331. Solve the equation $2\sqrt{2}x^4 = (\sqrt{3 - 1}) + i(\sqrt{3} + 1)$.

Solution: Putting $x^4 = \frac{\sqrt{3} - 1}{2\sqrt{2}} + i\frac{\sqrt{3} + 1}{2\sqrt{2}}$ in polar form we get

$x^4 = \cos\frac{5\pi}{12} + i \sin\frac{5\pi}{12} \therefore x = \cos\frac{(24r + 5)\pi}{48} + i \sin\frac{(24r + 5)\pi}{48}, r = 0, 1, 2, 4$.

332. If $z_r = \cos\frac{\pi}{3^r} + i\sin\frac{\pi}{3^r}$, prove that $z_1z_2z_3\ldots$ to $\infty = i$.

Solution: L.H.S. $= z_1z_2z_3\ldots = \left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\left(\cos\frac{\pi}{3^2} + i\sin\frac{\pi}{3^2}\right)\left(\cos\frac{\pi}{3^3} + i\sin\frac{\pi}{3^3}\right)\cdots$

$= \cos\left(\frac{\frac{\pi}{3}}{1 - \frac{1}{3}}\right) + i\sin\left(\frac{\frac{\pi}{3}}{1 - \frac{1}{3}}\right) = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = i =$ R.H.S.

333. If $\cos\theta + i\sin\theta$ is a solution of the equation $p_0x^n + p_1x^{n - 1} + p_2x^{n - 2} + \cdots + p_n = 0$, prove that $p_1\sin\theta + p_2\sin2\theta + \cdots + p_n = 0$ and $p_0 + p_2\cos\theta + \cdots + p_n\cos n\theta = 0, p_i\in\mathbb{R}, i=1, 2, 3, \ldots, n$.

Solution: Given $p_0x^n + p_1x^{n - 1} + p_2x^{n - 2} + \cdots + p_n = 0$, we have to prove that $p_1\sin\theta + p_2\sin2\theta + \cdots + p_n = 0 \Rightarrow p_0(\cos n\theta + i\sin n\theta) + p_1[\cos(n - 1)\theta + \sin(n - 1)\theta] + p_2[\cos(n - 2)\theta + i\sin(n - 2)\theta] + \cdots + p_n = 0\;[\because \cos\theta + i\sin\theta]$ is a solution.

Dividing both sides by $\cos n\theta + i\sin n\theta$, we have $p_0 + p_1(\cos\theta - i\sin\theta) + p_2(\cos2\theta - i\sin2\theta) + \cdots + p_n(\cos n\theta - i\sin n\theta) = 0$. Equating real and imaghinary parts we have required equations.

334. Show that $\left(\frac{1 + \cos\phi + i\sin\phi} {1 + \cos\phi - i\sin\phi}\right)^n = \cos n\phi + i\sin\phi$.

Solution: L.H.S. $= \left(\frac{1 + \cos \phi + i \sin\phi}{1 + \cos \phi - i\sin\phi}\right)^n = \left(\frac{(1 + \cos\phi + i \sin\phi)(1 + \cos\phi + i\sin\phi)}{(1 + \cos\phi)^2 + \sin^2\phi}\right)^n$

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

$= (\cos\phi + i \sin\phi^n) = \cos n\phi + i \sin n\phi =$ R.H.S.

335. If $2\cos\theta = x + \frac{1}{x}$ and $2\cos\phi = y + \frac{1}{y}$, then prove that

1. $\frac{x}{y} + \frac{y}{x} = 2\cos(\theta - \phi)$,
2. $xy + \frac{1}{xy} = 2\cos(\theta + \phi)$,
3. $x^my^n + \frac{1}{x^my^n} = 2\cos(m\theta + n\phi)$, and
4. $\frac{x^m}{y^n} + \frac{y^n}{x^m} = 2\cos(m\theta - n\phi)$.

Solution: Given $2\cos\theta = x + \frac{1}{x} \Rightarrow x^2 - 2\cos\theta x + 1 = 0 \Rightarrow x = \cos\theta\pm i\sin\theta$. Similarly, $y = \cos\phi\pm i\sin\phi$.

1. $\frac{x}{y} = \cos(\theta - \phi) \pm i\sin(\theta - \phi)$ and $\frac{y}{x} = \cos(\phi - \theta)\pm i\sin(\phi - \theta)$ $\therefore$ L.H.S. $= 2\cos(\theta - \phi) =$ R.H.S. $[\because \cos(-\theta) = \cos\theta, \sin(-\theta) = -\sin\theta]$
2. $xy = \cos(\theta + \phi) \pm i\sin(\theta + \phi), \frac{1}{xy} = \cos(\theta + \phi)\mp i\sin(\theta + \phi)$ $\therefore$ L.H.S. $= 2\cos(\theta + \phi) =$ R.H.S.
3. $x^my^n = (\cos m\theta \pm i\sin m\theta)(\cos n\phi \pm i\sin n\phi) = \cos(m\theta + n\phi)\pm i\sin(m\theta + n\phi)$ and $\frac{1}{x^my^n} = \cos(m\theta + n\phi)\mp i\sin(m\theta + n\phi)$ $\therefore$ L.H.S. $= 2\cos(m\theta + n\phi) =$ R.H.S.
4. $\frac{x^m}{y^n} = \cos(m\theta - n\phi)\pm i\sin(m\theta - n\phi)$ and $\frac{y^n}{x^m} = \cos(n\phi - m\theta)\pm i\sin(n\phi - m\theta)$ $\therefore$ L.H.S. $= 2\cos(m\theta - n\phi) =$ R.H.S.

336. If $\alpha, \beta$ are the roots of the equation $x^2 - 2x + 4 = 0$, prove that $\alpha^n + \beta^n = 2^{n + 1}\cos\frac{n\pi}{3}$.

Solution: Given equation is $x^2 - 2x + 4 = 0$ whose roots are $\alpha, \beta = 1\pm i\sqrt{3} = 2\left(\cos\frac{\pi}{3}\pm i\sin\frac{\pi}{3}\right)\Rightarrow \alpha^n, \beta^n = 2\left(\cos\frac{n\pi}{3}\pm i\sin\frac{n\pi}{3}\right)$

$\therefore \alpha^n + \beta^n = 2^{n + 1}\cos\frac{n\pi}{3} =$ R.H.S.

337. Find the equation whose roots are $n$th powers of the roots of the equation $x^2 - 2x\cos\theta + 1 = 0$.

Solution: Given equation is $x^2 - 2x\cos\theta + 1 = 0$, whose roots are $\cos\theta \pm i\sin\theta$, $n$th power of which are $\cos n\theta \pm i\sin n\theta$. Therefore, the equation having these roots is $x^2 - 2\cos n\theta + 1 = 0$.

338. Find the values of $A$ and $B$, where $Ae^{2i\theta} + Be^{-2i\theta} = 5\cos2\theta - 7\sin2\theta$.

Solution: L.H.S. $= A(\cos2\theta + i\sin2\theta) + B(\cos2\theta - i\sin2\theta) = 5\cos2\theta + 7i^2\sin2\theta$. $\Rightarrow A + B = 5, A - B = 7i \Rightarrow A = \frac{5 + 7i}{2}, B = \frac{5 - 7i}{2}$.

339. If $x = \cos\theta + i\sin\theta$ and $\sqrt{1 - c^2} = nc - 1$, prove that $1 + c\cos\theta = \frac{c}{2n}(1 + nx)(1 + \frac{n}{x})$.

Solution: Given $x = \cos\theta + i\sin\theta$ and $\sqrt{1 - c^2} = nc - 1$. Squaring the second equaiton $n^2c^2 + c^2 - 2nc = 0 \Rightarrow c = \frac{2n}{n^2 + 1}$. We have to prove that $1 + \cos\theta = \frac{c}{2n}(1 + nx)\left(1 + \frac{n}{x}\right)$.

R.H.S. $= \frac{1}{n^2 + 1}\left(1 + n^2 + 2n\cos\theta\right) = 1 + \frac{2n}{n^2 + 1}\cos\theta = 1 + c\cos\theta =$ L.H.S.

340. Show that the roots of equation $(1 + z)^n = (1 - z)^n$ are $i\tan\frac{r\pi}{n}, r = 0, 1, 2, \ldots, (n - 1)$ excluding the value when $n$ is even and $r = \frac{n}{2}$.

Solution: From the given equality, we have $\left(\frac{1+z}{1-z}\right)^n = 1\Rightarrow 1 + z = (1 - z)(\cos\frac{2r\pi}{n} +i \sin\frac{2r\pi}{n})$

Let $\frac{2r\pi}{n} = \theta$ then $1 + z = (1 - z)(\cos\theta + i \sin\theta) \Rightarrow z((1 + \cos\theta) + i \sin\theta) = (\cos\theta - 1) + i \sin\theta \Rightarrow z = \frac{(\cos\theta - 1) + i \sin\theta}{(1 + \cos\theta) + i \sin\theta}$

$z = i \tan\frac{\theta}{2} = i \tan\frac{2\pi}{n}, \; r = 0, 1, 2, ..., (n - 1)$

Clearly, the above equation is invalid if $n$ is even and $r = \frac{n}{2}$ as it will cause the value of $\tan$ function to reach infinity.