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Complex numbers

  1. Feb 14, 2006 #1
    i am to find the 4th roots of -16

    [tex](-16)^{1/4}=2i^{1/4}[/tex]
    [tex]i=e^{i \pi/2}[/tex]
    [tex]i^{1/4}=e^{i \pi/8}[/tex]
    [tex](-16)^{1/4}=2e^{i \pi/8}[/tex]
    or
    [tex](-16)^{1/4}=2e^{i 5\pi/8}[/tex]
    or
    [tex](-16)^{1/4}=2e^{i 9\pi/8}[/tex]

    is this correct?
     
  2. jcsd
  3. Feb 14, 2006 #2
    As far as it goes. Where is your fourth 4th root?
     
  4. Feb 14, 2006 #3
    the fourth root of -16 is [tex]2e^{i 9\pi/8}[/tex]

    didnt i show that?
     
  5. Feb 14, 2006 #4

    Hurkyl

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    If that's the fourth 4th root, then where's your third? You've only written three roots down, so you're missing at least one of them!
     
  6. Feb 14, 2006 #5
    I'm not sure I understand. I am to find the 4th root of -16, not the fourth 4th root. and (-16)^(1/4) is the fourth root of -16, so I'm not sure what else is needed since [tex](-16)^{1/4}=2e^{i 9\pi/8}[/tex]
     
  7. Feb 14, 2006 #6

    Hurkyl

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    In the original question, you said you're supposed to find the 4th roots of -16. As in all of them. How many 4th roots does -16 have? How many have you shown?


    I guess I'm not up on the convention for this stuff, but I would say that this is wrong. I would say the L.H.S. is multivalued, and denotes all fourth roots of -16, and the R.H.S. is a single value, denoting one fourth root of -16. Thus, it wouldn't be appropriate to write an equality there.
     
  8. Feb 14, 2006 #7
    I see, so they are equal but just not equal in showing ALL the fourth roots of -16 right?
     
  9. Feb 14, 2006 #8
    well that's like saying sin^-1(1/2) doesn't equal pi/6, because it also equals 5pi/6 and, well, so on
     
  10. Feb 14, 2006 #9

    Hurkyl

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    I agree. Lots of silly mistakes are made because people forget that the inverse of the sin function is multivalued.
     
  11. Feb 15, 2006 #10

    HallsofIvy

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    What??? "they are all equal but just not equal"???

    Any number has 4 distinct fourth (complex) roots. For example the fourth roots of 1 are 1, -1, i, and -i. You were asked to find all of the fourth roots of -16. ("i am to find the 4th roots of -16")
    You only showed three in your original post.

    Actually, your very first statement:
    [tex](-16)^{1/4}=2i^{1/4}[/tex]
    is wrong. The principle root of 16 is, of course, 2 but -1 is not equal to i!
    What you should have written was
    [tex](-16)^{1/4}= 2(-1)^{1/4}[/tex]
    Now, what are the 4 distinct fourth roots of -1?
     
  12. Feb 15, 2006 #11
    [tex](-1)^{1/4}=e^{i \pi /4}[/tex]
    [tex](-1)^{1/4}=e^{9i \pi /4}[/tex]
    [tex](-1)^{1/4}=e^{17i \pi /4}[/tex]
    [tex](-1)^{1/4}=e^{25i \pi /4}[/tex]

    right? so that:

    [tex](-16)^{1/4}= 2e^{i \pi /4}[/tex]
    or
    [tex](-16)^{1/4}= 2e^{9i \pi /4}[/tex]
    or
    [tex](-16)^{1/4}= 2e^{17i \pi /4}[/tex]
    or
    [tex](-16)^{1/4}= 2e^{25i \pi /4}[/tex]
     
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