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

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          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





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            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

            62.6k42262




            62.6k42262












            • $begingroup$
              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

                226k15157275




                226k15157275






























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