if $|z^2-3|=3|z|$ , then the max value of |z| is:












0












$begingroup$



if $|z^2-3|=3|z|$ , then the max value of |z| is:




I attempted this problem by subbing $z$ in as $re^{itheta}$



so



$$|r^2e^{i2theta}-3|= 3r$$



$$r=frac{3}{sqrt{(cos2theta-3)^2+sin^22theta}}$$



$$r=frac{3}{sqrt{10-6cos2theta}}$$



So we just have to find the minimum of the expression below in the denominator to find the maximum value of $r$ which is $|z|$ which is 2 and gives $|z|=3/2$.



On the other hand the author has used the following inequality




$$|z^2-3|-3|z|=0$$



$$|z^2|-3|z|-3<=0$$




Now, i can't figure out why is my answer not the same as the on which we get by solving the quadratic above $(3+sqrt21)/2$






proof-verification complex-numbers







share|cite|improve this question









$endgroup$

















    0












    $begingroup$



    if $|z^2-3|=3|z|$ , then the max value of |z| is:




    I attempted this problem by subbing $z$ in as $re^{itheta}$



    so



    $$|r^2e^{i2theta}-3|= 3r$$



    $$r=frac{3}{sqrt{(cos2theta-3)^2+sin^22theta}}$$



    $$r=frac{3}{sqrt{10-6cos2theta}}$$



    So we just have to find the minimum of the expression below in the denominator to find the maximum value of $r$ which is $|z|$ which is 2 and gives $|z|=3/2$.



    On the other hand the author has used the following inequality




    $$|z^2-3|-3|z|=0$$



    $$|z^2|-3|z|-3<=0$$




    Now, i can't figure out why is my answer not the same as the on which we get by solving the quadratic above $(3+sqrt21)/2$






    proof-verification complex-numbers







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      2



      $begingroup$



      if $|z^2-3|=3|z|$ , then the max value of |z| is:




      I attempted this problem by subbing $z$ in as $re^{itheta}$



      so



      $$|r^2e^{i2theta}-3|= 3r$$



      $$r=frac{3}{sqrt{(cos2theta-3)^2+sin^22theta}}$$



      $$r=frac{3}{sqrt{10-6cos2theta}}$$



      So we just have to find the minimum of the expression below in the denominator to find the maximum value of $r$ which is $|z|$ which is 2 and gives $|z|=3/2$.



      On the other hand the author has used the following inequality




      $$|z^2-3|-3|z|=0$$



      $$|z^2|-3|z|-3<=0$$




      Now, i can't figure out why is my answer not the same as the on which we get by solving the quadratic above $(3+sqrt21)/2$






      proof-verification complex-numbers







      share|cite|improve this question









      $endgroup$





      if $|z^2-3|=3|z|$ , then the max value of |z| is:




      I attempted this problem by subbing $z$ in as $re^{itheta}$



      so



      $$|r^2e^{i2theta}-3|= 3r$$



      $$r=frac{3}{sqrt{(cos2theta-3)^2+sin^22theta}}$$



      $$r=frac{3}{sqrt{10-6cos2theta}}$$



      So we just have to find the minimum of the expression below in the denominator to find the maximum value of $r$ which is $|z|$ which is 2 and gives $|z|=3/2$.



      On the other hand the author has used the following inequality




      $$|z^2-3|-3|z|=0$$



      $$|z^2|-3|z|-3<=0$$




      Now, i can't figure out why is my answer not the same as the on which we get by solving the quadratic above $(3+sqrt21)/2$






      proof-verification complex-numbers




      proof-verification complex-numbers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 27 '18 at 9:29









      CaptainQuestionCaptainQuestion

      1337




      1337






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Your method is wrong. You are treating $r$ and $theta$ as independent. $theta$ depends on $r$ so you cannot minimize over all values of $theta$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why aren't $r$ and $theta$ independent from what i gather $rcdot e^{itheta}$ over here r is the length/magnitude of the vector and $e^{itheta}$ is the unit direction vector?
            $endgroup$
            – CaptainQuestion
            Dec 27 '18 at 9:39












          • $begingroup$
            @CaptainQuestion The equation $r=frac 3 {sqrt {10-6cos 2theta}}$ shows that $r$ and $theta$ are not independent. You got $r=3/2$ by putting $cos (2theta)=1$ but then these values of $r$ and $theta$ do not satisfy the given equation!
            $endgroup$
            – Kavi Rama Murthy
            Dec 27 '18 at 9:44





















          2












          $begingroup$

          $z=r(cos t+isin t)$ where $rge0, t$ is real



          $$3r=sqrt{(r^2cos2t-3)^2+(r^2sin2t)^2}$$



          As $rge0,$ we safely take square on both sides:



          $$r^4-3r^2(2cos2t+3)+9=0$$



          Now as $t$ is real, $-1lecos2tle1$



          $implies-1ge-(2cos2t+3)ge-5$



          $implies r^4-3r^2+9ge r^4-3r^2(2cos2t+3)+9ge r^4-15r^2+9$



          $implies left(r^2-dfrac32right)^2+9-dfrac94ge0ge r^4-15r^2+9$



          First part is clearly true for real $r$



          As the roots of $r^4-15r^2+9=0$ are $dfrac{15pm3sqrt{21}}2$



          $0ge r^4-15r^2+9impliesdfrac{15-3sqrt{21}}2le r^2ledfrac{15+3sqrt{21}}2$



          Finally $$dfrac{15+3sqrt{21}}2=dfrac{3^2+(sqrt{21})^2+2cdot3cdotsqrt{21}}{2^2}=?$$






          share|cite|improve this answer









          $endgroup$













            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%2f3053740%2fif-z2-3-3z-then-the-max-value-of-z-is%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









            2












            $begingroup$

            Your method is wrong. You are treating $r$ and $theta$ as independent. $theta$ depends on $r$ so you cannot minimize over all values of $theta$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Why aren't $r$ and $theta$ independent from what i gather $rcdot e^{itheta}$ over here r is the length/magnitude of the vector and $e^{itheta}$ is the unit direction vector?
              $endgroup$
              – CaptainQuestion
              Dec 27 '18 at 9:39












            • $begingroup$
              @CaptainQuestion The equation $r=frac 3 {sqrt {10-6cos 2theta}}$ shows that $r$ and $theta$ are not independent. You got $r=3/2$ by putting $cos (2theta)=1$ but then these values of $r$ and $theta$ do not satisfy the given equation!
              $endgroup$
              – Kavi Rama Murthy
              Dec 27 '18 at 9:44


















            2












            $begingroup$

            Your method is wrong. You are treating $r$ and $theta$ as independent. $theta$ depends on $r$ so you cannot minimize over all values of $theta$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Why aren't $r$ and $theta$ independent from what i gather $rcdot e^{itheta}$ over here r is the length/magnitude of the vector and $e^{itheta}$ is the unit direction vector?
              $endgroup$
              – CaptainQuestion
              Dec 27 '18 at 9:39












            • $begingroup$
              @CaptainQuestion The equation $r=frac 3 {sqrt {10-6cos 2theta}}$ shows that $r$ and $theta$ are not independent. You got $r=3/2$ by putting $cos (2theta)=1$ but then these values of $r$ and $theta$ do not satisfy the given equation!
              $endgroup$
              – Kavi Rama Murthy
              Dec 27 '18 at 9:44
















            2












            2








            2





            $begingroup$

            Your method is wrong. You are treating $r$ and $theta$ as independent. $theta$ depends on $r$ so you cannot minimize over all values of $theta$.






            share|cite|improve this answer









            $endgroup$



            Your method is wrong. You are treating $r$ and $theta$ as independent. $theta$ depends on $r$ so you cannot minimize over all values of $theta$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 27 '18 at 9:36









            Kavi Rama MurthyKavi Rama Murthy

            60.5k42161




            60.5k42161












            • $begingroup$
              Why aren't $r$ and $theta$ independent from what i gather $rcdot e^{itheta}$ over here r is the length/magnitude of the vector and $e^{itheta}$ is the unit direction vector?
              $endgroup$
              – CaptainQuestion
              Dec 27 '18 at 9:39












            • $begingroup$
              @CaptainQuestion The equation $r=frac 3 {sqrt {10-6cos 2theta}}$ shows that $r$ and $theta$ are not independent. You got $r=3/2$ by putting $cos (2theta)=1$ but then these values of $r$ and $theta$ do not satisfy the given equation!
              $endgroup$
              – Kavi Rama Murthy
              Dec 27 '18 at 9:44




















            • $begingroup$
              Why aren't $r$ and $theta$ independent from what i gather $rcdot e^{itheta}$ over here r is the length/magnitude of the vector and $e^{itheta}$ is the unit direction vector?
              $endgroup$
              – CaptainQuestion
              Dec 27 '18 at 9:39












            • $begingroup$
              @CaptainQuestion The equation $r=frac 3 {sqrt {10-6cos 2theta}}$ shows that $r$ and $theta$ are not independent. You got $r=3/2$ by putting $cos (2theta)=1$ but then these values of $r$ and $theta$ do not satisfy the given equation!
              $endgroup$
              – Kavi Rama Murthy
              Dec 27 '18 at 9:44


















            $begingroup$
            Why aren't $r$ and $theta$ independent from what i gather $rcdot e^{itheta}$ over here r is the length/magnitude of the vector and $e^{itheta}$ is the unit direction vector?
            $endgroup$
            – CaptainQuestion
            Dec 27 '18 at 9:39






            $begingroup$
            Why aren't $r$ and $theta$ independent from what i gather $rcdot e^{itheta}$ over here r is the length/magnitude of the vector and $e^{itheta}$ is the unit direction vector?
            $endgroup$
            – CaptainQuestion
            Dec 27 '18 at 9:39














            $begingroup$
            @CaptainQuestion The equation $r=frac 3 {sqrt {10-6cos 2theta}}$ shows that $r$ and $theta$ are not independent. You got $r=3/2$ by putting $cos (2theta)=1$ but then these values of $r$ and $theta$ do not satisfy the given equation!
            $endgroup$
            – Kavi Rama Murthy
            Dec 27 '18 at 9:44






            $begingroup$
            @CaptainQuestion The equation $r=frac 3 {sqrt {10-6cos 2theta}}$ shows that $r$ and $theta$ are not independent. You got $r=3/2$ by putting $cos (2theta)=1$ but then these values of $r$ and $theta$ do not satisfy the given equation!
            $endgroup$
            – Kavi Rama Murthy
            Dec 27 '18 at 9:44













            2












            $begingroup$

            $z=r(cos t+isin t)$ where $rge0, t$ is real



            $$3r=sqrt{(r^2cos2t-3)^2+(r^2sin2t)^2}$$



            As $rge0,$ we safely take square on both sides:



            $$r^4-3r^2(2cos2t+3)+9=0$$



            Now as $t$ is real, $-1lecos2tle1$



            $implies-1ge-(2cos2t+3)ge-5$



            $implies r^4-3r^2+9ge r^4-3r^2(2cos2t+3)+9ge r^4-15r^2+9$



            $implies left(r^2-dfrac32right)^2+9-dfrac94ge0ge r^4-15r^2+9$



            First part is clearly true for real $r$



            As the roots of $r^4-15r^2+9=0$ are $dfrac{15pm3sqrt{21}}2$



            $0ge r^4-15r^2+9impliesdfrac{15-3sqrt{21}}2le r^2ledfrac{15+3sqrt{21}}2$



            Finally $$dfrac{15+3sqrt{21}}2=dfrac{3^2+(sqrt{21})^2+2cdot3cdotsqrt{21}}{2^2}=?$$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              $z=r(cos t+isin t)$ where $rge0, t$ is real



              $$3r=sqrt{(r^2cos2t-3)^2+(r^2sin2t)^2}$$



              As $rge0,$ we safely take square on both sides:



              $$r^4-3r^2(2cos2t+3)+9=0$$



              Now as $t$ is real, $-1lecos2tle1$



              $implies-1ge-(2cos2t+3)ge-5$



              $implies r^4-3r^2+9ge r^4-3r^2(2cos2t+3)+9ge r^4-15r^2+9$



              $implies left(r^2-dfrac32right)^2+9-dfrac94ge0ge r^4-15r^2+9$



              First part is clearly true for real $r$



              As the roots of $r^4-15r^2+9=0$ are $dfrac{15pm3sqrt{21}}2$



              $0ge r^4-15r^2+9impliesdfrac{15-3sqrt{21}}2le r^2ledfrac{15+3sqrt{21}}2$



              Finally $$dfrac{15+3sqrt{21}}2=dfrac{3^2+(sqrt{21})^2+2cdot3cdotsqrt{21}}{2^2}=?$$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                $z=r(cos t+isin t)$ where $rge0, t$ is real



                $$3r=sqrt{(r^2cos2t-3)^2+(r^2sin2t)^2}$$



                As $rge0,$ we safely take square on both sides:



                $$r^4-3r^2(2cos2t+3)+9=0$$



                Now as $t$ is real, $-1lecos2tle1$



                $implies-1ge-(2cos2t+3)ge-5$



                $implies r^4-3r^2+9ge r^4-3r^2(2cos2t+3)+9ge r^4-15r^2+9$



                $implies left(r^2-dfrac32right)^2+9-dfrac94ge0ge r^4-15r^2+9$



                First part is clearly true for real $r$



                As the roots of $r^4-15r^2+9=0$ are $dfrac{15pm3sqrt{21}}2$



                $0ge r^4-15r^2+9impliesdfrac{15-3sqrt{21}}2le r^2ledfrac{15+3sqrt{21}}2$



                Finally $$dfrac{15+3sqrt{21}}2=dfrac{3^2+(sqrt{21})^2+2cdot3cdotsqrt{21}}{2^2}=?$$






                share|cite|improve this answer









                $endgroup$



                $z=r(cos t+isin t)$ where $rge0, t$ is real



                $$3r=sqrt{(r^2cos2t-3)^2+(r^2sin2t)^2}$$



                As $rge0,$ we safely take square on both sides:



                $$r^4-3r^2(2cos2t+3)+9=0$$



                Now as $t$ is real, $-1lecos2tle1$



                $implies-1ge-(2cos2t+3)ge-5$



                $implies r^4-3r^2+9ge r^4-3r^2(2cos2t+3)+9ge r^4-15r^2+9$



                $implies left(r^2-dfrac32right)^2+9-dfrac94ge0ge r^4-15r^2+9$



                First part is clearly true for real $r$



                As the roots of $r^4-15r^2+9=0$ are $dfrac{15pm3sqrt{21}}2$



                $0ge r^4-15r^2+9impliesdfrac{15-3sqrt{21}}2le r^2ledfrac{15+3sqrt{21}}2$



                Finally $$dfrac{15+3sqrt{21}}2=dfrac{3^2+(sqrt{21})^2+2cdot3cdotsqrt{21}}{2^2}=?$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 27 '18 at 11:40









                lab bhattacharjeelab bhattacharjee

                225k15157275




                225k15157275






























                    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.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053740%2fif-z2-3-3z-then-the-max-value-of-z-is%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

                    Bressuire

                    Image
                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                    Read more

                    Cabo Verde

                    Image
                      Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                    Read more

                    Gyllenstierna

                    Image
                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                    Read more