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Complex Analysis - Proving an inequality

  1. Mar 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that if |z| = 10 then 497 ≤ |z[itex]^{3}[/itex] + 5iz[itex]^{2}[/itex] − 3| ≤ 1503.



    3. The attempt at a solution

    I'm not an entirely sure how to begin this one, or if what I'm doing is correct.

    If I sub in |z| = 10 into the equation; |1000 + 500i - 3| = 997 +500i

    Then the modulus of that is sqrt(997[itex]^{2}[/itex]+500[itex]^{2}[/itex]) = 1115.35

    497 ≤ 1115.35 ≤ 1503
     
  2. jcsd
  3. Mar 3, 2012 #2

    Office_Shredder

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    You subbed in z=10, but there are many other values of z with |z|=10 (for example, 10i, 5sqrt(2)+5sqrt(2)i, etc.)

    You just have to apply the triangle inequality (in a different form for each bound)
     
  4. Mar 4, 2012 #3
    I took a look at triangle inequalities for Complex Analysis, and there are essentially two types?

    One for the upper bound and one for the lower:

    |z1 + z2| ≤ |z1| + |z2|

    |z1 + z2| ≥ ||z1| − |z2||

    But I'm not too sure how that would fit in here.

    Would I need to find the roots of z as a starting point?
     
  5. Mar 4, 2012 #4

    Office_Shredder

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    For example, from this you can also show
    [tex]| z_1 + z_2 + z_3 | \leq |z_1| + |z_2| + |z_3| [/tex]
     
  6. Mar 4, 2012 #5
    It's really not clicking still.

    So I'm meant to show 497 ≤ |z1| + |z2| + |z3|, so would z1 = z^3, z2 = 5iz^2, and z3 = -3 ?
     
  7. Mar 4, 2012 #6

    Office_Shredder

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    That wouldn't prove anything, since 497 ≤ |z1| + |z2| + |z3| does not imply that 497 ≤ |z1 + z2+ z3|

    However, the other direction might be more fruitful
     
  8. Mar 4, 2012 #7
    OK.

    |z1 + z2 + z3| [itex]\geq[/itex] 497

    z1 = z^3
    z2 = 5iz^2
    z3 = -3

    z^3 + 5iz^2 [itex]\geq[/itex] 500
    |z| = 10, so |z|^2 = 100

    Simplifying equation to 1 + 5i [itex]\geq[/itex] 1/2

    Am I even remotely in the right direction?

    Really appreciate your help by the way. Thanks for that.
     
  9. Mar 4, 2012 #8
    I'm still stuck on this problem if anyone can help that'd be great =)
     
  10. Mar 5, 2012 #9
    Nvm, it is solved now! Can't believe it was actually very easy, once the tutor did it in class, I was like doh!
     
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