Q1:
a) Find the exact solutions in C for the equation z^{2}-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) = (z^{2} + 4)(z^{2}-2√3z + 4). Find the polynomial P(z) such that Q(z)P(z) = z^{6} + 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. z^{6} + 64 looks familiar, but I can't quite put my finger on it.
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.
Your factorization of Q(z) is incorrect.
Solving z^{2} - 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 z^{2} - 2√3z + 4 are also roots of z^{6} + 64.
We could try "polynomial division" to find P(z):
(z^{6} + 64)/Q(z) = (z^{6} + 12)/[(z^{2} + 4)(z^{2} - 2√3z + 4)] = (z^{6} + 64)/(z^{4} - 2√3z^{3} + 8z^{2} - 8√3z + 16).
Now Q(z)(z^{2}) = z^{6} - 2√3z^{5} + 8z^{4} - 8√3z^{3} + 16z^{2},
so z^{6} + 64 - Q(z)(z^{2}) = 2√3z^{5} - 8z^{4} + 8√3z^{3} - 16z^{2} + 64.
Since Q(z)(2√3z) = (z^{4} - 2√3z^{3} + 8z^{2} - 8√3z + 16)(2√3z) = 2√3z^{5} - 12z^{4} + 16√3z^{3} - 48z^{2} + 32√3z,
it follows that:
z^{6} + 64 - Q(z)(z^{2}) - Q(z)(2√3z) = 2√3z^{5} - 8z^{4} + 8√3z^{3} - 16z^{2} + 64 - (2√3z^{5} - 12z^{4} + 16√3z^{3} - 48z^{2} + 32√3z)
that is:
z^{6} + 64 - Q(z)(z^{2} + 2√3z) = 4z^{4} - 8√3z^{3} + 32z^{2} - 32√3z + 64 = 4(Q(z))
so that: z^{6} + 64 - Q(z)(z^{2} + 2√3z + 4) = 0, that is:
Q(z)(z^{2} + 2√3z + 4) = z^{6} + 64, so that:
P(z) = z^{2} + 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), z^{2} - 2√3z + 4, and z^{2} + 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:
x^{n} + 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:
x^{n} + a for a > 0, with n even.
We have:
x^{2n} - a^{2} = (x^{n} - a)(x^{n} + a).
The roots of x^{2n} - a^{2} are the 2n-th roots of unity times the positive real 2n-th root of a^{2}, 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 x^{n} - a, and the second "half" yields the roots of x^{n} + 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).