311. If $ax + cy + bz = X, cx + by + az = Y, bc + ay + cz = Z$, show that

1. $(a^2 + b^2 + c^2 - ab - bc - ca)(x^2 + y^2 + z^2 - xy - yz - zx) = X^2 + Y^2 + Z^2 - XY - YZ - ZX$
2. $(a^3 + b^3 + c^3 - 3abc)(x^3 + y^3 + z^3 - 3xyz) = X^3 + Y^3 + Z^3 - 3XYZ$

Solution: The solutions are given below:

1. $(a^2 + b^2 + c^2 - ab - bc - ca)(x^2 + y^2 + z^2 - xy - yz - zx) = (a + b\omega + c\omega^2)(a + b\omega^2 + c\omega)(x + y\omega + z\omega^2)(x + y\omega^2 + z\omega)$

   $= (a + b\omega + c\omega^2)(x + y\omega + z\omega^2)[(a + b\omega^2 + c\omega)(x + y\omega^2 + z\omega)]$

   $= (ax + cy\omega^3 + bz\omega^3 + cx\omega^2 + by\omega^2 + za\omega^2 + bx\omega + ay\omega + cz\omega^4)(ax + cy\omega^3 + bz\omega^3 + cx\omega + by\omega^4 + az\omega + bz\omega^2 + ay\omega^2 + cz\omega^2)$

   $= [(ax+cy+bz)(cx+by+az)\omega^2 + (bx+ay+cz)\omega][(ax+cy+bz)(cx+by+az)\omega + (bx+ay+cz)\omega^2]$ $= (X+Y\omega^2+Z\omega)(X+Y\omega+Z\omega^2) = (X^2+Y^2+Z^2 -YZ-ZX-XY)$.

2. We just introduce two new factors to previous problem $a + b + c$ and $x + y + z$ and then it is only a matter of simplification to obtain the result.

312. Prove that $\left(\frac{\cos\theta + i\sin\theta} {\sin\theta + i\cos\theta}\right)^4 = \cos8\theta + i\sin8\theta$.

Solution: L.H.S. $= \left(\frac{\cos\theta + i\sin\theta} {\sin\theta + i\cos\theta}\right)^4 = \left(\frac{\cos\theta + i\sin\theta}{i(\cos\theta - i\sin\theta)}\right)^4 = \frac{e^{i4\theta}}{e^{-i4\theta}} = e^{i8\theta} = \cos8\theta + i\sin8\theta =$ R.H.S.

313. If $z^2 - 2z\cos\theta + 1 = 0$, show that $z^2 + z^{-2} = 2\cos2\theta$.

Solution: Roots of the quadratic equation $z^2 - 2z\cos\theta + 1 = 0$ are given by $z = \cos\theta\pm i\sin\theta$.

$\Rightarrow z^2 + z^{-2} = \cos2\theta \pm i\sin2\theta + \cos2\theta \mp i\sin2\theta = 2\cos2\theta =$ R.H.S.

314. Prove that $(1 + i)^n + (1 - i)^n = 2^{n/2 + 1}\cos\frac{n\pi}{4}$.

Solution: $1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$ and $(1 - i) = \sqrt{2}\left(\cos\frac{\pi}{4} - i\sin\frac{\pi}{4}\right)$

L.H.S. $= (1 + i)^n + (1 - i)^n = (\sqrt{2})^n.2\cos\frac{n\pi}{4} = 2^{\frac{n}{2} + 1}.cos\frac{n\pi}{4} =$ R.H.S.

315. Show that the value of $\displaystyle\sum_{k = 1}^6\left(\sin\frac{2\pi k}{7} - i\cos\frac{2\pi k}{7}\right)$ is $i$.

Solution: $\displaystyle\sum_{k = 1}^6\left(sin\frac{2\pi k}{7} -icos\frac{2\pi k}{7}\right) = -i \sum_{k = 1}^6\left(cos\frac{2\pi k}{7} + isin\frac{2\pi k}{7}\right)$

$\displaystyle= -i \sum_{k = 1}^6 e^{\frac{i2\pi k}{7}} = -i \left[e^{\frac{i2\pi}{7}} + e^{\frac{i4\pi}{7}} + \cdots + e^{\frac{i12\pi}{7}}\right] = -i \left[\left(\frac{1 - e^{2\pi}}{1 - e^{\frac{i2\pi}{7}}}\right) - 1\right] = -i [0 - 1] = i$.

316. Show that $e^{2mi\cot^{-1}p}\left(\frac{pi + 1}{pi - 1}\right)^m = 1$.

Solution: Let $cot^{-1}p = \theta$, then $cot\theta = p$. Now, L. H. S. is $e^{2mi\theta}\left(\frac{icot\theta + 1}{icot\theta - 1}\right)^m = e^{2mi\theta}\left[\frac{i(cot\theta - i)}{i(cot\theta + i)}\right]^m$

$= e^{2mi\theta}\left(\frac{cos\theta - isin\theta}{cos\theta + isin\theta}\right)^m$ $= e^{2mi\theta}\left(\frac{e^{-i\theta}}{e^i\theta}\right)^m = e^{2mi\theta} . e^{-2mi\theta} = e^0 = 1 =$ R.H.S.

317. Prove that $\left(\frac{1 + \sin\phi + i\cos\phi} {1 + \sin\phi - i\cos\phi}\right)^n = \cos\left(\frac{n\pi}{2} - n\phi\right) + i\sin\left(\frac{n\pi}{2} - n\phi\right)$.

Solution: Let $1 + \sin\phi + i \cos\phi = r(\cos\theta + i \sin\theta) \therefore 1 + \sin\phi = r\cos\theta$ and $\cos\phi = r\sin\theta$

Now $(1 + \sin\phi + i \cos\phi)^n = r^n(\cos n\theta + i\sin n\theta)$. Taking conjugates, we get $(1 + \sin\phi - i \cos\phi)^n = r^n(\cos n\theta - i\sin n\theta)$

From these two, we get $\left(\frac{1 + \sin\phi + i \cos\phi}{1 + \sin\phi - i \cos\phi}\right)^n = \frac{\cos n\theta + i\sin n\theta}{\cos n\theta - i\sin n\theta} = \frac{e^{in\theta}}{e^{-in\theta}}$ $= e^{2in\theta} = \cos 2n\theta + \sin 2n\theta$

$\tan \theta = \frac{\cos \phi}{1 + \sin \phi} = \frac{\cos^2\frac{\phi}{2} - \sin^2\frac{\phi}{2}}{\left(\cos\frac{\phi}{2} + \sin\frac{\phi}{2}\right)^2} = \frac{\cos\frac{\phi}{2} - \sin\frac{\phi}{2}}{\cos\frac{\phi}{2} + \sin\frac{\phi}{2}} = \frac{1 - \tan\frac{\phi}{2}}{1 + \tan\frac{\phi}{2}} = \tan\left(\frac{\pi}{4} - \frac{\phi}{2}\right)$

$\therefore \theta = \frac{\pi}{4} - \frac{\phi}{2}\;\therefore 2n\theta = \left(\frac{n\pi}{2} - n\phi\right)$. Hence, proved.

318. If $\sin\alpha + \sin\beta + \sin\gamma = \cos\alpha + \cos\beta + \cos\gamma = 0$, show that $\cos3\alpha + \cos3\beta + \cos3\gamma = 3\cos(\alpha + \beta + \gamma)$ and $\sin3\alpha + \sin3\beta + \sin3\gamma = 3\sin(\alpha + \beta + \gamma)$.

Solution: Let $a = \cos\alpha + i \sin\alpha, b = \cos\beta + i \sin\beta, c = \cos\gamma + i \sin\gamma$

Now, $a + b + c = (\cos\alpha + \cos\beta + \cos\gamma) + i(\sin\alpha + \sin\beta + \sin\gamma) = 0 + i.0 = 0$

Now, $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$ $[\because a + b + c = 0]$

$\therefore a^3 + b^3 + c^3 = 3abc\therefore \cos3\alpha + \cos3\beta + \cos3\gamma = 3\cos(\alpha + \beta + \gamma)$ and $\sin3\alpha + \sin3\beta + \sin3\gamma = 3\sin(\alpha + \beta + \gamma)$.

319. If $\sin\alpha + \sin\beta + \sin\gamma = \cos\alpha + \cos\beta + \cos\gamma = 0$, show that $\cos2\alpha + \cos2\beta + \cos2\gamma = \sin2\alpha + \sin2\beta + \sin2\gamma = 0$.

Solution: Proceeding similarly as last problem and with an extra calculation we have

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = (\cos\alpha + \cos\beta + \cos\gamma) - i(\sin\alpha + \sin\beta + \sin\gamma) = 0$

$\therefore a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = (a + b + c)^2 - 2 abc\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$

$\Rightarrow 0^2 - 2abc.0 = 0\therefore L.H.S. = (\cos2\alpha + \cos2\beta + \cos2\gamma) + i(\sin2\alpha + \sin2\beta + \sin2\gamma) = 0$

Equating real and imaginary parts we have our desired result.

320. If $\alpha, \beta$ are the roots of the equation $t^2 - 2t + 2 = 0,$ show that a value of $x$, satisfying $\frac{(x + \alpha)^n - (x + \beta)^n}{\alpha - \beta} = \frac{\sin\theta}{\sin^n\theta}$ is $x = \cot\theta - 1$.

Solution: $t^2 -2t + 2 = 0 \Leftrightarrow t = \frac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i$

Let $\alpha = 1+ i$ and $\beta = 1 - i\;\therefore x + \alpha = (x + 1) + i, x + \beta = (x + 1) - i$ and $\alpha - \beta = 2i$

Let $x + 1 = r\cos\phi$ and $1 = r\sin\phi$. We have, $\frac{(x + \alpha)^n - (x + \beta)^n}{(\alpha - \beta)} = \frac{\sin\theta}{\sin^n\theta}$

$ \Leftrightarrow \frac{r^n(\cos n\phi + i \sin n\phi) - r^n(\cos n\phi - i\sin n\phi)}{2i} = \frac{\sin\theta}{\sin^n\theta} \Leftrightarrow r^n \sin n\phi = \frac{\sin\theta}{\sin^n\theta}$

$\Leftrightarrow \frac{\sin n\phi}{\sin^n\phi} = \frac{\sin\theta}{\sin^n\theta} \Leftrightarrow$ one of the values of $\phi$ is $\theta$. $\left[\because r\sin\phi = 1\Rightarrow r^n = \frac{1}{\sin^n\phi}\right]$

$\therefore x + 1 = r \cos\theta$ and $1 = r \sin\theta$. Dividing and evaluating we get $x = \cot\theta - 1$.