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Triangle inequality

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Use the triangle inequality to show:
    [tex]\left|z^2+3\right|[/tex] [tex]\leq(12)[/tex] for [tex]\left|z\right|[/tex]=3
    where z is a complex number

    2. Relevant equations

    triangle inequality: [tex]\left|z_1+z_2\right|[/tex][tex]\leq[/tex] [tex]\left|z_1\right|[/tex]+[tex]\left|z_2\right|[/tex]

    3. The attempt at a solution

    I understand the triangle inequality but i cant seem to do the question.
     
  2. jcsd
  3. Oct 2, 2009 #2
    Simply substitute z1 with z^2 and z2 with 3.
     
  4. Oct 2, 2009 #3
    but how do i use the information: 'on abs(z)=3' ?
     
  5. Oct 2, 2009 #4
    oh i see:
    like this:
    [tex]\left|z^2+3\right|[/tex][tex]\leq[/tex][tex]\left|z^2\right|[/tex]+[tex]\left|3\right|[/tex]
    [tex]\left|z^2+3\right|[/tex][tex]\leq[/tex]9+3=12
    am i missing something?
     
  6. Oct 2, 2009 #5
    Thats right. And note that |z^2|= |z|^2
     
  7. Oct 2, 2009 #6
    thank you.
     
  8. Oct 3, 2009 #7
    I'm working on a similar question but this time, i must show that:
    [tex]\left|z^2(2+i)+1\right|[/tex] [tex]\geq[/tex] 1 for [tex]\left|z\right|[/tex]=1

    this time we can use the triangle inequality:
    [tex]\left|z_1-z_2\right|[/tex] [tex]\geq[/tex] [tex]\left|abs(z_1)-abs(z_2)\right|[/tex]

    do i substitute: z1 with z2(2+i) and z2 with -1 ?
    because when i do this, i get:
    [tex]\left|z^2(2+i)+1\right|[/tex] [tex]\geq[/tex] [tex]\left|z^2(2+i)-1\right|\right|\left|[/tex]
    I think im making a mistake
     
  9. Oct 3, 2009 #8
    Your substitution is correct, but evaluate the absolute values on the right hand side
     
  10. Oct 3, 2009 #9
    after evaluating the absolute values, the right hand side would look like:
    2z^2 +iz^2-1
    so somehow i must show that this is 1
    ?
     
  11. Oct 3, 2009 #10
    The absolute value a complex number a+ib is
    Sqrt(a^2+b^2), and thus a real number. Further you know abs(z)=1. And further abs(xy) = abs(x)abs(y). See what you can do with that
     
  12. Oct 3, 2009 #11
    oh right, thanks, so i will get:
    right hand side:
    abs(1*abs(2+i)-1)=abs(4+i^2-1)=abs(sqrt(3)-1)=sqrt(3+1)=2
    ?? not 1 ?
     
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