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.