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Just a complex Complex product

  1. Aug 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Find x
    [itex] x = (1 + \frac{1+i}{2})(1 + (\frac{1+i}{2})^{2})(1 + (\frac{1+i}{2})^{2^{2}})...(1 + (\frac{1+i}{2})^{2^{n}}) [/itex]


    2. Relevant equations
    Complex algebra equations


    3. The attempt at a solution
    Me again and another olympic question. I've tried some trigonometric substitutions, even expand; yet, nothing
     
    Last edited: Aug 3, 2014
  2. jcsd
  3. Aug 3, 2014 #2

    SammyS

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    How about showing some of those results?
     
  4. Aug 3, 2014 #3

    Ray Vickson

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    For ##z = (1+i)/2##, look at ##z^2##. What does this tell you about ##z^4, z^8, \ldots##?
     
  5. Aug 3, 2014 #4
    I tried it, though, but I couldn't really solve the recurrency because of the +1 on each factor:
    Resolving powers
    (1 + (1+i)/2)(1 + i/2)(1 - 1/4)(1 + 1/8)(1 + 1/64)(1 + 1/4096)...
    Let's say
    a = (1 + (1+i)/2)(1 + i/2)(1 - 1/4)
    and
    b = (1 + 1/8)(1 + 1/64)(1 + 1/4096)...
    We can calculate b's denominator which is
    (2^3)[2^(3.2)][2^(3.2.2)][2^(3.2.2.2)]...[2^(3.2.2.2...(n-3))] = 2^[3.2^(n-2) -1]
    but b's numerator, I don't know
    (9)(65)(4097)...
     
  6. Aug 3, 2014 #5

    SammyS

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    Did you do what Ray V. suggested?

    He did not suggest expanding out each factor, just parts of a few specific factors.
     
    Last edited: Aug 4, 2014
  7. Aug 4, 2014 #6

    Ray Vickson

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    The simple relation
    [tex] \frac{w-1}{w}\times \left(1+\frac{1}{w} \right) = 1-\frac{1}{w^2} \equiv \frac{w^2-1}{w^2}[/tex]
    is helpful here.
     
  8. Aug 4, 2014 #7
    Thanks, Ray V and SammyS!!
     
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