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











up vote
4
down vote

favorite












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















up vote
4
down vote

favorite












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













up vote
4
down vote

favorite









up vote
4
down vote

favorite











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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 16:29









steven gregory

17.6k32257




17.6k32257










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

votes

















up vote
5
down vote



accepted










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


















up vote
1
down vote













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





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011692%2fhow-to-show-that-p1-textis-real-iff-p-1-textis-real%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    5
    down vote



    accepted










    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















    up vote
    5
    down vote



    accepted










    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













    up vote
    5
    down vote



    accepted







    up vote
    5
    down vote



    accepted






    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

    275k21268495




    275k21268495












    • 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










    up vote
    1
    down vote













    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

























      up vote
      1
      down vote













      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























        up vote
        1
        down vote










        up vote
        1
        down vote









        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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011692%2fhow-to-show-that-p1-textis-real-iff-p-1-textis-real%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Willebadessen

            Image
            Wappen Deutschlandkarte 51.633055555556 9.0330555555556 295 Koordinaten: 51° 38′  N , 9° 2′  O Basisdaten Bundesland: Nordrhein-Westfalen Regierungsbezirk: Detmold Kreis: Höxter Höhe: 295 m ü. NHN Fläche: 128,41 km 2 Einwohner: 8227 (31. Dez. 2017) [1] Bevölkerungsdichte: 64 Einwohner je km 2 Postleitzahlen: 34435, 34437, 34439 Vorwahlen: 05644, 05646, 05642 Kfz-Kennzeichen: HX Gemeindeschlüssel: 05 7 62 040 Stadtgliederung: 13 Stadtteile Adresse der Stadtverwaltung: Abdinghofweg 1 34439 Willebadessen Website: www.willebadessen.de Bürgermeister: Hans Hermann Bluhm (CDU) Lage der Stadt Willebadessen im Kreis Höxter Willebadessen ist eine Kleinstadt in Nordrhein-Westfalen und gehört zum Kreis Höxter im Regierungsbezirk Detmold. Die Stadt liegt im Osten des Regierungsbezirkes mitten im Naturpark Teutoburger Wald/Eggegebirge. Für den heu...
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
            Read more

            Residenzschloss Arolsen

            Image
            Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
            Read more