How to show that $p(1) text{is real} iff p(-1) text{is real}$












4














I have been working on this problem and I cannot figure it out! I spent hours of time on it with no use. Can anybody help? The question is:



Suppose $p(x)$ is a polynomial with complex coefficients and even degree($n=2k$). All zeros of $p$ are non-real and with length equal to $1$. prove $$p(1)inmathbb{R} ;;Longleftrightarrow;; p(-1)inmathbb{R} $$






linear-algebra functions polynomials







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  • What have you tried?
    – Tito Eliatron
    Nov 24 at 15:45
















4














I have been working on this problem and I cannot figure it out! I spent hours of time on it with no use. Can anybody help? The question is:



Suppose $p(x)$ is a polynomial with complex coefficients and even degree($n=2k$). All zeros of $p$ are non-real and with length equal to $1$. prove $$p(1)inmathbb{R} ;;Longleftrightarrow;; p(-1)inmathbb{R} $$






linear-algebra functions polynomials







share|cite|improve this question
























  • What have you tried?
    – Tito Eliatron
    Nov 24 at 15:45














4












4








4







I have been working on this problem and I cannot figure it out! I spent hours of time on it with no use. Can anybody help? The question is:



Suppose $p(x)$ is a polynomial with complex coefficients and even degree($n=2k$). All zeros of $p$ are non-real and with length equal to $1$. prove $$p(1)inmathbb{R} ;;Longleftrightarrow;; p(-1)inmathbb{R} $$






linear-algebra functions polynomials







share|cite|improve this question















I have been working on this problem and I cannot figure it out! I spent hours of time on it with no use. Can anybody help? The question is:



Suppose $p(x)$ is a polynomial with complex coefficients and even degree($n=2k$). All zeros of $p$ are non-real and with length equal to $1$. prove $$p(1)inmathbb{R} ;;Longleftrightarrow;; p(-1)inmathbb{R} $$






linear-algebra functions polynomials




linear-algebra functions polynomials






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edited Nov 24 at 16:29









steven gregory

17.7k32257




17.7k32257










asked Nov 24 at 15:38









Ali.Maleky7997

233




233












  • What have you tried?
    – Tito Eliatron
    Nov 24 at 15:45


















  • What have you tried?
    – Tito Eliatron
    Nov 24 at 15:45
















What have you tried?
– Tito Eliatron
Nov 24 at 15:45




What have you tried?
– Tito Eliatron
Nov 24 at 15:45










2 Answers
2






active

oldest

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5














Note that (assuming $p(-1)ne 0$ to begin with
$$ frac{p(1)}{p(-1)}=prod_{j=1}^{2k}frac{1-w_j}{-1-w_j}$$
where the $w_j$ run over the complex roots (with multiplicity).
For a single factor,
$$frac{1-w}{-1-w}=-frac{(1-w)(1+bar w)}{|1+w|^2}=frac{|w|^2-1+(w-bar w)}{|1+w|^2}. $$
As we are given that $|w|=1$ for all roots, this is the purely imaginary number $frac{2operatorname{im} w}{|1+w|^2}i$. The product of an even number of imaginaries is real.






share|cite|improve this answer





















  • Thank you for your precise and nice answer
    – Ali.Maleky7997
    Nov 24 at 16:29










  • @Ali.Maleky7997 You said all zeros are non-real.
    – SinTan1729
    Nov 24 at 16:37



















1














We have that



$$p(z)=prod_{r=1}^{2k}left(z-e^{itheta_r}right))$$



then assuming $p(1)neq 0$



$$p(1)=prod_{r=1}^{2k}(1-e^{itheta_r})=prod_{r=1}^{2k}(1-e^{-itheta_r})=prod_{r=1}^{2k}frac{1-e^{itheta_r}}{e^{itheta_r}} iff prod_{r=1}^{2k}frac{1}{e^{itheta_r}}=1$$



$$p(-1)=prod_{r=1}^{2k}(-1-e^{itheta_r})=prod_{r=1}^{2k}(-1-e^{-itheta_r}) iff prod_{r=1}^{2k}frac{1}{e^{itheta_r}}=1$$






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    5














    Note that (assuming $p(-1)ne 0$ to begin with
    $$ frac{p(1)}{p(-1)}=prod_{j=1}^{2k}frac{1-w_j}{-1-w_j}$$
    where the $w_j$ run over the complex roots (with multiplicity).
    For a single factor,
    $$frac{1-w}{-1-w}=-frac{(1-w)(1+bar w)}{|1+w|^2}=frac{|w|^2-1+(w-bar w)}{|1+w|^2}. $$
    As we are given that $|w|=1$ for all roots, this is the purely imaginary number $frac{2operatorname{im} w}{|1+w|^2}i$. The product of an even number of imaginaries is real.






    share|cite|improve this answer





















    • Thank you for your precise and nice answer
      – Ali.Maleky7997
      Nov 24 at 16:29










    • @Ali.Maleky7997 You said all zeros are non-real.
      – SinTan1729
      Nov 24 at 16:37
















    5














    Note that (assuming $p(-1)ne 0$ to begin with
    $$ frac{p(1)}{p(-1)}=prod_{j=1}^{2k}frac{1-w_j}{-1-w_j}$$
    where the $w_j$ run over the complex roots (with multiplicity).
    For a single factor,
    $$frac{1-w}{-1-w}=-frac{(1-w)(1+bar w)}{|1+w|^2}=frac{|w|^2-1+(w-bar w)}{|1+w|^2}. $$
    As we are given that $|w|=1$ for all roots, this is the purely imaginary number $frac{2operatorname{im} w}{|1+w|^2}i$. The product of an even number of imaginaries is real.






    share|cite|improve this answer





















    • Thank you for your precise and nice answer
      – Ali.Maleky7997
      Nov 24 at 16:29










    • @Ali.Maleky7997 You said all zeros are non-real.
      – SinTan1729
      Nov 24 at 16:37














    5












    5








    5






    Note that (assuming $p(-1)ne 0$ to begin with
    $$ frac{p(1)}{p(-1)}=prod_{j=1}^{2k}frac{1-w_j}{-1-w_j}$$
    where the $w_j$ run over the complex roots (with multiplicity).
    For a single factor,
    $$frac{1-w}{-1-w}=-frac{(1-w)(1+bar w)}{|1+w|^2}=frac{|w|^2-1+(w-bar w)}{|1+w|^2}. $$
    As we are given that $|w|=1$ for all roots, this is the purely imaginary number $frac{2operatorname{im} w}{|1+w|^2}i$. The product of an even number of imaginaries is real.






    share|cite|improve this answer












    Note that (assuming $p(-1)ne 0$ to begin with
    $$ frac{p(1)}{p(-1)}=prod_{j=1}^{2k}frac{1-w_j}{-1-w_j}$$
    where the $w_j$ run over the complex roots (with multiplicity).
    For a single factor,
    $$frac{1-w}{-1-w}=-frac{(1-w)(1+bar w)}{|1+w|^2}=frac{|w|^2-1+(w-bar w)}{|1+w|^2}. $$
    As we are given that $|w|=1$ for all roots, this is the purely imaginary number $frac{2operatorname{im} w}{|1+w|^2}i$. The product of an even number of imaginaries is real.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 24 at 15:55









    Hagen von Eitzen

    276k21268495




    276k21268495












    • Thank you for your precise and nice answer
      – Ali.Maleky7997
      Nov 24 at 16:29










    • @Ali.Maleky7997 You said all zeros are non-real.
      – SinTan1729
      Nov 24 at 16:37


















    • Thank you for your precise and nice answer
      – Ali.Maleky7997
      Nov 24 at 16:29










    • @Ali.Maleky7997 You said all zeros are non-real.
      – SinTan1729
      Nov 24 at 16:37
















    Thank you for your precise and nice answer
    – Ali.Maleky7997
    Nov 24 at 16:29




    Thank you for your precise and nice answer
    – Ali.Maleky7997
    Nov 24 at 16:29












    @Ali.Maleky7997 You said all zeros are non-real.
    – SinTan1729
    Nov 24 at 16:37




    @Ali.Maleky7997 You said all zeros are non-real.
    – SinTan1729
    Nov 24 at 16:37











    1














    We have that



    $$p(z)=prod_{r=1}^{2k}left(z-e^{itheta_r}right))$$



    then assuming $p(1)neq 0$



    $$p(1)=prod_{r=1}^{2k}(1-e^{itheta_r})=prod_{r=1}^{2k}(1-e^{-itheta_r})=prod_{r=1}^{2k}frac{1-e^{itheta_r}}{e^{itheta_r}} iff prod_{r=1}^{2k}frac{1}{e^{itheta_r}}=1$$



    $$p(-1)=prod_{r=1}^{2k}(-1-e^{itheta_r})=prod_{r=1}^{2k}(-1-e^{-itheta_r}) iff prod_{r=1}^{2k}frac{1}{e^{itheta_r}}=1$$






    share|cite|improve this answer


























      1














      We have that



      $$p(z)=prod_{r=1}^{2k}left(z-e^{itheta_r}right))$$



      then assuming $p(1)neq 0$



      $$p(1)=prod_{r=1}^{2k}(1-e^{itheta_r})=prod_{r=1}^{2k}(1-e^{-itheta_r})=prod_{r=1}^{2k}frac{1-e^{itheta_r}}{e^{itheta_r}} iff prod_{r=1}^{2k}frac{1}{e^{itheta_r}}=1$$



      $$p(-1)=prod_{r=1}^{2k}(-1-e^{itheta_r})=prod_{r=1}^{2k}(-1-e^{-itheta_r}) iff prod_{r=1}^{2k}frac{1}{e^{itheta_r}}=1$$






      share|cite|improve this answer
























        1












        1








        1






        We have that



        $$p(z)=prod_{r=1}^{2k}left(z-e^{itheta_r}right))$$



        then assuming $p(1)neq 0$



        $$p(1)=prod_{r=1}^{2k}(1-e^{itheta_r})=prod_{r=1}^{2k}(1-e^{-itheta_r})=prod_{r=1}^{2k}frac{1-e^{itheta_r}}{e^{itheta_r}} iff prod_{r=1}^{2k}frac{1}{e^{itheta_r}}=1$$



        $$p(-1)=prod_{r=1}^{2k}(-1-e^{itheta_r})=prod_{r=1}^{2k}(-1-e^{-itheta_r}) iff prod_{r=1}^{2k}frac{1}{e^{itheta_r}}=1$$






        share|cite|improve this answer












        We have that



        $$p(z)=prod_{r=1}^{2k}left(z-e^{itheta_r}right))$$



        then assuming $p(1)neq 0$



        $$p(1)=prod_{r=1}^{2k}(1-e^{itheta_r})=prod_{r=1}^{2k}(1-e^{-itheta_r})=prod_{r=1}^{2k}frac{1-e^{itheta_r}}{e^{itheta_r}} iff prod_{r=1}^{2k}frac{1}{e^{itheta_r}}=1$$



        $$p(-1)=prod_{r=1}^{2k}(-1-e^{itheta_r})=prod_{r=1}^{2k}(-1-e^{-itheta_r}) iff prod_{r=1}^{2k}frac{1}{e^{itheta_r}}=1$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 at 16:01









        gimusi

        1




        1






























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