# 28. Complex Numbers Solutions Part 6

1. This can be solved as problem no. 250.

2. Rewriting the equality of the 248 in the following way

$x^{2n - 2} + x^{2n - 4} + ... + x^2 + 1 = \prod_{s=1}^{n-1}\left(x^2 - 2x \cos\frac{s\pi}{n} + 1\right)$

Putting in this identity $$x = 1$$. We have

$n = \prod_{s=1}^{n-1}\left(2 - 2\cos\frac{s\pi}{n}\right) = \prod_{s=1}^{n-1}4\sin^2\frac{s\pi}{n} = 2^{2(n-1)}\sin^2\frac{\pi}{n}.\sin^2\frac{2\pi}{n} ...\sin^2\frac{(n-1)\pi}{n}$
$\therefore \sin\frac{\pi}{n}\sin\frac{2\pi}{n} ... \sin\frac{(n-1)\pi}{n} = \frac{\sqrt{n}}{2^{n-1}}$
3. This is solved like 252.

4. Since $$\cos\alpha + i \sin\alpha$$ is the root of the given equation, we have

$\sum_{k=0}^n p_k(\cos\alpha + i \sin\alpha)^{n - k} = 0 (p_0 = 1)$

or

$(\cos\alpha + i \sin\alpha)^n\sum_{k=0}^np_k(\cos\alpha + i \sin\alpha)^{-k} = 0$

But $$(\cos\alpha + i \sin\alpha)^{-1} = \cos\alpha - i \sin\alpha$$

$\therefore \sum_{k=0}^np_k(\cos\alpha - i \sin\alpha)^k = 0, \sum_{k=0}^np_k(\cos k\alpha - i \sin k\alpha)$
$\therefore \sum_{k=0}^np_k \sin k\alpha = p_1\sin\alpha + p_2\sin 2\alpha + ... + p_n \sin n\alpha = 0$
5. The roots of the equation $$x^7 = 1$$ are

$\cos \frac{2k\pi}{7} + i \sin\frac{2k\pi}{7}~~~~~(k = 0, 1, 2, ..., 6)$

Therefore the roots of the equation $$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = 0$$ will be

$x_k = \cos \frac{2k\pi}{7} + i \sin\frac{2k\pi}{7}~~~~~(k = 1, 2, ..., 6)$

Putting $$x + \frac{1}{x} = y$$, then

$x^2 + \frac{1}{x^2} = y^2 - 2, x^3 - \frac{1}{x^3} = y^3 - 3$

The above equation may be rewritten in the following way

$\left(x^3 + \frac{1}{x^3}\right) + \left(x^2 + \frac{1}{x^2}\right) + \left(x + \frac{1}{x}\right) + 1 = 0$

It is clear that $$x_1 = \overline{x_6}, x_2=\overline{x_4}, x_3 = \overline{x_4}, x_k + \frac{1}{x_k} = x_k + \overline{x_k} = 2cos\frac{2k]pi}{7}$$

Hence, we conclude that the quantities

$2\cos\frac{2\pi}{7}, 2cos\frac{4\pi}{7}, 2cos\frac{8\pi}{7}$

are the root of $$y^3 + y^2 - 2y - 1 = 0$$.

Let us set up an equation with the following roots

$\sqrt[3]{2\cos\frac{2\pi}{7}}, \sqrt[3]{2cos\frac{4\pi}{7}}, \sqrt[3]{2cos\frac{8\pi}{7}}$

Let the roots of a certain cubic equation $$x^3 - ax^2 + bx -c = 0$$ be $$\alpha, \beta, \gamma$$.

We then have

$$\alpha + \beta + \gamma = a, \alpha\beta + \alpha\gamma + \beta\gamma = b, \alpha\beta\gamma = c$$.

Let the equation, whose roots are the quantities $$\sqrt[3]{\alpha}, \sqrt[3]{\beta}, \sqrt[3]{\gamma}$$ be $$x^3 - Ax^2 + Bx - C = 0$$.

Then, $$\sqrt[3]{\alpha} + \sqrt[3]{\beta} + \sqrt[3]{\gamma} = A$$, $$\sqrt[3]{\alpha}\sqrt[3]{\beta} + \sqrt[3]{\alpha}\sqrt[3]{\gamma} + \sqrt[3]{beta}\sqrt[3]{\gamma} = B$$, $$\sqrt[3]{\alpha\beta\gamma} = C$$.

Let us make use of the follwing identity

$(m + p + q)^3 = m^3 + p^3 + q^3 + 3(m + p + q)(mp + mq + pq) - 3mpq.$

Thus,

$A^3 = a + 3AB - 3C, B^3 = 3AB - 5$

Multiplying these equations and putting $$AV = z$$, we find

$z^3 - 9z^2 + 27z -20 = 0, (z - 3)^3 + 7 = 0, z = 3 - \sqrt[3]{7}$

But

$A^3 = 3z - 4 = 5 - 3\sqrt[3]{7}, A = \sqrt[3]{5 - 3\sqrt[3]{7}}$
$\therefore \sqrt[3]{\alpha} + \sqrt[3]{\beta} + \sqrt[3]{\gamma} = \sqrt[3]{2\cos\frac{2\pi}{7}} + \sqrt[3]{2cos\frac{4\pi}{7}} + \sqrt[3]{2cos\frac{8\pi}{7}} = \sqrt[3]{5 - 3\sqrt[3]{7}}.$
6. This question can be solved like previous problem.