Greatest value of $|z|$ given that $left|z- frac 4z right| = 2$. Why does using triangle inequality in this...












2












$begingroup$



Greatest value of $|z|$ given that $left|z- dfrac 4z right| = 2$ is?




This has been asked here before but my question is about a specific method that doesn't seem to work.



The method is:



$left|z- dfrac 4z right|le |z|+ left|dfrac{4}{z}right|implies 2 le |z|+ dfrac{4}{|z|}$



$implies |z|^2 - 2|z| +4ge 0$ since $|z| > 0$ (here)



which is true for all values of $|z|$.



Hence $|z| in (0, infty)$.



Then why is the maximum given by $|z| =1 + sqrt 5$, in other words, what's the fault in this method ?



A user has also asked that in one of the comments but hasn't received a reply there.






algebra-precalculus proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    the fault of the method is that the triangle inequality gives you something that could be bigger. It doesn't help us to say a hundred million godzillion might be bigger than the maximum.
    $endgroup$
    – fleablood
    Dec 12 '18 at 20:50










  • $begingroup$
    "which is true for all values of $|z|$." Which doesn't tell you anything, because what you've basically done is just confirming that the triangle inequality holds.
    $endgroup$
    – Arthur
    Dec 12 '18 at 20:51






  • 1




    $begingroup$
    Basically that shows $|z| le 1+sqrt{5} le infty$. that's perfectly true. And you did nothing wrong. It just doesn't help us in the least bit.
    $endgroup$
    – fleablood
    Dec 12 '18 at 20:52






  • 1




    $begingroup$
    Suppose several people were to guess my brother's weight. One person puts him on a scale and says "He is 178 lbs and 3 oz". Another person weighs him against a cat, then a dog then a sheep then a seal. And says "He weighs between 120 lbs and 250 lbs". A third person weighs him against a battleship and says "He weighs less then 300,000 tons". What was wrong with the battleship method? Technically not a thing.
    $endgroup$
    – fleablood
    Dec 12 '18 at 20:59
















2












$begingroup$



Greatest value of $|z|$ given that $left|z- dfrac 4z right| = 2$ is?




This has been asked here before but my question is about a specific method that doesn't seem to work.



The method is:



$left|z- dfrac 4z right|le |z|+ left|dfrac{4}{z}right|implies 2 le |z|+ dfrac{4}{|z|}$



$implies |z|^2 - 2|z| +4ge 0$ since $|z| > 0$ (here)



which is true for all values of $|z|$.



Hence $|z| in (0, infty)$.



Then why is the maximum given by $|z| =1 + sqrt 5$, in other words, what's the fault in this method ?



A user has also asked that in one of the comments but hasn't received a reply there.






algebra-precalculus proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    the fault of the method is that the triangle inequality gives you something that could be bigger. It doesn't help us to say a hundred million godzillion might be bigger than the maximum.
    $endgroup$
    – fleablood
    Dec 12 '18 at 20:50










  • $begingroup$
    "which is true for all values of $|z|$." Which doesn't tell you anything, because what you've basically done is just confirming that the triangle inequality holds.
    $endgroup$
    – Arthur
    Dec 12 '18 at 20:51






  • 1




    $begingroup$
    Basically that shows $|z| le 1+sqrt{5} le infty$. that's perfectly true. And you did nothing wrong. It just doesn't help us in the least bit.
    $endgroup$
    – fleablood
    Dec 12 '18 at 20:52






  • 1




    $begingroup$
    Suppose several people were to guess my brother's weight. One person puts him on a scale and says "He is 178 lbs and 3 oz". Another person weighs him against a cat, then a dog then a sheep then a seal. And says "He weighs between 120 lbs and 250 lbs". A third person weighs him against a battleship and says "He weighs less then 300,000 tons". What was wrong with the battleship method? Technically not a thing.
    $endgroup$
    – fleablood
    Dec 12 '18 at 20:59














2












2








2





$begingroup$



Greatest value of $|z|$ given that $left|z- dfrac 4z right| = 2$ is?




This has been asked here before but my question is about a specific method that doesn't seem to work.



The method is:



$left|z- dfrac 4z right|le |z|+ left|dfrac{4}{z}right|implies 2 le |z|+ dfrac{4}{|z|}$



$implies |z|^2 - 2|z| +4ge 0$ since $|z| > 0$ (here)



which is true for all values of $|z|$.



Hence $|z| in (0, infty)$.



Then why is the maximum given by $|z| =1 + sqrt 5$, in other words, what's the fault in this method ?



A user has also asked that in one of the comments but hasn't received a reply there.






algebra-precalculus proof-verification







share|cite|improve this question











$endgroup$





Greatest value of $|z|$ given that $left|z- dfrac 4z right| = 2$ is?




This has been asked here before but my question is about a specific method that doesn't seem to work.



The method is:



$left|z- dfrac 4z right|le |z|+ left|dfrac{4}{z}right|implies 2 le |z|+ dfrac{4}{|z|}$



$implies |z|^2 - 2|z| +4ge 0$ since $|z| > 0$ (here)



which is true for all values of $|z|$.



Hence $|z| in (0, infty)$.



Then why is the maximum given by $|z| =1 + sqrt 5$, in other words, what's the fault in this method ?



A user has also asked that in one of the comments but hasn't received a reply there.






algebra-precalculus proof-verification




algebra-precalculus proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 21:30









Lorenzo B.

1,8402520




1,8402520










asked Dec 12 '18 at 20:45









AbcdAbcd

3,03831235




3,03831235












  • $begingroup$
    the fault of the method is that the triangle inequality gives you something that could be bigger. It doesn't help us to say a hundred million godzillion might be bigger than the maximum.
    $endgroup$
    – fleablood
    Dec 12 '18 at 20:50










  • $begingroup$
    "which is true for all values of $|z|$." Which doesn't tell you anything, because what you've basically done is just confirming that the triangle inequality holds.
    $endgroup$
    – Arthur
    Dec 12 '18 at 20:51






  • 1




    $begingroup$
    Basically that shows $|z| le 1+sqrt{5} le infty$. that's perfectly true. And you did nothing wrong. It just doesn't help us in the least bit.
    $endgroup$
    – fleablood
    Dec 12 '18 at 20:52






  • 1




    $begingroup$
    Suppose several people were to guess my brother's weight. One person puts him on a scale and says "He is 178 lbs and 3 oz". Another person weighs him against a cat, then a dog then a sheep then a seal. And says "He weighs between 120 lbs and 250 lbs". A third person weighs him against a battleship and says "He weighs less then 300,000 tons". What was wrong with the battleship method? Technically not a thing.
    $endgroup$
    – fleablood
    Dec 12 '18 at 20:59


















  • $begingroup$
    the fault of the method is that the triangle inequality gives you something that could be bigger. It doesn't help us to say a hundred million godzillion might be bigger than the maximum.
    $endgroup$
    – fleablood
    Dec 12 '18 at 20:50










  • $begingroup$
    "which is true for all values of $|z|$." Which doesn't tell you anything, because what you've basically done is just confirming that the triangle inequality holds.
    $endgroup$
    – Arthur
    Dec 12 '18 at 20:51






  • 1




    $begingroup$
    Basically that shows $|z| le 1+sqrt{5} le infty$. that's perfectly true. And you did nothing wrong. It just doesn't help us in the least bit.
    $endgroup$
    – fleablood
    Dec 12 '18 at 20:52






  • 1




    $begingroup$
    Suppose several people were to guess my brother's weight. One person puts him on a scale and says "He is 178 lbs and 3 oz". Another person weighs him against a cat, then a dog then a sheep then a seal. And says "He weighs between 120 lbs and 250 lbs". A third person weighs him against a battleship and says "He weighs less then 300,000 tons". What was wrong with the battleship method? Technically not a thing.
    $endgroup$
    – fleablood
    Dec 12 '18 at 20:59
















$begingroup$
the fault of the method is that the triangle inequality gives you something that could be bigger. It doesn't help us to say a hundred million godzillion might be bigger than the maximum.
$endgroup$
– fleablood
Dec 12 '18 at 20:50




$begingroup$
the fault of the method is that the triangle inequality gives you something that could be bigger. It doesn't help us to say a hundred million godzillion might be bigger than the maximum.
$endgroup$
– fleablood
Dec 12 '18 at 20:50












$begingroup$
"which is true for all values of $|z|$." Which doesn't tell you anything, because what you've basically done is just confirming that the triangle inequality holds.
$endgroup$
– Arthur
Dec 12 '18 at 20:51




$begingroup$
"which is true for all values of $|z|$." Which doesn't tell you anything, because what you've basically done is just confirming that the triangle inequality holds.
$endgroup$
– Arthur
Dec 12 '18 at 20:51




1




1




$begingroup$
Basically that shows $|z| le 1+sqrt{5} le infty$. that's perfectly true. And you did nothing wrong. It just doesn't help us in the least bit.
$endgroup$
– fleablood
Dec 12 '18 at 20:52




$begingroup$
Basically that shows $|z| le 1+sqrt{5} le infty$. that's perfectly true. And you did nothing wrong. It just doesn't help us in the least bit.
$endgroup$
– fleablood
Dec 12 '18 at 20:52




1




1




$begingroup$
Suppose several people were to guess my brother's weight. One person puts him on a scale and says "He is 178 lbs and 3 oz". Another person weighs him against a cat, then a dog then a sheep then a seal. And says "He weighs between 120 lbs and 250 lbs". A third person weighs him against a battleship and says "He weighs less then 300,000 tons". What was wrong with the battleship method? Technically not a thing.
$endgroup$
– fleablood
Dec 12 '18 at 20:59




$begingroup$
Suppose several people were to guess my brother's weight. One person puts him on a scale and says "He is 178 lbs and 3 oz". Another person weighs him against a cat, then a dog then a sheep then a seal. And says "He weighs between 120 lbs and 250 lbs". A third person weighs him against a battleship and says "He weighs less then 300,000 tons". What was wrong with the battleship method? Technically not a thing.
$endgroup$
– fleablood
Dec 12 '18 at 20:59










2 Answers
2






active

oldest

votes


















4












$begingroup$

You can apply the triangle inequality this way:
$2 geq |z-frac{4}{z}| geq |z|-|frac{4}{z}|$, whence $|z|^2-2|z|-4 leq 0$ whence $|z| leq 1 + sqrt{5}$ and note(very important!) that equality is indeed attained for $z=1+sqrt{5}$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    I was solving a similar problem in this question



    edit



    Assume $zneq 0.;$
    $left|z-frac{4}{z}right|$ is the distance of points representing $z$ and $frac{4}{z}.$ The maximum of $|z|$ is achieved when $0, z, frac 4z$ are collinear (see bellow) and we have to take the difference, not the sum of absolute values.



    Solution




    • If $z$ is real, then $z, frac 4z$ lie on the same half-line starting in $0$ and we have
      $$left|z-frac 4z right|=|z|-frac {4}{|z|}quad text{or} quad left| z-frac 4z right|=frac{4}{|z|}-|z|,$$
      and so $$2=|z|-frac {4}{|z|} quad text{or} quad 2=frac{4}{|z|}-|z|.$$


    Multiplying by $|z|$ and solving the quadratic equations gives positive solutions $|z|=sqrt 5 +1;$ from the first and $|z|=sqrt 5 -1;$ from the second one. Note that $sqrt 5 -1=frac{4}{sqrt 5 +1}.$




    • If $z=ib, bin mathbb{R},$ then $0$ lies on the segment with bounds $z, frac 4z$. The equation to solve is then $2=|z|+frac {4}{|z|}$ and doesn't have solution.


    • In all other cases, by triangle inequality is $|z|<sqrt 5 +1.$



    The maximal value of $|z|$ is $sqrt5 + 1,$ the only convenient numbers are $z=pm(sqrt 5 +1).$






    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

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      4












      $begingroup$

      You can apply the triangle inequality this way:
      $2 geq |z-frac{4}{z}| geq |z|-|frac{4}{z}|$, whence $|z|^2-2|z|-4 leq 0$ whence $|z| leq 1 + sqrt{5}$ and note(very important!) that equality is indeed attained for $z=1+sqrt{5}$






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        You can apply the triangle inequality this way:
        $2 geq |z-frac{4}{z}| geq |z|-|frac{4}{z}|$, whence $|z|^2-2|z|-4 leq 0$ whence $|z| leq 1 + sqrt{5}$ and note(very important!) that equality is indeed attained for $z=1+sqrt{5}$






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          You can apply the triangle inequality this way:
          $2 geq |z-frac{4}{z}| geq |z|-|frac{4}{z}|$, whence $|z|^2-2|z|-4 leq 0$ whence $|z| leq 1 + sqrt{5}$ and note(very important!) that equality is indeed attained for $z=1+sqrt{5}$






          share|cite|improve this answer









          $endgroup$



          You can apply the triangle inequality this way:
          $2 geq |z-frac{4}{z}| geq |z|-|frac{4}{z}|$, whence $|z|^2-2|z|-4 leq 0$ whence $|z| leq 1 + sqrt{5}$ and note(very important!) that equality is indeed attained for $z=1+sqrt{5}$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 21:05









          Sorin TircSorin Tirc

          1,755213




          1,755213























              1












              $begingroup$

              I was solving a similar problem in this question



              edit



              Assume $zneq 0.;$
              $left|z-frac{4}{z}right|$ is the distance of points representing $z$ and $frac{4}{z}.$ The maximum of $|z|$ is achieved when $0, z, frac 4z$ are collinear (see bellow) and we have to take the difference, not the sum of absolute values.



              Solution




              • If $z$ is real, then $z, frac 4z$ lie on the same half-line starting in $0$ and we have
                $$left|z-frac 4z right|=|z|-frac {4}{|z|}quad text{or} quad left| z-frac 4z right|=frac{4}{|z|}-|z|,$$
                and so $$2=|z|-frac {4}{|z|} quad text{or} quad 2=frac{4}{|z|}-|z|.$$


              Multiplying by $|z|$ and solving the quadratic equations gives positive solutions $|z|=sqrt 5 +1;$ from the first and $|z|=sqrt 5 -1;$ from the second one. Note that $sqrt 5 -1=frac{4}{sqrt 5 +1}.$




              • If $z=ib, bin mathbb{R},$ then $0$ lies on the segment with bounds $z, frac 4z$. The equation to solve is then $2=|z|+frac {4}{|z|}$ and doesn't have solution.


              • In all other cases, by triangle inequality is $|z|<sqrt 5 +1.$



              The maximal value of $|z|$ is $sqrt5 + 1,$ the only convenient numbers are $z=pm(sqrt 5 +1).$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                I was solving a similar problem in this question



                edit



                Assume $zneq 0.;$
                $left|z-frac{4}{z}right|$ is the distance of points representing $z$ and $frac{4}{z}.$ The maximum of $|z|$ is achieved when $0, z, frac 4z$ are collinear (see bellow) and we have to take the difference, not the sum of absolute values.



                Solution




                • If $z$ is real, then $z, frac 4z$ lie on the same half-line starting in $0$ and we have
                  $$left|z-frac 4z right|=|z|-frac {4}{|z|}quad text{or} quad left| z-frac 4z right|=frac{4}{|z|}-|z|,$$
                  and so $$2=|z|-frac {4}{|z|} quad text{or} quad 2=frac{4}{|z|}-|z|.$$


                Multiplying by $|z|$ and solving the quadratic equations gives positive solutions $|z|=sqrt 5 +1;$ from the first and $|z|=sqrt 5 -1;$ from the second one. Note that $sqrt 5 -1=frac{4}{sqrt 5 +1}.$




                • If $z=ib, bin mathbb{R},$ then $0$ lies on the segment with bounds $z, frac 4z$. The equation to solve is then $2=|z|+frac {4}{|z|}$ and doesn't have solution.


                • In all other cases, by triangle inequality is $|z|<sqrt 5 +1.$



                The maximal value of $|z|$ is $sqrt5 + 1,$ the only convenient numbers are $z=pm(sqrt 5 +1).$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  I was solving a similar problem in this question



                  edit



                  Assume $zneq 0.;$
                  $left|z-frac{4}{z}right|$ is the distance of points representing $z$ and $frac{4}{z}.$ The maximum of $|z|$ is achieved when $0, z, frac 4z$ are collinear (see bellow) and we have to take the difference, not the sum of absolute values.



                  Solution




                  • If $z$ is real, then $z, frac 4z$ lie on the same half-line starting in $0$ and we have
                    $$left|z-frac 4z right|=|z|-frac {4}{|z|}quad text{or} quad left| z-frac 4z right|=frac{4}{|z|}-|z|,$$
                    and so $$2=|z|-frac {4}{|z|} quad text{or} quad 2=frac{4}{|z|}-|z|.$$


                  Multiplying by $|z|$ and solving the quadratic equations gives positive solutions $|z|=sqrt 5 +1;$ from the first and $|z|=sqrt 5 -1;$ from the second one. Note that $sqrt 5 -1=frac{4}{sqrt 5 +1}.$




                  • If $z=ib, bin mathbb{R},$ then $0$ lies on the segment with bounds $z, frac 4z$. The equation to solve is then $2=|z|+frac {4}{|z|}$ and doesn't have solution.


                  • In all other cases, by triangle inequality is $|z|<sqrt 5 +1.$



                  The maximal value of $|z|$ is $sqrt5 + 1,$ the only convenient numbers are $z=pm(sqrt 5 +1).$






                  share|cite|improve this answer











                  $endgroup$



                  I was solving a similar problem in this question



                  edit



                  Assume $zneq 0.;$
                  $left|z-frac{4}{z}right|$ is the distance of points representing $z$ and $frac{4}{z}.$ The maximum of $|z|$ is achieved when $0, z, frac 4z$ are collinear (see bellow) and we have to take the difference, not the sum of absolute values.



                  Solution




                  • If $z$ is real, then $z, frac 4z$ lie on the same half-line starting in $0$ and we have
                    $$left|z-frac 4z right|=|z|-frac {4}{|z|}quad text{or} quad left| z-frac 4z right|=frac{4}{|z|}-|z|,$$
                    and so $$2=|z|-frac {4}{|z|} quad text{or} quad 2=frac{4}{|z|}-|z|.$$


                  Multiplying by $|z|$ and solving the quadratic equations gives positive solutions $|z|=sqrt 5 +1;$ from the first and $|z|=sqrt 5 -1;$ from the second one. Note that $sqrt 5 -1=frac{4}{sqrt 5 +1}.$




                  • If $z=ib, bin mathbb{R},$ then $0$ lies on the segment with bounds $z, frac 4z$. The equation to solve is then $2=|z|+frac {4}{|z|}$ and doesn't have solution.


                  • In all other cases, by triangle inequality is $|z|<sqrt 5 +1.$



                  The maximal value of $|z|$ is $sqrt5 + 1,$ the only convenient numbers are $z=pm(sqrt 5 +1).$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 13 '18 at 12:11

























                  answered Dec 12 '18 at 21:08









                  user376343user376343

                  3,6683827




                  3,6683827






























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