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Complex numbers question

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    (i) Find the roots of the equation z^2 + (2 sqrt 3)z +4 = 0, giving your answers in the form x + iy, where x and y are real.
    (ii) State the modulus and argument of each root.
    (iii) Showing all your working, verify that each root also satisfies the equation
    z^6 = −64.


    How do I do the third question without expanding the roots of z to the power 6?
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    (Original post by bmqib)
    (i) Find the roots of the equation z^2 + (2 sqrt 3)z +4 = 0, giving your answers in the form x + iy, where x and y are real.
    (ii) State the modulus and argument of each root.
    (iii) Showing all your working, verify that each root also satisfies the equation
    z^6 = −64.


    How do I do the third question without expanding the roots of z to the power 6?
    Do you know that Mod(z^6)=(modz)^6 and that arg(z^6) = 6x(arg z)?
    (or put another way, De Moivre's theorem)
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    z^6+64=(z^2+4)(z^4-4z^2+16)

    Assuming OP doesn't know De Moivre. Split the quartic into the product of two quadratics - you know what one of them is going to be.
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    (Original post by tiny hobbit)
    Do you know that Mod(z^6)=(modz)^6 and that arg(z^6) = 6x(arg z)?
    (or put another way, De Moivre's theorem)
    Yeah but that's not in the curriculum so there could be a different way to do it... do I have to find z in the r(sin theta + i cos theta) form then?
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    (Original post by Mr M)
    z^6+64=(z^2+4)(z^4+4z^2+16)

    Assuming OP doesn't know De Moivre. Solve the disguised quadratic.
    But how do I factorise it that quickly? and the right hand side is not equal to lhs?
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    (Original post by bmqib)
    But how do I factorise it that quickly? and the right hand side is not equal to lhs?
    It is now I corrected the wrong sign!

    http://www.purplemath.com/modules/specfact2.htm
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    (Original post by bmqib)
    (i) Find the roots of the equation z^2 + (2 sqrt 3)z +4 = 0, giving your answers in the form x + iy, where x and y are real.
    (ii) State the modulus and argument of each root.
    (iii) Showing all your working, verify that each root also satisfies the equation
    z^6 = −64.


    How do I do the third question without expanding the roots of z to the power 6?
    Lol, I just did the same paper myself. Wanted to post the same question. CIE sucks.

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