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Homework Help: Roots of Complex Numbers, Help!

  1. Dec 13, 2003 #1
    I'm trying to find the: 4 4th roots of [itex] {\sqrt{3}} [/tex] + i .

    So I made a Cartesian plane and graphed radical 3 and 1.. but these numbers can be in 2 quadrants, 1st and 3rd.

    r=2 ==> 2([itex] {\sqrt{3}} [/tex] + i) ===> 2(cos 30+isin 30)
    =4rad2(cos 7.5+isin 7.5)

    Would I also use 2(cos 210+isin210) ??? Thanks..
     
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  3. Dec 13, 2003 #2

    Hurkyl

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    Recall that, for instance, [itex]\cos 30 = \cos 390[/itex]. Does that give you any ideas about how to proceed?


    P.S., the symbols you're looking for are:

    [itex]\sqrt{3}[/itex] (click it to see the LaTeX source)

    and "& radic ; 3" (the stuff between the quotes, with no spaces), gives √3
     
    Last edited: Dec 13, 2003
  4. Dec 13, 2003 #3
    kinda. i know that to get the 4 4th roots of this equation are
    30/4 = 7.5 ===>
    (cos 7.5 + isin 7.5)
    +30 (cos 37.5 + isin 37.5)
    +30 (cos 67.5 + isin 67.5)
    +30 (cos 97.5 + isin 97.5)

    but there is also a second set which you could start from 210.
    210/4 = 52.5 ===> (cos 52.5 + isin 52.5)
    +210 (cos 262.5 + isin 262.5)
    +210 (cos 472.5 + isin 472.5)
    +210 (cos 682.5 + isin 682.5)

    Would there be 2 sets of answers to this root?
     
  5. Dec 13, 2003 #4

    Hurkyl

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    I agree with the first of these four. What justification do you have for the other three? It seems you are simply adding 30 degrees... why?

    (For the record, I agree with the last one as well, but not the middle two)


    Also, why would you use 210 anyplace?

    [tex]2(\cos 210 + i \sin 210) \neq 2(\cos 30 + i \sin 30)[/tex]


    However, note that

    [tex]2(\cos 390 + i \sin 390) = 2(\cos 30 + i \sin 30)[/tex]
     
    Last edited: Dec 13, 2003
  6. Dec 13, 2003 #5
    you use 30 to find all of the roots of that particular equation.
    substituting k =0,1,2,3..etc.

    k=0 for general solution (cos 30 + isin30)
    =2[1/2 + i[itex]/sqrt{3}[/itex] /2]
    = [itex] /sqrt{3} [/itex] + i

    k=1 you add 30 because 30 is the reference angle.
    k=2 add 30
    k=3 add 30.

    haha im going to stop im confusing myself
     
  7. Dec 13, 2003 #6

    Hurkyl

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    Can you write out what the general solution is supposed to be?

    (P.S. You're using the wrong slash in your LaTeX; it should be \ not /)
     
  8. Dec 13, 2003 #7
    i wish my teacher would go by the book, he solves it differently than how the book does, which is very confusing. this is homework so i don't have the solution yet.

    but i think i understand your point. i use k=90, because 360/4 = 90. if i wanted the 6th roots, i would use 60 as k?
     
  9. Dec 13, 2003 #8

    Hurkyl

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    Right (I think):

    You know that:

    [tex]\sqrt{3} + i = 2(\cos (30 + 360n) + i \sin(30 + 360n))
    \quad n \in \mathbb{Z}[/tex]

    So the fourth roots would be

    [itex]
    2 ( \cos \frac{30 + 360n}{4} + i \sin \frac{30 + 360n}{4})
    = 2 (\cos (7.5 + 90n) + i \sin (7.5 + 90))
    [/itex]


    edit: trying to get the formats right
     
    Last edited: Dec 13, 2003
  10. Dec 13, 2003 #9
    ok thanks. so i would use k=0,1,2,3.
    so the 210 angle has nothing to do with the equation?
    in essence would it be the same as [tex]2(\cos \frac{30 + 360n}{4} + i \sin \frac{30 + 360n}{4}) = 2 (\cos (7.5 + 90n} + i \sin (7.5 + 90))[/tex]

    Thanks for the help I appreciate it much!
     
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