{intheta}}{e^{intheta}left( e^{-intheta} + e^{intheta}right)} = frac{1}{ e^{-intheta} + e^{intheta}} = frac{1}{2cos(ntheta)} in mathbb{R}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 24 at 20:48









                  TheSilverDoeTheSilverDoe

                  5,598316




                  5,598316















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