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Question about imaginary numbers

  1. Jan 19, 2012 #1
    1. The problem statement, all variables and given/known data

    For a real number x, √√(-x) equals :

    a) +x b) -x c) complex d) pure imaginary

    2. Relevant equations

    √-1 = i

    3. The attempt at a solution

    Here is what i did:
    If x is a positive real number then the answer comes out to be x^0.25 * √i (now what is square root of i equal to?)

    But if x is a negative real number then the solution would be :
    x^0.25

    I don't know what to do next, any help would be appreciated.
     
  2. jcsd
  3. Jan 19, 2012 #2
    Have you learnt how to take the square roots of complex numbers?
     
  4. Jan 19, 2012 #3
    do you know about the old [itex]re^{i x} = r Cos[x]+r i Sin[x][/itex]?
    if so, I'd use that
     
  5. Jan 19, 2012 #4

    HallsofIvy

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    If [itex](a+bi)^2= i[/itex] then [itex]a^2- b^2+ (2ab)i= 0+ 1(i)[/itex] so you must have [itex]a^2- b^2= 0[/itex] and 2ab= 1. Solve for the values of a and b.
     
    Last edited: Jan 20, 2012
  6. Jan 19, 2012 #5

    vela

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    You mean |x|1/4.

    Since this is a multiple-choice question, you should be able to deduce what the correct answer is. I hope you can see that you can rule out the first two choices, so the answer has to be complex or purely imaginary. Can you rule out one or the other?
     
  7. Jan 20, 2012 #6
    No, can teach me that ?
    It would really help me

    If [itex](a+bi)^2= i[/itex]
    ^^how is that TRUE ?
     
  8. Jan 20, 2012 #7
    I know about that identity, but I can't figure out how would ot apply to this situation
     
  9. Jan 20, 2012 #8

    Mentallic

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    What is the mod-arg form for i? Now, how can [tex]\sqrt{re^{i\theta}}[/tex] be simplified?
     
  10. Jan 20, 2012 #9
    Alright, so for x=1
    The answer is e^[2∏i]
    Or another one, and perhaps i should have opened with this one, is 1


    So the answer is probably
    Complex



    What do you think, am i right ?
     
  11. Jan 20, 2012 #10

    HallsofIvy

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    You asked for the square root of i, didn't you? If a+ bi is the square root of i, then [itex](a+ bi)^2[/itex] is equal to i.
     
  12. Jan 21, 2012 #11
    Alright, understood.

    I have a concern however, why do I get two different answers by
    Using the two methods
    1.which was stated by hallsofivy
    2.the one which uses Taylor series representation to represent
    a complex number in e^(ix) form.
     
  13. Jan 21, 2012 #12

    HallsofIvy

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    Well, what answers did you get?
     
  14. Jan 23, 2012 #13
    Alright, this one is by the taylor series representation method:

    i = e^(i*π/2)
    so, √i = e^(i*π/4)

    now representing the complex number in the form x+yi :

    cos(π/4)+isin(π/4) = √i

    so, the answer is 1/√2 + i1/√2

    and the answer by the other method :

    let i = (a + bi)^2

    is exactly the same as before.
    My bad on post no. 11

    anyways, thanks
     
  15. Jan 23, 2012 #14

    SammyS

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    Of course, if x ≤ 0, then [itex]\sqrt{\sqrt{-x}\ }[/itex] is a real number.
     
  16. Jan 23, 2012 #15

    HallsofIvy

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    In the complex numbers, every number has two square roots.
    [tex]\sqrt{i}= \frac{1}{\sqrt{2}}\pm i\frac{1}{\sqrt{2}}[/tex]

     
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