Find the minimum of $|z+frac{1}{z}|$ for $|z|ge2$












0












$begingroup$


Find the least value assumed by the function $w=|z+frac{1}{z}|$ where $,|z|ge2$




Attempt

By triangular inequality,$|z+frac{1}{z}|ge ||z|-|frac{1}{z}||$

$|z|ge2......................(1)$
$|frac{1}{z}|lefrac{1}{2}$
$-|frac{1}{z}|gefrac{-1}{2}.........(2)$

$|z|-|frac{1}{z}|ge2-frac{1}{2}$

$|z|-|frac{1}{z}|gefrac{3}{2}$

so $|z+frac{1}{z}|ge frac{3}{2}$

But the answer given in my book is $-frac{1}{8}$






complex-numbers







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  • 1




    $begingroup$
    $-frac 18$ or $+frac 18$? Absolute value cannot be negative.
    $endgroup$
    – user376343
    Dec 18 '18 at 14:01










  • $begingroup$
    A negative answer must be wrong, and I find your calculations persuasive: $z=pm2i$ will give $frac32$
    $endgroup$
    – Henry
    Dec 18 '18 at 14:01
















0












$begingroup$


Find the least value assumed by the function $w=|z+frac{1}{z}|$ where $,|z|ge2$




Attempt

By triangular inequality,$|z+frac{1}{z}|ge ||z|-|frac{1}{z}||$

$|z|ge2......................(1)$
$|frac{1}{z}|lefrac{1}{2}$
$-|frac{1}{z}|gefrac{-1}{2}.........(2)$

$|z|-|frac{1}{z}|ge2-frac{1}{2}$

$|z|-|frac{1}{z}|gefrac{3}{2}$

so $|z+frac{1}{z}|ge frac{3}{2}$

But the answer given in my book is $-frac{1}{8}$






complex-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $-frac 18$ or $+frac 18$? Absolute value cannot be negative.
    $endgroup$
    – user376343
    Dec 18 '18 at 14:01










  • $begingroup$
    A negative answer must be wrong, and I find your calculations persuasive: $z=pm2i$ will give $frac32$
    $endgroup$
    – Henry
    Dec 18 '18 at 14:01














0












0








0





$begingroup$


Find the least value assumed by the function $w=|z+frac{1}{z}|$ where $,|z|ge2$




Attempt

By triangular inequality,$|z+frac{1}{z}|ge ||z|-|frac{1}{z}||$

$|z|ge2......................(1)$
$|frac{1}{z}|lefrac{1}{2}$
$-|frac{1}{z}|gefrac{-1}{2}.........(2)$

$|z|-|frac{1}{z}|ge2-frac{1}{2}$

$|z|-|frac{1}{z}|gefrac{3}{2}$

so $|z+frac{1}{z}|ge frac{3}{2}$

But the answer given in my book is $-frac{1}{8}$






complex-numbers







share|cite|improve this question











$endgroup$




Find the least value assumed by the function $w=|z+frac{1}{z}|$ where $,|z|ge2$




Attempt

By triangular inequality,$|z+frac{1}{z}|ge ||z|-|frac{1}{z}||$

$|z|ge2......................(1)$
$|frac{1}{z}|lefrac{1}{2}$
$-|frac{1}{z}|gefrac{-1}{2}.........(2)$

$|z|-|frac{1}{z}|ge2-frac{1}{2}$

$|z|-|frac{1}{z}|gefrac{3}{2}$

so $|z+frac{1}{z}|ge frac{3}{2}$

But the answer given in my book is $-frac{1}{8}$






complex-numbers




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 14:58









Did

249k23226466




249k23226466










asked Dec 18 '18 at 13:58









user984325user984325

246112




246112








  • 1




    $begingroup$
    $-frac 18$ or $+frac 18$? Absolute value cannot be negative.
    $endgroup$
    – user376343
    Dec 18 '18 at 14:01










  • $begingroup$
    A negative answer must be wrong, and I find your calculations persuasive: $z=pm2i$ will give $frac32$
    $endgroup$
    – Henry
    Dec 18 '18 at 14:01














  • 1




    $begingroup$
    $-frac 18$ or $+frac 18$? Absolute value cannot be negative.
    $endgroup$
    – user376343
    Dec 18 '18 at 14:01










  • $begingroup$
    A negative answer must be wrong, and I find your calculations persuasive: $z=pm2i$ will give $frac32$
    $endgroup$
    – Henry
    Dec 18 '18 at 14:01








1




1




$begingroup$
$-frac 18$ or $+frac 18$? Absolute value cannot be negative.
$endgroup$
– user376343
Dec 18 '18 at 14:01




$begingroup$
$-frac 18$ or $+frac 18$? Absolute value cannot be negative.
$endgroup$
– user376343
Dec 18 '18 at 14:01












$begingroup$
A negative answer must be wrong, and I find your calculations persuasive: $z=pm2i$ will give $frac32$
$endgroup$
– Henry
Dec 18 '18 at 14:01




$begingroup$
A negative answer must be wrong, and I find your calculations persuasive: $z=pm2i$ will give $frac32$
$endgroup$
– Henry
Dec 18 '18 at 14:01










1 Answer
1






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$begingroup$

Let $r = |z| ge 2$ then



$w=|z+frac{1}{z}| = |r e^{i phi}+frac{1}{r} e^{-i phi}|= r |1 +frac{1}{r^2} e^{-2 i phi}|$.



Now the minimum w.r.t. $phi$ is obtained for $phi = pi/2$ and we get



$w ge r |1 -frac{1}{r^2}| = r - 1/r$.



Now for $r ge 2$, the last expression takes its lowest value at the boundary, i.e. at $r=2$, which gives $3/2$. This is tight, since $z = 2 i $ gives exactly that result.






share|cite|improve this answer











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    4












    $begingroup$

    Let $r = |z| ge 2$ then



    $w=|z+frac{1}{z}| = |r e^{i phi}+frac{1}{r} e^{-i phi}|= r |1 +frac{1}{r^2} e^{-2 i phi}|$.



    Now the minimum w.r.t. $phi$ is obtained for $phi = pi/2$ and we get



    $w ge r |1 -frac{1}{r^2}| = r - 1/r$.



    Now for $r ge 2$, the last expression takes its lowest value at the boundary, i.e. at $r=2$, which gives $3/2$. This is tight, since $z = 2 i $ gives exactly that result.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      Let $r = |z| ge 2$ then



      $w=|z+frac{1}{z}| = |r e^{i phi}+frac{1}{r} e^{-i phi}|= r |1 +frac{1}{r^2} e^{-2 i phi}|$.



      Now the minimum w.r.t. $phi$ is obtained for $phi = pi/2$ and we get



      $w ge r |1 -frac{1}{r^2}| = r - 1/r$.



      Now for $r ge 2$, the last expression takes its lowest value at the boundary, i.e. at $r=2$, which gives $3/2$. This is tight, since $z = 2 i $ gives exactly that result.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        Let $r = |z| ge 2$ then



        $w=|z+frac{1}{z}| = |r e^{i phi}+frac{1}{r} e^{-i phi}|= r |1 +frac{1}{r^2} e^{-2 i phi}|$.



        Now the minimum w.r.t. $phi$ is obtained for $phi = pi/2$ and we get



        $w ge r |1 -frac{1}{r^2}| = r - 1/r$.



        Now for $r ge 2$, the last expression takes its lowest value at the boundary, i.e. at $r=2$, which gives $3/2$. This is tight, since $z = 2 i $ gives exactly that result.






        share|cite|improve this answer











        $endgroup$



        Let $r = |z| ge 2$ then



        $w=|z+frac{1}{z}| = |r e^{i phi}+frac{1}{r} e^{-i phi}|= r |1 +frac{1}{r^2} e^{-2 i phi}|$.



        Now the minimum w.r.t. $phi$ is obtained for $phi = pi/2$ and we get



        $w ge r |1 -frac{1}{r^2}| = r - 1/r$.



        Now for $r ge 2$, the last expression takes its lowest value at the boundary, i.e. at $r=2$, which gives $3/2$. This is tight, since $z = 2 i $ gives exactly that result.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 19 '18 at 7:36

























        answered Dec 18 '18 at 14:44









        AndreasAndreas

        8,3961137




        8,3961137






























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