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Math Help - Complex Numbers :D

  1. #1
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    Complex Numbers :D

    The question says,

    Find all splutions of the following equations, expressing your answers in Cartesian form (i.e as z=x+iy)

    (a)  z^4 - 1 = i\sqrt3

    Do i simple open it out and simplify?.. is there anything else that needs to be done?
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  2. #2
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    Find the four fourth roots of 1 + i\sqrt 3.
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  3. #3
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    Oh ok, And how do i go about starting that?
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  4. #4
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    Quote Originally Posted by brd_7 View Post
    Oh ok, And how do i go about starting that?
    I favor putting the complex number into polar form:
    1 + i \sqrt{3} = re^{i \theta}

    Then go from there.

    -Dan
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  5. #5
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     z^4 = 1 + i \sqrt{3}

    Represent 1 + i \sqrt{3} in the form  r(\cos \theta + i \sin \theta)

    I got  z^4 = 2 ( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} )

    Let  z = r (\cos \theta + i \sin \theta)

    Then  z^4 = r^4 ( cos \theta + i \sin \theta)^4

    So  r^4 ( cos \theta + i \sin \theta)^4 = 2 ( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} )

    And then-
    z^4 = 2.{cos[(1/3 + 2n)pi] + isin[(1/3 + 2n)pi]}. And so
    z = 2^[1/4]{cos[(1/12 + n/2)pi] + isin[(1/12 + n/2)pi]].

    Is this all right?.. What do i do now?.. How do i change it back to x +iy?
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  6. #6
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    Yes it is correct if this is what you mean.
    z = \sqrt[4]{2}\left[ {\cos \left( {\frac{\pi }{{12}} + \frac{{k\pi }}{2}} \right) + i\sin \left( {\frac{\pi }{{12}} + \frac{{k\pi }}{2}} \right)} \right]\,,\,k = 0,1,2,3<br />
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  7. #7
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    and so how do i show it in terms of x+iy?

    Thanks for all your help btw

    Also, how do i begin this question -
     (z +1)^4 + (z-1)^4 = 0
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  8. #8
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    The first step is to find, \sqrt[4]{1+i\sqrt{3}} = |1+i\sqrt{3}| \exp (i \arg (1+i\sqrt{3})/4 )
    Which is,
    \sqrt{1+3} \cdot e^{i\pi/ 12} = 2 \cos \frac{\pi}{12} + i 2\sin \frac{\pi}{12} = \left( 1 + \cos \frac{\pi}{6} \right) + i \left( 1 - \cos \frac{\pi}{6}\right)
    Which is,
    a = \left( 1 + \frac{\sqrt{3}}{2} \right) + i \left( 1 - \frac{\sqrt{3}}{2} \right)
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by brd_7 View Post
    and so how do i show it in terms of x+iy?

    Thanks for all your help btw

    Also, how do i begin this question -

     (z +1)^4 + (z-1)^4 = 0
    Tsk! New questions go in new threads.

    I would expand:
    (z^4 + 4z^3 + 6z^2 + 4z + 1) + (z^4 - 4z^3 + 6z^2 - 4z + 1) = 0

    2z^4 + 12z^2 + 2 = 0
    which is a biquadratic and easy to solve.

    -Dan
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  10. #10
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    Quote Originally Posted by topsquark View Post
    I favor putting the complex number into polar form:
    1 + i \sqrt{3} = re^{i \theta}

    Then go from there.

    -Dan
    I have a preference cutting straight to the chase and using 1 + i \sqrt{3} = re^{i (\theta + 2 n \pi)}, \, n \in \mathbb{Z}.
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mr fantastic View Post
    I have a preference cutting straight to the chase and using 1 + i \sqrt{3} = re^{i (\theta + 2 n \pi)}, \, n \in \mathbb{Z}.
    Well... If I want to do it the hard way, then let me!

    -Dan
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  12. #12
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    I remember when I first learned de Moivre's theorem I did not like solving the equation x^n = a because of the setting up that is involved. But here is my perferred method which turns this problem into an automatic problem.
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  13. #13
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    The two roots that i gained were  +- \frac{12i}{2} + \frac{\sqrt(8)sqrt2}{2}
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