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Homework Help: Prove of complex numbers

  1. May 20, 2008 #1
    1. The problem statement, all variables and given/known data

    Hello! :smile:

    a)Prove that [tex](a+bi)^n[/tex] and [tex](a-bi)^n[/tex], n [tex]\in \mathbb{N}[/tex] are conjugate complex numbers;

    b)Prove that quotient of any two numbers from the set of [tex]\sqrt[n]{1}[/tex] is again number from the set of [tex]\sqrt[n]{1}[/tex]

    c)Prove that reciproca value of any number from the set of [tex]\sqrt[n]{1}[/tex], is again number from the set [tex]\sqrt[n]{1}[/tex]

    2. Relevant equations

    [tex]z=r(cos\alpha+isin\alpha)[/tex]

    [tex]\bar{z}=r(cos\alpha-isin\alpha)[/tex]

    [tex]w_k=\sqrt[n]{r}(cos\frac{\alpha + 2k\pi}{n}+isin{\alpha + 2k\pi}{n}) ; k=0,1,2,...,n-1[/tex]

    3. The attempt at a solution

    a) [tex](a+bi)=r(cos\alpha+isin\alpha)[/tex]

    [tex] (a+bi)^n=r^n(cosn\alpha+isinn\alpha)[/tex]

    [tex] (a-bi)=r(cos\alpha-sin\alpha)[/tex]

    [tex] (a-bi)^n=r^n(cosn\alpha-sinn\alpha)[/tex]

    Is this enough to prove that they are conjugate complex numbers?

    b)[tex]E_k=\sqrt[n]{1}=cos\frac{2k\pi}{n}+isin{2k\pi}{n} , k=0,1,2,3,...,n-1[/tex]

    Should I make like this?

    [tex]E_n_-_1=\sqrt[n]{1}=cos\frac{2(n-1)\pi}{n}+isin\frac{2(n-1)\pi}{n}[/tex]

    What should I do next?

    c)If I know how to prove b) I will prove c)

    In this case just [tex](\sqrt[n]{1})^-^1=\sqrt[n]{1}[/tex], right?
     
  2. jcsd
  3. May 20, 2008 #2

    Defennder

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    Multiply eqns 2 and 4 and you're done.

    For b,c you don't have to resort to any fanciful complex number manipulation. Suppose a complex number [tex]w = \sqrt[n]{1}[/tex] That implies [tex]w^n = 1[/tex] Similarly consider a different nth complex root of unity, say z. z^n = 1. Quotient = w/z. How can we tell if w/z is also an nth root of unity?

    c)

    Again suppose [tex]w = \sqrt[n]{1}[/tex]. It's reciprocal is [tex]\frac{1}{w}[/tex]. What is [tex]\left (\frac{1}{w}\right)^n[/tex]
     
  4. May 20, 2008 #3
    But I need to prove like I wrote at b)

    First, let me apologise, I have some typo error in b), it should look like:

    [tex]E_k=\sqrt[n]{1}=cos\frac{2k\pi}{n}+isin\frac{2k\pi}{n} , k=0,1,2,3,...,n-1[/tex]

    In a) why I need to multiply them ?
     
  5. May 20, 2008 #4

    Defennder

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    Well, okay I misread your answer for a). It's fine.

    For b,c, what kind of proof are you looking for? Is there something unacceptable about the suggestion I gave?
     
  6. May 20, 2008 #5
    I think your way of thinking is good, but probably I should get two general solutions from the set of solutions of [tex]\sqrt[n]{1}[/tex] and prove...
     
  7. May 20, 2008 #6
    For b), switching to exponential form simplifies a lot.
     
  8. May 20, 2008 #7
    I need to prove with complex numbers like the [tex]E_k[/tex] formula above. Please help!
     
  9. May 21, 2008 #8
    Any help?
     
  10. May 21, 2008 #9

    HallsofIvy

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    Suppose a and b are both "from the set [itex]^n\sqrt{1}[/tex]. What is (ab)n?
     
  11. May 22, 2008 #10
    Maybe again [tex](ab)^n[/tex] is from the set of [tex]\sqrt[n]{1}[/tex]. I am not sure.
     
  12. May 22, 2008 #11

    tiny-tim

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    Hint: [itex](ab)^n[/itex] = ababab … ababab = … ? :smile:
     
  13. May 22, 2008 #12
    [tex]a^nb^n[/tex] ? :smile:
     
  14. May 22, 2008 #13

    tiny-tim

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    Yes … but … why the question-mark?

    Theofilius , you keep answering questions with a question … even when, as in this case, it's actually the complete answer.

    ok … and [tex]a^nb^n[/tex] = … ?

    :smile: … and don't answer with a question … ! :smile:
     
  15. May 22, 2008 #14
    [tex]a^nb^n=[/tex]? I honesly don't know.. :smile: I don't even know what are you giving me that question for?
     
  16. May 22, 2008 #15

    tiny-tim

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    Because of:
     
  17. May 22, 2008 #16
    [tex](ab)^n[/tex] is complex or real number. :smile:
     
  18. May 23, 2008 #17
    I don't know, we are just going in circle.
     
  19. May 23, 2008 #18

    tiny-tim

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    So [itex]a^n\,=\,b^n\,=\,1[/itex]
     
  20. May 24, 2008 #19
    Is that the proof?
     
  21. May 24, 2008 #20

    tiny-tim

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    Well, it needs to be put together into one continuous piece, and the actual question was about a/b rather than ab, but apart from that … yes! :smile:
     
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