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Math Help - Complex Numbers

  1. #1
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    Complex Numbers

    Q1:
    a) Find the exact solutions in C for the equation z2-2√3z + 4 = 0, writing your solutions in cartesian form.
    b) i) Plot the two solutions from part a on an Argand diagram.
    ii) Find the equation of the circle, with centre the origin, which passes through these two points.
    iii) Find the value of a Z such that the circle passes through (0, a).
    iv) Let Q(z) = (z2 + 4)(z2-2√3z + 4). Find the polynomial P(z) such that Q(z)P(z) = z6 + 64 and explain the significance of the result.

    Parts a and b i, ii and iii were straight forward, but I've hit a wall with iv. z6 + 64 looks familiar, but I can't quite put my finger on it.
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  2. #2
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    Re: Complex Numbers

    Quote Originally Posted by Fratricide View Post
    Q1:
    a) Find the exact solutions in C for the equation z2-2√3z + 4 = 0, writing your solutions in cartesian form.
    b) i) Plot the two solutions from part a on an Argand diagram.
    ii) Find the equation of the circle, with centre the origin, which passes through these two points.
    iii) Find the value of a Z such that the circle passes through (0, a).
    iv) Let Q(z) = (z2 + 4)(z2-2√3z + 4). Find the polynomial P(z) such that Q(z)P(z) = z6 + 64 and explain the significance of the result.

    Parts a and b i, ii and iii were straight forward, but I've hit a wall with iv. z6 + 64 looks familiar, but I can't quite put my finger on it.
    find the roots of Q(z) and of $z^6+64$

    There will be two roots of $z^6+64$ not appearing in Q(z). Their product will be P(z)

    Hint: the roots of $z^6 + 64$ will be 6 evenly spaced numbers on a circle of radius 8 in the complex plane.
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    Re: Complex Numbers

    So Q(z) = (z2 + 4)(z2-2√3z + 4) = (z + 2i)(z - 2i)(√3 + i)(√3 - i)

    But I can't recall how to solve z6 + 64.
    z6 + 64 = 0
    z6 = -64
    z = 6√-64
    ?
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  4. #4
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    Re: Complex Numbers

    Notice that $\displaystyle \begin{align*} z^6 = \left( z^2 \right) ^3 \end{align*}$ and $\displaystyle \begin{align*} 64 = 4^3 \end{align*}$, so that means $\displaystyle \begin{align*} z^6 + 64 = \left( z^2 \right) ^3 + 4^3 \end{align*}$. Factorise using the sum of two cubes rule.
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    Re: Complex Numbers

    Quote Originally Posted by Fratricide View Post
    So Q(z) = (z2 + 4)(z2-2√3z + 4) = (z + 2i)(z - 2i)(√3 + i)(√3 - i)

    But I can't recall how to solve z6 + 64.
    z6 + 64 = 0
    z6 = -64
    z = 6√-64
    ?
    $z^6=-64=2e^{\imath \pi(2k+1)}$

    $\large z=2e^{\imath \pi{\frac{2k+1}{6}}}~~ k=0,5$
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    Re: Complex Numbers

    Quote Originally Posted by romsek View Post
    $z^6=-64=2e^{\imath \pi(2k+1)}$

    $\large z=2e^{\imath \pi{\frac{2k+1}{6}}}~~ k=0,5$
    this is obviously wrong. What I meant was

    $z^6=-64=2^6 e^{\imath \pi(2k+1)}$ then

    $\large z=2e^{\imath \pi{\frac{2k+1}{6}}}~~ k=0,5$
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    Re: Complex Numbers

    Quote Originally Posted by Fratricide View Post
    So Q(z) = (z2 + 4)(z2-2√3z + 4) = (z + 2i)(z - 2i)(√3 + i)(√3 - i)

    But I can't recall how to solve z6 + 64.
    z6 + 64 = 0
    z6 = -64
    z = 6√-64
    ?
    Your factorization of Q(z) is incorrect.

    Solving z2 - 2√3z + 4 using the quadratic formula gives:

    z = √3 i so Q(z) = (z + 2i)(z - 2i)(z - (√3 + i))(z - (√3 - i)).

    Note that (√3 - i)6 = [(√3 - i)2]3 = (3 - 2√3i - 1)3 = (2 - 2√3i)3 = 8(1 - √3i)3 = 8(1 - √3i)2(1 - √3i) = -8(2 + 2√3i)(1 - √3i)

    = -8(2 + 6) + -8(-2√3 + 2√3)i = -8(8) = -64, (and (√3 + i) = (√3 - i)*, so (√3 + i)6 = [(√3 - i)*]6 = [(√3 - i)6]* = -64* = -64),

    so the roots of z2 - 2√3z + 4 are also roots of z6 + 64.

    We could try "polynomial division" to find P(z):

    (z6 + 64)/Q(z) = (z6 + 12)/[(z2 + 4)(z2 - 2√3z + 4)] = (z6 + 64)/(z4 - 2√3z3 + 8z2 - 8√3z + 16).

    Now Q(z)(z2) = z6 - 2√3z5 + 8z4 - 8√3z3 + 16z2,

    so z6 + 64 - Q(z)(z2) = 2√3z5 - 8z4 + 8√3z3 - 16z2 + 64.

    Since Q(z)(2√3z) = (z4 - 2√3z3 + 8z2 - 8√3z + 16)(2√3z) = 2√3z5 - 12z4 + 16√3z3 - 48z2 + 32√3z,

    it follows that:

    z6 + 64 - Q(z)(z2) - Q(z)(2√3z) = 2√3z5 - 8z4 + 8√3z3 - 16z2 + 64 - (2√3z5 - 12z4 + 16√3z3 - 48z2 + 32√3z)

    that is:

    z6 + 64 - Q(z)(z2 + 2√3z) = 4z4 - 8√3z3 + 32z2 - 32√3z + 64 = 4(Q(z))

    so that: z6 + 64 - Q(z)(z2 + 2√3z + 4) = 0, that is:

    Q(z)(z2 + 2√3z + 4) = z6 + 64, so that:

    P(z) = z2 + 2√3z + 4.

    It should be clear the roots of P(z) are: z = -√3 i, and it might be helpful to plot the roots of P(z), z2 - 2√3z + 4, and z2 + 4 on a piece of graph paper (draw a circle of radius 2 first).

    In general, for future reference, the roots of the REAL polynomial:

    xn + a = 0, for even n and a > 0, will be "every other 2n-th root of unity" times the (real positive) n-th root of a (in other words, we're looking for n-th roots of -a).

    You can look at it this way:

    Suppose we have the polynomial:

    xn + a for a > 0, with n even.

    We have:

    x2n - a2 = (xn - a)(xn + a).

    The roots of x2n - a2 are the 2n-th roots of unity times the positive real 2n-th root of a2, which is the real positive n-th root of a.

    Half the 2n-th roots of unity when raised to the n-th power will be 1, the other half will be -1. The first "half" yields the roots of xn - a, and the second "half" yields the roots of xn + a.

    (I urge you to consider what this means for the two easiest cases: n = 2, and n = 4. This problem is an example of what happens when n = 6, and involves (even if you didn't see it) finding 12-th roots of unity, although we are only interested in the 6 which are 6-th roots of -1, two of which happen to be i and -i. These 12-th roots can be found using trigonometry somewhat faster than using polynomials).
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