Verify that $sqrt{2}left| z right| ge left|Re(z)right| + left|Im(z)right|$












2












$begingroup$


Verify that $sqrt{2}left| z right| ge left|Re(z)right| + left|Im(z)right|.$



I started off noting that $z=x+iy$ and that $Re(z)=x$ and $Im(z)=y$



Then I know that I have to square both sides, giving me



$$2(x^2 +y^2) ge x^2+2|xy|+y^2$$



I'm not sure what to do after this.






inequality complex-numbers







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    $$iff(x-y)^2ge0$$
    $endgroup$
    – lab bhattacharjee
    Sep 3 '14 at 18:06










  • $begingroup$
    @labbhattacharjee, using that property, would I then be able to say that $(x-y)^2$ is equal to $x^2 + y^2 -2xy$. and then maybe adding $x^2 + y^2$ to both sides??
    $endgroup$
    – cele
    Sep 3 '14 at 18:09






  • 1




    $begingroup$
    @cele Your Mathjax code does not work. I was about to fix it but I don't understand the point of these lefts and rights. Could you rewrite the code yourself, for example following this tutorial?
    $endgroup$
    – JiK
    Sep 3 '14 at 18:10










  • $begingroup$
    @JiK its suppose to be absolute value, but let me work on fixing it
    $endgroup$
    – cele
    Sep 3 '14 at 18:11










  • $begingroup$
    @cele You can get the absolute value by left| x right|, or simply by |x| if there's no need to get | signs of different size.
    $endgroup$
    – JiK
    Sep 3 '14 at 18:13
















2












$begingroup$


Verify that $sqrt{2}left| z right| ge left|Re(z)right| + left|Im(z)right|.$



I started off noting that $z=x+iy$ and that $Re(z)=x$ and $Im(z)=y$



Then I know that I have to square both sides, giving me



$$2(x^2 +y^2) ge x^2+2|xy|+y^2$$



I'm not sure what to do after this.






inequality complex-numbers







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    $$iff(x-y)^2ge0$$
    $endgroup$
    – lab bhattacharjee
    Sep 3 '14 at 18:06










  • $begingroup$
    @labbhattacharjee, using that property, would I then be able to say that $(x-y)^2$ is equal to $x^2 + y^2 -2xy$. and then maybe adding $x^2 + y^2$ to both sides??
    $endgroup$
    – cele
    Sep 3 '14 at 18:09






  • 1




    $begingroup$
    @cele Your Mathjax code does not work. I was about to fix it but I don't understand the point of these lefts and rights. Could you rewrite the code yourself, for example following this tutorial?
    $endgroup$
    – JiK
    Sep 3 '14 at 18:10










  • $begingroup$
    @JiK its suppose to be absolute value, but let me work on fixing it
    $endgroup$
    – cele
    Sep 3 '14 at 18:11










  • $begingroup$
    @cele You can get the absolute value by left| x right|, or simply by |x| if there's no need to get | signs of different size.
    $endgroup$
    – JiK
    Sep 3 '14 at 18:13














2












2








2


0



$begingroup$


Verify that $sqrt{2}left| z right| ge left|Re(z)right| + left|Im(z)right|.$



I started off noting that $z=x+iy$ and that $Re(z)=x$ and $Im(z)=y$



Then I know that I have to square both sides, giving me



$$2(x^2 +y^2) ge x^2+2|xy|+y^2$$



I'm not sure what to do after this.






inequality complex-numbers







share|cite|improve this question











$endgroup$




Verify that $sqrt{2}left| z right| ge left|Re(z)right| + left|Im(z)right|.$



I started off noting that $z=x+iy$ and that $Re(z)=x$ and $Im(z)=y$



Then I know that I have to square both sides, giving me



$$2(x^2 +y^2) ge x^2+2|xy|+y^2$$



I'm not sure what to do after this.






inequality complex-numbers




inequality complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 3 '14 at 18:17









Jack D'Aurizio

292k33284672




292k33284672










asked Sep 3 '14 at 18:01









celecele

63211029




63211029








  • 4




    $begingroup$
    $$iff(x-y)^2ge0$$
    $endgroup$
    – lab bhattacharjee
    Sep 3 '14 at 18:06










  • $begingroup$
    @labbhattacharjee, using that property, would I then be able to say that $(x-y)^2$ is equal to $x^2 + y^2 -2xy$. and then maybe adding $x^2 + y^2$ to both sides??
    $endgroup$
    – cele
    Sep 3 '14 at 18:09






  • 1




    $begingroup$
    @cele Your Mathjax code does not work. I was about to fix it but I don't understand the point of these lefts and rights. Could you rewrite the code yourself, for example following this tutorial?
    $endgroup$
    – JiK
    Sep 3 '14 at 18:10










  • $begingroup$
    @JiK its suppose to be absolute value, but let me work on fixing it
    $endgroup$
    – cele
    Sep 3 '14 at 18:11










  • $begingroup$
    @cele You can get the absolute value by left| x right|, or simply by |x| if there's no need to get | signs of different size.
    $endgroup$
    – JiK
    Sep 3 '14 at 18:13














  • 4




    $begingroup$
    $$iff(x-y)^2ge0$$
    $endgroup$
    – lab bhattacharjee
    Sep 3 '14 at 18:06










  • $begingroup$
    @labbhattacharjee, using that property, would I then be able to say that $(x-y)^2$ is equal to $x^2 + y^2 -2xy$. and then maybe adding $x^2 + y^2$ to both sides??
    $endgroup$
    – cele
    Sep 3 '14 at 18:09






  • 1




    $begingroup$
    @cele Your Mathjax code does not work. I was about to fix it but I don't understand the point of these lefts and rights. Could you rewrite the code yourself, for example following this tutorial?
    $endgroup$
    – JiK
    Sep 3 '14 at 18:10










  • $begingroup$
    @JiK its suppose to be absolute value, but let me work on fixing it
    $endgroup$
    – cele
    Sep 3 '14 at 18:11










  • $begingroup$
    @cele You can get the absolute value by left| x right|, or simply by |x| if there's no need to get | signs of different size.
    $endgroup$
    – JiK
    Sep 3 '14 at 18:13








4




4




$begingroup$
$$iff(x-y)^2ge0$$
$endgroup$
– lab bhattacharjee
Sep 3 '14 at 18:06




$begingroup$
$$iff(x-y)^2ge0$$
$endgroup$
– lab bhattacharjee
Sep 3 '14 at 18:06












$begingroup$
@labbhattacharjee, using that property, would I then be able to say that $(x-y)^2$ is equal to $x^2 + y^2 -2xy$. and then maybe adding $x^2 + y^2$ to both sides??
$endgroup$
– cele
Sep 3 '14 at 18:09




$begingroup$
@labbhattacharjee, using that property, would I then be able to say that $(x-y)^2$ is equal to $x^2 + y^2 -2xy$. and then maybe adding $x^2 + y^2$ to both sides??
$endgroup$
– cele
Sep 3 '14 at 18:09




1




1




$begingroup$
@cele Your Mathjax code does not work. I was about to fix it but I don't understand the point of these lefts and rights. Could you rewrite the code yourself, for example following this tutorial?
$endgroup$
– JiK
Sep 3 '14 at 18:10




$begingroup$
@cele Your Mathjax code does not work. I was about to fix it but I don't understand the point of these lefts and rights. Could you rewrite the code yourself, for example following this tutorial?
$endgroup$
– JiK
Sep 3 '14 at 18:10












$begingroup$
@JiK its suppose to be absolute value, but let me work on fixing it
$endgroup$
– cele
Sep 3 '14 at 18:11




$begingroup$
@JiK its suppose to be absolute value, but let me work on fixing it
$endgroup$
– cele
Sep 3 '14 at 18:11












$begingroup$
@cele You can get the absolute value by left| x right|, or simply by |x| if there's no need to get | signs of different size.
$endgroup$
– JiK
Sep 3 '14 at 18:13




$begingroup$
@cele You can get the absolute value by left| x right|, or simply by |x| if there's no need to get | signs of different size.
$endgroup$
– JiK
Sep 3 '14 at 18:13










2 Answers
2






active

oldest

votes


















0












$begingroup$

Let $z=x+iy$, We have $(x-y)^2geq 0$, therefore
begin{align}
2|z|^2=2(|x|^2+|y|^2)&geq x^2+y^2+2|x||y|\
&=(|x|+|y|)^2\
&=(|Rez|+|Imz|)^2
end{align}
so we have that
$$sqrt{2}|z|geq|Rez|+|Imz|$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint:



    Let $z = a + ib$ Then we have:



    $$sqrt{2}sqrt{a^2+b^2} geq |a| + |b| $$
    $$sqrt{2(a^2+b^2)} geq |a| + |b| $$
    $$ 2(a^2+b^2) geq (|a| + |b|)^2$$



    Try to simplify it so LHS is an expression, and the right hand side is $0$. Notice what the left hand side is and confirm the equality.






    share|cite|improve this answer











    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Let $z=x+iy$, We have $(x-y)^2geq 0$, therefore
      begin{align}
      2|z|^2=2(|x|^2+|y|^2)&geq x^2+y^2+2|x||y|\
      &=(|x|+|y|)^2\
      &=(|Rez|+|Imz|)^2
      end{align}
      so we have that
      $$sqrt{2}|z|geq|Rez|+|Imz|$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Let $z=x+iy$, We have $(x-y)^2geq 0$, therefore
        begin{align}
        2|z|^2=2(|x|^2+|y|^2)&geq x^2+y^2+2|x||y|\
        &=(|x|+|y|)^2\
        &=(|Rez|+|Imz|)^2
        end{align}
        so we have that
        $$sqrt{2}|z|geq|Rez|+|Imz|$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Let $z=x+iy$, We have $(x-y)^2geq 0$, therefore
          begin{align}
          2|z|^2=2(|x|^2+|y|^2)&geq x^2+y^2+2|x||y|\
          &=(|x|+|y|)^2\
          &=(|Rez|+|Imz|)^2
          end{align}
          so we have that
          $$sqrt{2}|z|geq|Rez|+|Imz|$$






          share|cite|improve this answer









          $endgroup$



          Let $z=x+iy$, We have $(x-y)^2geq 0$, therefore
          begin{align}
          2|z|^2=2(|x|^2+|y|^2)&geq x^2+y^2+2|x||y|\
          &=(|x|+|y|)^2\
          &=(|Rez|+|Imz|)^2
          end{align}
          so we have that
          $$sqrt{2}|z|geq|Rez|+|Imz|$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 3 '14 at 20:47









          user62498user62498

          1,968714




          1,968714























              0












              $begingroup$

              Hint:



              Let $z = a + ib$ Then we have:



              $$sqrt{2}sqrt{a^2+b^2} geq |a| + |b| $$
              $$sqrt{2(a^2+b^2)} geq |a| + |b| $$
              $$ 2(a^2+b^2) geq (|a| + |b|)^2$$



              Try to simplify it so LHS is an expression, and the right hand side is $0$. Notice what the left hand side is and confirm the equality.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Hint:



                Let $z = a + ib$ Then we have:



                $$sqrt{2}sqrt{a^2+b^2} geq |a| + |b| $$
                $$sqrt{2(a^2+b^2)} geq |a| + |b| $$
                $$ 2(a^2+b^2) geq (|a| + |b|)^2$$



                Try to simplify it so LHS is an expression, and the right hand side is $0$. Notice what the left hand side is and confirm the equality.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint:



                  Let $z = a + ib$ Then we have:



                  $$sqrt{2}sqrt{a^2+b^2} geq |a| + |b| $$
                  $$sqrt{2(a^2+b^2)} geq |a| + |b| $$
                  $$ 2(a^2+b^2) geq (|a| + |b|)^2$$



                  Try to simplify it so LHS is an expression, and the right hand side is $0$. Notice what the left hand side is and confirm the equality.






                  share|cite|improve this answer











                  $endgroup$



                  Hint:



                  Let $z = a + ib$ Then we have:



                  $$sqrt{2}sqrt{a^2+b^2} geq |a| + |b| $$
                  $$sqrt{2(a^2+b^2)} geq |a| + |b| $$
                  $$ 2(a^2+b^2) geq (|a| + |b|)^2$$



                  Try to simplify it so LHS is an expression, and the right hand side is $0$. Notice what the left hand side is and confirm the equality.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Sep 4 '14 at 6:19

























                  answered Sep 4 '14 at 6:11









                  DairDair

                  1,88811124




                  1,88811124






























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