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Inequality of a complex number

  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose that w is a complex number which is not both real and [itex]\left\lfloor[/itex]w[itex]\right\rfloor[/itex][itex]\geq[/itex]1 (the absolute value of w).
    Verify that Re[(1-w[itex]^{2}[/itex])[itex]^{1/2}[/itex]+iw]>0.

    2. Relevant equations



    3. The attempt at a solution

    I attempted to solve this problem by dividing it into three cases; Im(w)>0, w[itex]\in[/itex](-1,1), and Im(w)<0. I could make the conclusion in the case of w[itex]\in[/itex](-1,1).
    But, I don't have any idea how to approach in the cases of Im(w)>0 and Im(w)<0.
    Could you give me a hint??
     
  2. jcsd
  3. Oct 11, 2011 #2

    lanedance

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    Homework Helper

    haven't worked it through fully but here's some stuff to get you going

    rather than using cases, i'd consider the following reperesentations
    [tex] |w|>1 \ \to w=re^{i\theta} \ , \ r>1[/tex]
    [tex] |w|>1 \ \to w=a+ib \ , \ a^2 + b^2 >1[/tex]

    can you convince yourself
    [tex] Re\{(1-w^2)^{1/2}+iw\} = Re\{(1-w^2)^{1/2}\} +Re\{iw\} [/tex]

    an then you have
    [tex] Re\{(1-w^2)^{1/2}\} +Re\{iw\}> 0[/tex]

    this is just a normal equality on the reals, so you do the normally allowed operations
    [tex] Re\{(1-w^2)^{1/2}\} > -Re\{iw\}= Re\{-iw\}[/tex]

    then note
    [tex] -Re\{iw\}= Re\{-iw\}= Im\{w\}[/tex]
     
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