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How to prove triangle inequality for euclidean norm on complex number?


Cauchy–Schwarz inequality for complex numbersProving the reverse triangle inequality of the complex numbersz and w are two complex numbers prove the relationshipElementary-Looking Inequality on n Complex NumbersTriangle Inequality about complex numbers, special caseInequality in the complex spaceTriangle inequality for complex numbersGeneralize standard deviation to vectorsA third triangle inequality?How the prove that the equality in the triangle inequality holds only if $z = tw$ or $w = cz$ for some $t,c in mathbbR^nn$













0












$begingroup$


We were asked to show that when:
$displaystyle Vert ZVert = left(sum_k=1^n (x_k+iy_k)(x_k-iy_k)right)^1/2$ that $Vert Z+WVert leq Vert ZVert+Vert WVert$ whenever $Z$ and $W$ are vectors in complex numbers holds.



Can someone help me to prove this please?










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  • $begingroup$
    Welcome to Math.SE! Please consider editing your question to include your thoughts on the problem, and/or showing what you have already tried.
    $endgroup$
    – Peter Woolfitt
    Mar 21 '15 at 0:58















0












$begingroup$


We were asked to show that when:
$displaystyle Vert ZVert = left(sum_k=1^n (x_k+iy_k)(x_k-iy_k)right)^1/2$ that $Vert Z+WVert leq Vert ZVert+Vert WVert$ whenever $Z$ and $W$ are vectors in complex numbers holds.



Can someone help me to prove this please?










share|cite|improve this question











$endgroup$











  • $begingroup$
    Welcome to Math.SE! Please consider editing your question to include your thoughts on the problem, and/or showing what you have already tried.
    $endgroup$
    – Peter Woolfitt
    Mar 21 '15 at 0:58













0












0








0





$begingroup$


We were asked to show that when:
$displaystyle Vert ZVert = left(sum_k=1^n (x_k+iy_k)(x_k-iy_k)right)^1/2$ that $Vert Z+WVert leq Vert ZVert+Vert WVert$ whenever $Z$ and $W$ are vectors in complex numbers holds.



Can someone help me to prove this please?










share|cite|improve this question











$endgroup$




We were asked to show that when:
$displaystyle Vert ZVert = left(sum_k=1^n (x_k+iy_k)(x_k-iy_k)right)^1/2$ that $Vert Z+WVert leq Vert ZVert+Vert WVert$ whenever $Z$ and $W$ are vectors in complex numbers holds.



Can someone help me to prove this please?







complex-numbers euclidean-geometry






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edited Mar 21 '15 at 1:17









TravisJ

6,40831830




6,40831830










asked Mar 21 '15 at 0:55









Zhiyue JiangZhiyue Jiang

11




11











  • $begingroup$
    Welcome to Math.SE! Please consider editing your question to include your thoughts on the problem, and/or showing what you have already tried.
    $endgroup$
    – Peter Woolfitt
    Mar 21 '15 at 0:58
















  • $begingroup$
    Welcome to Math.SE! Please consider editing your question to include your thoughts on the problem, and/or showing what you have already tried.
    $endgroup$
    – Peter Woolfitt
    Mar 21 '15 at 0:58















$begingroup$
Welcome to Math.SE! Please consider editing your question to include your thoughts on the problem, and/or showing what you have already tried.
$endgroup$
– Peter Woolfitt
Mar 21 '15 at 0:58




$begingroup$
Welcome to Math.SE! Please consider editing your question to include your thoughts on the problem, and/or showing what you have already tried.
$endgroup$
– Peter Woolfitt
Mar 21 '15 at 0:58










2 Answers
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$begingroup$

Hint: the Cauchy-Schwarz inequality.






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    0












    $begingroup$

    Assuming Cauchy-Schwarz's inequality $|langle z,wrangle| leq |z||w|$, you have: $$beginalign |z+w|^2 &= (z+w)overline(z+w) \ &= (z+w)(overlinez+overlinew) \ &= zoverlinez+zoverlinew+overlinezw+woverlinew \ &= |z|^2 + 2,rm Re(zoverlinew) + |w|^2 \ &leq |z|^2+2|langle z,wrangle|+|w|^2 \ &leq |z|^2 + 2|z||w|+|w|^2 \ &= (|z|+|w|)^2, endalign$$ and take roots.






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      2 Answers
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      2 Answers
      2






      active

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      active

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      active

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      0












      $begingroup$

      Hint: the Cauchy-Schwarz inequality.






      share|cite|improve this answer









      $endgroup$

















        0












        $begingroup$

        Hint: the Cauchy-Schwarz inequality.






        share|cite|improve this answer









        $endgroup$















          0












          0








          0





          $begingroup$

          Hint: the Cauchy-Schwarz inequality.






          share|cite|improve this answer









          $endgroup$



          Hint: the Cauchy-Schwarz inequality.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 21 '15 at 0:58









          Fan ZhengFan Zheng

          1,312617




          1,312617





















              0












              $begingroup$

              Assuming Cauchy-Schwarz's inequality $|langle z,wrangle| leq |z||w|$, you have: $$beginalign |z+w|^2 &= (z+w)overline(z+w) \ &= (z+w)(overlinez+overlinew) \ &= zoverlinez+zoverlinew+overlinezw+woverlinew \ &= |z|^2 + 2,rm Re(zoverlinew) + |w|^2 \ &leq |z|^2+2|langle z,wrangle|+|w|^2 \ &leq |z|^2 + 2|z||w|+|w|^2 \ &= (|z|+|w|)^2, endalign$$ and take roots.






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                Assuming Cauchy-Schwarz's inequality $|langle z,wrangle| leq |z||w|$, you have: $$beginalign |z+w|^2 &= (z+w)overline(z+w) \ &= (z+w)(overlinez+overlinew) \ &= zoverlinez+zoverlinew+overlinezw+woverlinew \ &= |z|^2 + 2,rm Re(zoverlinew) + |w|^2 \ &leq |z|^2+2|langle z,wrangle|+|w|^2 \ &leq |z|^2 + 2|z||w|+|w|^2 \ &= (|z|+|w|)^2, endalign$$ and take roots.






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  Assuming Cauchy-Schwarz's inequality $|langle z,wrangle| leq |z||w|$, you have: $$beginalign |z+w|^2 &= (z+w)overline(z+w) \ &= (z+w)(overlinez+overlinew) \ &= zoverlinez+zoverlinew+overlinezw+woverlinew \ &= |z|^2 + 2,rm Re(zoverlinew) + |w|^2 \ &leq |z|^2+2|langle z,wrangle|+|w|^2 \ &leq |z|^2 + 2|z||w|+|w|^2 \ &= (|z|+|w|)^2, endalign$$ and take roots.






                  share|cite|improve this answer









                  $endgroup$



                  Assuming Cauchy-Schwarz's inequality $|langle z,wrangle| leq |z||w|$, you have: $$beginalign |z+w|^2 &= (z+w)overline(z+w) \ &= (z+w)(overlinez+overlinew) \ &= zoverlinez+zoverlinew+overlinezw+woverlinew \ &= |z|^2 + 2,rm Re(zoverlinew) + |w|^2 \ &leq |z|^2+2|langle z,wrangle|+|w|^2 \ &leq |z|^2 + 2|z||w|+|w|^2 \ &= (|z|+|w|)^2, endalign$$ and take roots.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 21 '15 at 1:01









                  Ivo TerekIvo Terek

                  46.7k954146




                  46.7k954146



























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