Complex inequality $|a+b|le |a|+|b|$












2












$begingroup$


My textbook says if $a$ and $b$ are two complex numbers, then $$|a+b|le |a|+|b|,$$ and the equality holds if and only if $abar{b} ge 0$.



How can we say the equality holds if and only if $abar b ge 0$? I think $abar b$ is a complex number and complex numbers do not have order.



If we square both sides and cancel some terms, then we can see that the equality holds if Re$(abar b) = |a||b|$.



It is on page 9 of Ahlfors' Complex Analysis.






complex-analysis proof-verification complex-numbers proof-explanation







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What they mean with "$abar bgeqslant0$" is that $abar b$ should be a nonnegative real number. Yes, this is sloppy writing.
    $endgroup$
    – Did
    Dec 31 '18 at 21:33












  • $begingroup$
    See the parenthesis in Ahlfors immediately after this point: it is convenient to let $c>0$ indicate that $c$ is real and positive.
    $endgroup$
    – GEdgar
    Dec 31 '18 at 21:53












  • $begingroup$
    We can show $abar b$ is nonnegative real number by expanding $abar b$ and notice that its imaginary part is zero since we have the relation Re$(abar b) = |a||b|$.
    $endgroup$
    – user398843
    Dec 31 '18 at 21:59
















2












$begingroup$


My textbook says if $a$ and $b$ are two complex numbers, then $$|a+b|le |a|+|b|,$$ and the equality holds if and only if $abar{b} ge 0$.



How can we say the equality holds if and only if $abar b ge 0$? I think $abar b$ is a complex number and complex numbers do not have order.



If we square both sides and cancel some terms, then we can see that the equality holds if Re$(abar b) = |a||b|$.



It is on page 9 of Ahlfors' Complex Analysis.






complex-analysis proof-verification complex-numbers proof-explanation







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What they mean with "$abar bgeqslant0$" is that $abar b$ should be a nonnegative real number. Yes, this is sloppy writing.
    $endgroup$
    – Did
    Dec 31 '18 at 21:33












  • $begingroup$
    See the parenthesis in Ahlfors immediately after this point: it is convenient to let $c>0$ indicate that $c$ is real and positive.
    $endgroup$
    – GEdgar
    Dec 31 '18 at 21:53












  • $begingroup$
    We can show $abar b$ is nonnegative real number by expanding $abar b$ and notice that its imaginary part is zero since we have the relation Re$(abar b) = |a||b|$.
    $endgroup$
    – user398843
    Dec 31 '18 at 21:59














2












2








2





$begingroup$


My textbook says if $a$ and $b$ are two complex numbers, then $$|a+b|le |a|+|b|,$$ and the equality holds if and only if $abar{b} ge 0$.



How can we say the equality holds if and only if $abar b ge 0$? I think $abar b$ is a complex number and complex numbers do not have order.



If we square both sides and cancel some terms, then we can see that the equality holds if Re$(abar b) = |a||b|$.



It is on page 9 of Ahlfors' Complex Analysis.






complex-analysis proof-verification complex-numbers proof-explanation







share|cite|improve this question









$endgroup$




My textbook says if $a$ and $b$ are two complex numbers, then $$|a+b|le |a|+|b|,$$ and the equality holds if and only if $abar{b} ge 0$.



How can we say the equality holds if and only if $abar b ge 0$? I think $abar b$ is a complex number and complex numbers do not have order.



If we square both sides and cancel some terms, then we can see that the equality holds if Re$(abar b) = |a||b|$.



It is on page 9 of Ahlfors' Complex Analysis.






complex-analysis proof-verification complex-numbers proof-explanation




complex-analysis proof-verification complex-numbers proof-explanation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 31 '18 at 21:26









user398843user398843

718316




718316








  • 1




    $begingroup$
    What they mean with "$abar bgeqslant0$" is that $abar b$ should be a nonnegative real number. Yes, this is sloppy writing.
    $endgroup$
    – Did
    Dec 31 '18 at 21:33












  • $begingroup$
    See the parenthesis in Ahlfors immediately after this point: it is convenient to let $c>0$ indicate that $c$ is real and positive.
    $endgroup$
    – GEdgar
    Dec 31 '18 at 21:53












  • $begingroup$
    We can show $abar b$ is nonnegative real number by expanding $abar b$ and notice that its imaginary part is zero since we have the relation Re$(abar b) = |a||b|$.
    $endgroup$
    – user398843
    Dec 31 '18 at 21:59














  • 1




    $begingroup$
    What they mean with "$abar bgeqslant0$" is that $abar b$ should be a nonnegative real number. Yes, this is sloppy writing.
    $endgroup$
    – Did
    Dec 31 '18 at 21:33












  • $begingroup$
    See the parenthesis in Ahlfors immediately after this point: it is convenient to let $c>0$ indicate that $c$ is real and positive.
    $endgroup$
    – GEdgar
    Dec 31 '18 at 21:53












  • $begingroup$
    We can show $abar b$ is nonnegative real number by expanding $abar b$ and notice that its imaginary part is zero since we have the relation Re$(abar b) = |a||b|$.
    $endgroup$
    – user398843
    Dec 31 '18 at 21:59








1




1




$begingroup$
What they mean with "$abar bgeqslant0$" is that $abar b$ should be a nonnegative real number. Yes, this is sloppy writing.
$endgroup$
– Did
Dec 31 '18 at 21:33






$begingroup$
What they mean with "$abar bgeqslant0$" is that $abar b$ should be a nonnegative real number. Yes, this is sloppy writing.
$endgroup$
– Did
Dec 31 '18 at 21:33














$begingroup$
See the parenthesis in Ahlfors immediately after this point: it is convenient to let $c>0$ indicate that $c$ is real and positive.
$endgroup$
– GEdgar
Dec 31 '18 at 21:53






$begingroup$
See the parenthesis in Ahlfors immediately after this point: it is convenient to let $c>0$ indicate that $c$ is real and positive.
$endgroup$
– GEdgar
Dec 31 '18 at 21:53














$begingroup$
We can show $abar b$ is nonnegative real number by expanding $abar b$ and notice that its imaginary part is zero since we have the relation Re$(abar b) = |a||b|$.
$endgroup$
– user398843
Dec 31 '18 at 21:59




$begingroup$
We can show $abar b$ is nonnegative real number by expanding $abar b$ and notice that its imaginary part is zero since we have the relation Re$(abar b) = |a||b|$.
$endgroup$
– user398843
Dec 31 '18 at 21:59










3 Answers
3






active

oldest

votes


















1












$begingroup$

Note that
$$
operatorname{Re}(abar b)=|a||b|=|ab|=|abar b|
$$
if and only if $operatorname{Im}(abar b)=0$ and $abar bge 0$, so $abar b$ has to be real.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    If $a=1$, $b=-1$, then $operatorname{Im}(abar b)=0$, but $operatorname{Re}(abar b)ne|abar b|$.
    $endgroup$
    – Martin Argerami
    Dec 31 '18 at 21:33










  • $begingroup$
    @MartinArgerami That's true, thanks.
    $endgroup$
    – A.Γ.
    Dec 31 '18 at 21:36



















0












$begingroup$

The inequality holds if and only if $|a+b|^2le(|a|+|b|)^2$ holds, which becomes
$$
|a|^2+abar{b}+bar{a}b+|b|^2le |a|^2+2|a|,|b|+|b|^2
$$
We can cancel real terms from both sides and still get an equivalent inequality:
$$
abar{b}+bar{a}ble2|a|,|b| tag{*}
$$



This surely holds and is strict if $abar{b}+bar{a}b<0$ (following Ahlfors's convention that $c>0$ means that $c$ is real and positive).



Let's suppose $abar{b}+bar{a}bge0$. Then squaring is allowed and yields an equivalent inequality:
$$
a^2bar{b}^2+2abar{a}bbar{b}+bar{a}^2b^2le 4abar{a}bbar{b}
$$
that is, moving the (real) right-hand side to the left
$$
(abar{b}-bar{a}b)^2le0
$$
which is true, because $abar{b}-bar{a}b$ is purely imaginary.



A necessary condition for having an equality is that $abar{b}=bar{a}b$, that is $abar{b}$ is real. Together with (*) we obtain $abar{b}=2|a|,|b|ge0$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Think of $a$ and $b$ as being planar vectors. Draw the parallelogram and apply the parallelogram law. You will see the geometric meaning of this immediately.






    share|cite|improve this answer









    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Note that
      $$
      operatorname{Re}(abar b)=|a||b|=|ab|=|abar b|
      $$
      if and only if $operatorname{Im}(abar b)=0$ and $abar bge 0$, so $abar b$ has to be real.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        If $a=1$, $b=-1$, then $operatorname{Im}(abar b)=0$, but $operatorname{Re}(abar b)ne|abar b|$.
        $endgroup$
        – Martin Argerami
        Dec 31 '18 at 21:33










      • $begingroup$
        @MartinArgerami That's true, thanks.
        $endgroup$
        – A.Γ.
        Dec 31 '18 at 21:36
















      1












      $begingroup$

      Note that
      $$
      operatorname{Re}(abar b)=|a||b|=|ab|=|abar b|
      $$
      if and only if $operatorname{Im}(abar b)=0$ and $abar bge 0$, so $abar b$ has to be real.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        If $a=1$, $b=-1$, then $operatorname{Im}(abar b)=0$, but $operatorname{Re}(abar b)ne|abar b|$.
        $endgroup$
        – Martin Argerami
        Dec 31 '18 at 21:33










      • $begingroup$
        @MartinArgerami That's true, thanks.
        $endgroup$
        – A.Γ.
        Dec 31 '18 at 21:36














      1












      1








      1





      $begingroup$

      Note that
      $$
      operatorname{Re}(abar b)=|a||b|=|ab|=|abar b|
      $$
      if and only if $operatorname{Im}(abar b)=0$ and $abar bge 0$, so $abar b$ has to be real.






      share|cite|improve this answer











      $endgroup$



      Note that
      $$
      operatorname{Re}(abar b)=|a||b|=|ab|=|abar b|
      $$
      if and only if $operatorname{Im}(abar b)=0$ and $abar bge 0$, so $abar b$ has to be real.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 31 '18 at 21:34

























      answered Dec 31 '18 at 21:31









      A.Γ.A.Γ.

      22.9k32656




      22.9k32656








      • 1




        $begingroup$
        If $a=1$, $b=-1$, then $operatorname{Im}(abar b)=0$, but $operatorname{Re}(abar b)ne|abar b|$.
        $endgroup$
        – Martin Argerami
        Dec 31 '18 at 21:33










      • $begingroup$
        @MartinArgerami That's true, thanks.
        $endgroup$
        – A.Γ.
        Dec 31 '18 at 21:36














      • 1




        $begingroup$
        If $a=1$, $b=-1$, then $operatorname{Im}(abar b)=0$, but $operatorname{Re}(abar b)ne|abar b|$.
        $endgroup$
        – Martin Argerami
        Dec 31 '18 at 21:33










      • $begingroup$
        @MartinArgerami That's true, thanks.
        $endgroup$
        – A.Γ.
        Dec 31 '18 at 21:36








      1




      1




      $begingroup$
      If $a=1$, $b=-1$, then $operatorname{Im}(abar b)=0$, but $operatorname{Re}(abar b)ne|abar b|$.
      $endgroup$
      – Martin Argerami
      Dec 31 '18 at 21:33




      $begingroup$
      If $a=1$, $b=-1$, then $operatorname{Im}(abar b)=0$, but $operatorname{Re}(abar b)ne|abar b|$.
      $endgroup$
      – Martin Argerami
      Dec 31 '18 at 21:33












      $begingroup$
      @MartinArgerami That's true, thanks.
      $endgroup$
      – A.Γ.
      Dec 31 '18 at 21:36




      $begingroup$
      @MartinArgerami That's true, thanks.
      $endgroup$
      – A.Γ.
      Dec 31 '18 at 21:36











      0












      $begingroup$

      The inequality holds if and only if $|a+b|^2le(|a|+|b|)^2$ holds, which becomes
      $$
      |a|^2+abar{b}+bar{a}b+|b|^2le |a|^2+2|a|,|b|+|b|^2
      $$
      We can cancel real terms from both sides and still get an equivalent inequality:
      $$
      abar{b}+bar{a}ble2|a|,|b| tag{*}
      $$



      This surely holds and is strict if $abar{b}+bar{a}b<0$ (following Ahlfors's convention that $c>0$ means that $c$ is real and positive).



      Let's suppose $abar{b}+bar{a}bge0$. Then squaring is allowed and yields an equivalent inequality:
      $$
      a^2bar{b}^2+2abar{a}bbar{b}+bar{a}^2b^2le 4abar{a}bbar{b}
      $$
      that is, moving the (real) right-hand side to the left
      $$
      (abar{b}-bar{a}b)^2le0
      $$
      which is true, because $abar{b}-bar{a}b$ is purely imaginary.



      A necessary condition for having an equality is that $abar{b}=bar{a}b$, that is $abar{b}$ is real. Together with (*) we obtain $abar{b}=2|a|,|b|ge0$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        The inequality holds if and only if $|a+b|^2le(|a|+|b|)^2$ holds, which becomes
        $$
        |a|^2+abar{b}+bar{a}b+|b|^2le |a|^2+2|a|,|b|+|b|^2
        $$
        We can cancel real terms from both sides and still get an equivalent inequality:
        $$
        abar{b}+bar{a}ble2|a|,|b| tag{*}
        $$



        This surely holds and is strict if $abar{b}+bar{a}b<0$ (following Ahlfors's convention that $c>0$ means that $c$ is real and positive).



        Let's suppose $abar{b}+bar{a}bge0$. Then squaring is allowed and yields an equivalent inequality:
        $$
        a^2bar{b}^2+2abar{a}bbar{b}+bar{a}^2b^2le 4abar{a}bbar{b}
        $$
        that is, moving the (real) right-hand side to the left
        $$
        (abar{b}-bar{a}b)^2le0
        $$
        which is true, because $abar{b}-bar{a}b$ is purely imaginary.



        A necessary condition for having an equality is that $abar{b}=bar{a}b$, that is $abar{b}$ is real. Together with (*) we obtain $abar{b}=2|a|,|b|ge0$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          The inequality holds if and only if $|a+b|^2le(|a|+|b|)^2$ holds, which becomes
          $$
          |a|^2+abar{b}+bar{a}b+|b|^2le |a|^2+2|a|,|b|+|b|^2
          $$
          We can cancel real terms from both sides and still get an equivalent inequality:
          $$
          abar{b}+bar{a}ble2|a|,|b| tag{*}
          $$



          This surely holds and is strict if $abar{b}+bar{a}b<0$ (following Ahlfors's convention that $c>0$ means that $c$ is real and positive).



          Let's suppose $abar{b}+bar{a}bge0$. Then squaring is allowed and yields an equivalent inequality:
          $$
          a^2bar{b}^2+2abar{a}bbar{b}+bar{a}^2b^2le 4abar{a}bbar{b}
          $$
          that is, moving the (real) right-hand side to the left
          $$
          (abar{b}-bar{a}b)^2le0
          $$
          which is true, because $abar{b}-bar{a}b$ is purely imaginary.



          A necessary condition for having an equality is that $abar{b}=bar{a}b$, that is $abar{b}$ is real. Together with (*) we obtain $abar{b}=2|a|,|b|ge0$.






          share|cite|improve this answer









          $endgroup$



          The inequality holds if and only if $|a+b|^2le(|a|+|b|)^2$ holds, which becomes
          $$
          |a|^2+abar{b}+bar{a}b+|b|^2le |a|^2+2|a|,|b|+|b|^2
          $$
          We can cancel real terms from both sides and still get an equivalent inequality:
          $$
          abar{b}+bar{a}ble2|a|,|b| tag{*}
          $$



          This surely holds and is strict if $abar{b}+bar{a}b<0$ (following Ahlfors's convention that $c>0$ means that $c$ is real and positive).



          Let's suppose $abar{b}+bar{a}bge0$. Then squaring is allowed and yields an equivalent inequality:
          $$
          a^2bar{b}^2+2abar{a}bbar{b}+bar{a}^2b^2le 4abar{a}bbar{b}
          $$
          that is, moving the (real) right-hand side to the left
          $$
          (abar{b}-bar{a}b)^2le0
          $$
          which is true, because $abar{b}-bar{a}b$ is purely imaginary.



          A necessary condition for having an equality is that $abar{b}=bar{a}b$, that is $abar{b}$ is real. Together with (*) we obtain $abar{b}=2|a|,|b|ge0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 31 '18 at 22:07









          egregegreg

          186k1486209




          186k1486209























              0












              $begingroup$

              Think of $a$ and $b$ as being planar vectors. Draw the parallelogram and apply the parallelogram law. You will see the geometric meaning of this immediately.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Think of $a$ and $b$ as being planar vectors. Draw the parallelogram and apply the parallelogram law. You will see the geometric meaning of this immediately.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Think of $a$ and $b$ as being planar vectors. Draw the parallelogram and apply the parallelogram law. You will see the geometric meaning of this immediately.






                  share|cite|improve this answer









                  $endgroup$



                  Think of $a$ and $b$ as being planar vectors. Draw the parallelogram and apply the parallelogram law. You will see the geometric meaning of this immediately.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 31 '18 at 23:52









                  ncmathsadistncmathsadist

                  43.2k261103




                  43.2k261103






























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