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Help with simple math problem

  1. Oct 17, 2005 #1
    i forgot how to do this type of problem and i i was wondering if you guys can help me out.


    (1+i)^n-1=i_a im trying find i given i_a. n is the number of terms which i also have. i know u have to add the negative one to the other side. but no clue as to go from there.
     
  2. jcsd
  3. Oct 17, 2005 #2
    anyone help?
     
  4. Oct 17, 2005 #3

    James R

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    Is this a complex number problem? What is i_a? Is it just a constant?
     
  5. Oct 18, 2005 #4

    HallsofIvy

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    You won't get much help if you don't state the problem more clearly. Can I take it that i and i_a are just numbers- If you are given i_a, solve for i?
    Even then is (i+1)^n-1 supposed to be (i+1)n-1 or (i+1)n-1?

    In either case solving the equation is just taking a root.
    If [itex](i+1)^{n-1}= i_a[/itex] then [itex]i+1= ^{n+1}\sqrt{i_a}[/itex] so [itex]i= ^{n+1}\sqrt{i_a}-1[/itex].
    If [itex](i+1)^n-1= i_a[/itex] then [itex](i+1)^n= i_a+1[/itex] so
    [itex]i+1= ^n\sqrt{i_a+1}[/itex] and [itex]i= ^n\sqrt{i_a+1}-1[/itex].

    ([itex]^n\sqrt{}[/itex] is the nth root.)
     
    Last edited: Oct 18, 2005
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