Find the maximum of the $| left( w + 2 right) ^3 left( w - 3 right)^2|$ with $|w|=1$

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Let $w in mathbbC$, and $left | w right | = 1$. Find the maximum of the function $| left( w + 2 right) ^3 left( w - 3 right)^2|$



Since $$|(w+2)^3(w-3)^2|=|w^5-15w^3-10w^2+60w+72|$$



Let $w=cos x+i sin x$. Then we have an ugly form







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    up vote
    3
    down vote

    favorite












    Let $w in mathbbC$, and $left | w right | = 1$. Find the maximum of the function $| left( w + 2 right) ^3 left( w - 3 right)^2|$



    Since $$|(w+2)^3(w-3)^2|=|w^5-15w^3-10w^2+60w+72|$$



    Let $w=cos x+i sin x$. Then we have an ugly form







    share|cite|improve this question
























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      Let $w in mathbbC$, and $left | w right | = 1$. Find the maximum of the function $| left( w + 2 right) ^3 left( w - 3 right)^2|$



      Since $$|(w+2)^3(w-3)^2|=|w^5-15w^3-10w^2+60w+72|$$



      Let $w=cos x+i sin x$. Then we have an ugly form







      share|cite|improve this question














      Let $w in mathbbC$, and $left | w right | = 1$. Find the maximum of the function $| left( w + 2 right) ^3 left( w - 3 right)^2|$



      Since $$|(w+2)^3(w-3)^2|=|w^5-15w^3-10w^2+60w+72|$$



      Let $w=cos x+i sin x$. Then we have an ugly form









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 14 at 13:49









      z100

      48848




      48848










      asked Aug 14 at 8:56









      communnites

      1,204431




      1,204431




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          Hint:



          $$sqrt[5]w+2leqdfracw+25$$



          the equality occurs if $|w+2|=|w-3|$






          share|cite|improve this answer





























            up vote
            0
            down vote













            Calling



            $$
            z_1=rho_1 e^iphi_1 = sqrt(cosphi+2)^2+sin^2phi = sqrt5+4cosphi\
            z_2=rho_2 e^iphi_2 = sqrt(cosphi-2)^2+sin^2phi = sqrt10-6cosphi
            $$



            so



            $$
            |z_1^3z_2^2| = rho_1^3rho_2^2 = left(5+4cosphiright)^frac 32|10-6cosphi|
            $$



            now



            $$
            fracddphileft( left(5+4cosphiright)^frac 32(10-6cosphi)right) = 6 sinphi (4 cosphi+5)^3/2-6 sinphi (10-6 cosphi) sqrt4 cosphi+5=0
            $$



            for $phi in 0,pmfracpi3$ so the maximum is



            $$
            108=max_w| left( w + 2 right) ^3 left( w - 3 right)^2|;;mboxs. t.;;|w| = 1
            $$



            for $phi = 0$






            share|cite|improve this answer






















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              4
              down vote



              accepted










              Hint:



              $$sqrt[5]w+2leqdfracw+25$$



              the equality occurs if $|w+2|=|w-3|$






              share|cite|improve this answer


























                up vote
                4
                down vote



                accepted










                Hint:



                $$sqrt[5]w+2leqdfracw+25$$



                the equality occurs if $|w+2|=|w-3|$






                share|cite|improve this answer
























                  up vote
                  4
                  down vote



                  accepted







                  up vote
                  4
                  down vote



                  accepted






                  Hint:



                  $$sqrt[5]w+2leqdfracw+25$$



                  the equality occurs if $|w+2|=|w-3|$






                  share|cite|improve this answer














                  Hint:



                  $$sqrt[5]w+2leqdfracw+25$$



                  the equality occurs if $|w+2|=|w-3|$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 14 at 9:06









                  Jack D'Aurizio♦

                  271k31266632




                  271k31266632










                  answered Aug 14 at 9:05









                  lab bhattacharjee

                  215k14152264




                  215k14152264




















                      up vote
                      0
                      down vote













                      Calling



                      $$
                      z_1=rho_1 e^iphi_1 = sqrt(cosphi+2)^2+sin^2phi = sqrt5+4cosphi\
                      z_2=rho_2 e^iphi_2 = sqrt(cosphi-2)^2+sin^2phi = sqrt10-6cosphi
                      $$



                      so



                      $$
                      |z_1^3z_2^2| = rho_1^3rho_2^2 = left(5+4cosphiright)^frac 32|10-6cosphi|
                      $$



                      now



                      $$
                      fracddphileft( left(5+4cosphiright)^frac 32(10-6cosphi)right) = 6 sinphi (4 cosphi+5)^3/2-6 sinphi (10-6 cosphi) sqrt4 cosphi+5=0
                      $$



                      for $phi in 0,pmfracpi3$ so the maximum is



                      $$
                      108=max_w| left( w + 2 right) ^3 left( w - 3 right)^2|;;mboxs. t.;;|w| = 1
                      $$



                      for $phi = 0$






                      share|cite|improve this answer


























                        up vote
                        0
                        down vote













                        Calling



                        $$
                        z_1=rho_1 e^iphi_1 = sqrt(cosphi+2)^2+sin^2phi = sqrt5+4cosphi\
                        z_2=rho_2 e^iphi_2 = sqrt(cosphi-2)^2+sin^2phi = sqrt10-6cosphi
                        $$



                        so



                        $$
                        |z_1^3z_2^2| = rho_1^3rho_2^2 = left(5+4cosphiright)^frac 32|10-6cosphi|
                        $$



                        now



                        $$
                        fracddphileft( left(5+4cosphiright)^frac 32(10-6cosphi)right) = 6 sinphi (4 cosphi+5)^3/2-6 sinphi (10-6 cosphi) sqrt4 cosphi+5=0
                        $$



                        for $phi in 0,pmfracpi3$ so the maximum is



                        $$
                        108=max_w| left( w + 2 right) ^3 left( w - 3 right)^2|;;mboxs. t.;;|w| = 1
                        $$



                        for $phi = 0$






                        share|cite|improve this answer
























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          Calling



                          $$
                          z_1=rho_1 e^iphi_1 = sqrt(cosphi+2)^2+sin^2phi = sqrt5+4cosphi\
                          z_2=rho_2 e^iphi_2 = sqrt(cosphi-2)^2+sin^2phi = sqrt10-6cosphi
                          $$



                          so



                          $$
                          |z_1^3z_2^2| = rho_1^3rho_2^2 = left(5+4cosphiright)^frac 32|10-6cosphi|
                          $$



                          now



                          $$
                          fracddphileft( left(5+4cosphiright)^frac 32(10-6cosphi)right) = 6 sinphi (4 cosphi+5)^3/2-6 sinphi (10-6 cosphi) sqrt4 cosphi+5=0
                          $$



                          for $phi in 0,pmfracpi3$ so the maximum is



                          $$
                          108=max_w| left( w + 2 right) ^3 left( w - 3 right)^2|;;mboxs. t.;;|w| = 1
                          $$



                          for $phi = 0$






                          share|cite|improve this answer














                          Calling



                          $$
                          z_1=rho_1 e^iphi_1 = sqrt(cosphi+2)^2+sin^2phi = sqrt5+4cosphi\
                          z_2=rho_2 e^iphi_2 = sqrt(cosphi-2)^2+sin^2phi = sqrt10-6cosphi
                          $$



                          so



                          $$
                          |z_1^3z_2^2| = rho_1^3rho_2^2 = left(5+4cosphiright)^frac 32|10-6cosphi|
                          $$



                          now



                          $$
                          fracddphileft( left(5+4cosphiright)^frac 32(10-6cosphi)right) = 6 sinphi (4 cosphi+5)^3/2-6 sinphi (10-6 cosphi) sqrt4 cosphi+5=0
                          $$



                          for $phi in 0,pmfracpi3$ so the maximum is



                          $$
                          108=max_w| left( w + 2 right) ^3 left( w - 3 right)^2|;;mboxs. t.;;|w| = 1
                          $$



                          for $phi = 0$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Aug 14 at 13:23









                          z100

                          48848




                          48848










                          answered Aug 14 at 13:05









                          Cesareo

                          5,8802412




                          5,8802412






















                               

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