# 26. Complex Numbers Solutions Part 5

1. 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

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

$$= e^{2in\theta} = \cos 2n\theta + \sin 2n\theta$$

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

Hence, proved.

2. 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 $$\sin3\alpha + \sin3\beta + \sin3\gamma = 3\sin(\alpha + \beta + \gamma)$$

3. 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.

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

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,

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

$$\Leftrightarrow$$ one of the values of $$\phi$$ is $$\theta.$$

$$\therefore x + 1 = r \cos\theta$$ and $$1 = r \sin\theta$$

Dividing and evaluating we get $$x = \cot\theta - 1.$$

5. Given, $$(1 + x)^n = p_0 + p_1x + p_2x^2 + ... + p_nx^n$$

putting $$x = i$$, we get

$$(1 + i)^n = p_0 + p_1i + p_2i^2 + ... + p_ni^n$$

$$(p_0 - p_2 + p_4 - ...) + i(p_1 - p_3 + p_5 - ...)$$

or $$\left[\sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)\right]^n = (p_0 - p_2 + p_4 - ...) + i(p_1 - p_3 + p_5 - ...)$$

Equating real and imaginary parts, we have

$$p_0 - p_2 + p_4 ... = 2^{\frac{n}{2}}\cos\frac{n\pi}{4}$$ and

$$p_1 - p_3 + p_5 + ... = 2^{\frac{n}{2}}\sin\frac{n\pi}{4}.$$

6. Given, $$(1 - x + x^2)^n = a_0 + a_1 + a_2x^2 + ... a_{2n}x^{2n}$$.

Putting $$x = 1, \omega$$ and $$\omega^2$$, we get

$$1 = a_0 + a_1 + a_2 + ... + a_{2n}$$

$$(-2\omega)^n = a_0 + a1\omega + a_2\omega^2 + ... + a_{2n}\omega^{2n}$$

$$(-2\omega^2)^n = a_0 + a_1\omega^2 + a_2\omega^4 + ... + a_{2n}\omega^{4n}$$

$$3(a_0 + a_3 + a_6 + ...) = 1 + (-2)^n(\omega^n + \omega^{2n})$$

Now $$\omega = \frac{-1 + \sqrt{3}i}{2} = \left(\cos\frac{2\pi}{3} + i\sin \frac{2\pi}{3}\right)$$

$$\omega^n = \cos\frac{2n\pi}{3} + i\sin\frac{2n\pi}{3}$$

Now $$\omega^2 = \frac{-1 - \sqrt{3}i}{2} = \left(\cos\frac{2\pi}{3} - i\sin \frac{2\pi}{3}\right)$$

$$\therefore \omega^n + \omega^{2n} = 2\cos\frac{2n\pi}{3} = 2\cos\left(n\pi - \frac{n\pi}{3}\right)$$

$$= 2(-1)^n\cos\frac{n\pi}{3}$$

Thus, $$3(a_0 + a_3 a_6 + ...) = 1 + (-2)^n2(-1)^n\cos\frac{n\pi}{3} = 1 +2^{n + 1}\cos\frac{n\pi}{3}.$$

$$a_0 + a_3 + a_6 + ... = \frac{1}{3}\left(1 + 2^{n+1}\cos\frac{n\pi}{3}\right)$$

7. Given, $$(1 + x)^n = c_0 + c_1x + c_2x^2 + ... + c_nx^n$$

Putting $$x = 1$$ and $$x = -1$$, we get

$$2^n = c_0 + c_1 + c_2 + ... + c_n$$

and $$0 = c_0 - c_1 + c_2 - ... + (-1)^nc_n$$

$$2^n = c_0 + c_2 + c_4 + ...$$

or $$c_0 + c_2 + c_4 + ... = 2^{n - 1}$$

Putting $$x = i$$, we get

$$(1 + i)^n = c_0 + c_1i + c_2i^2 + c_3i^3 + ... + c_ni^n$$

$$\left[\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\right]^n = (c_0 -c_2 + c_4 - ...) + i(c_1 - c_3 + ...)$$

or $$2^{\frac{n}{2}}\left(\cos\frac{n\pi}{4}+i\sin\frac{i\pi}{4}\right) = (c_0 -c_2 + c_4 - ...) + i(c_1 - c_3 + ...)$$

Equating real parts, we get

$$c_0 - c_2 + c_4 - ... = 2^\frac{n}{2}\cos\frac{n\pi}{4}$$

Adding this result with the one obtained previously, we have

$$2[c_0 + c_4 + c_8 + ...] = 2^{n - 1} + 2^\frac{n}{2}cos\frac{n\pi}{4}$$

Hence, we have our desired result.

8. $$z^8 + 1 = 0 \Rightarrow z^8 = -1 = \cos\pi + i \sin\pi$$

$$\therefore z = (\cos\pi + i \sin\pi)^\frac{1}{8} = \cos\frac{2r\pi + \pi}{8} + i \sin\frac{2r\pi + \pi}{8}, r = 0, 1, 2, ..., 7$$

$$\therefore z = \cos\frac{\pi}{8} \pm \sin\frac{\pi}{8}, \cos\frac{3\pi}{8} \pm \sin\frac{3\pi}{8}, \cos\frac{5\pi}{8} \pm \sin\frac{5\pi}{8}, \cos\frac{7\pi}{8} \pm \sin\frac{7\pi}{8}$$

These are the roots of the above equation.

Now, quadratic equation whose roots are $$\cos\frac{\pi}{8} \pm \sin\frac{\pi}{8},$$ is

$$z^2 - 2\cos\frac{\pi}{8}z + 1 = 0$$

Similarly, we can find the quadratic equations for remaining three pairs of roots. Thus,

$$z^8 + 1 = \left(z^2 - 2\cos\frac{\pi}{8}z + 1\right)\left(z^2 - 2\cos\frac{3\pi}{8}z + 1\right)\left(z^2 - 2\cos\frac{5\pi}{8}z + 1\right)\left(z^2 - 2\cos\frac{7\pi}{8}z + 1\right)$$

Dividing both sides by $$z^4$$, we get

$$z^4 + \frac{1}{z^4} = \left(z + \frac{1}{z} - 2\cos\frac{\pi}{8}\right)\left(z + \frac{1}{z} - 2\cos\frac{3\pi}{8}\right)\left(z + \frac{1}{z} - 2\cos\frac{5\pi}{8}\right)\left(z + \frac{1}{z} - 2\cos\frac{7\pi}{8}\right)$$

Putting $$z = \cos\theta + i \sin\theta$$, so that $$z^n + \frac{1}{z^n} = 2n\cos n\theta$$, we get

$$2\cos 4\theta = 2\left(\cos \theta - \cos\frac{\pi}{8}\right)2\left(\cos \theta - \cos\frac{3\pi}{8}\right)2\left(\cos \theta - \cos\frac{5\pi}{8}\right)2\left(\cos \theta - \cos\frac{5\pi}{8}\right)$$

$$\therefore \cos 4\theta = 8\left(\cos \theta - \cos\frac{\pi}{8}\right)\left(\cos \theta - \cos\frac{3\pi}{8}\right)\left(\cos \theta - \cos\frac{5\pi}{8}\right)\left(\cos \theta - \cos\frac{7\pi}{8}\right)$$

9. Let $$z = \cos\theta + i \sin\theta$$, then $$z^7 = \cos 7\theta + i \sin 7\theta$$

If

$\theta = \frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \frac{7\pi}{7}, \frac{9\pi}{7}, \frac{11\pi}{7}, \frac{13\pi}{7}$

then $$z^7 = \cos 7\theta + i \sin 7\theta = 1$$ or $$z^7 + 1 =0$$

Thus, z = costheta + i sintheta`, where

$\theta = \frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \frac{7\pi}{7}, \frac{9\pi}{7}, \frac{11\pi}{7}, \frac{13\pi}{7}$

are the roots of the equation.

Also, when $$\theta = \pi, z = -1$$

Now, $$z^7 + 1 = 0 \Rightarrow (z + 1)(z^6 - z^5 + z^4 - z^3 + z^2 - z + 1) = 0$$

Root of equation $$z + 1 = 0$$ is $$\cos \theta + i \sin \theta,$$ where $$\theta = \pi$$

Roots of equation $$z^6 - z^5 + z^4 - z^3 + z^2 - z + 1 = 0~~~~~(1)$$

are $$\cos \theta + i \sin \theta,$$ where

$\theta = \frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \frac{7\pi}{7}, \frac{9\pi}{7}, \frac{11\pi}{7}, \frac{13\pi}{7}$

Let $$x = \cos \theta$$, then

$z + \frac{1}{z} = \cos \theta + i \sin \theta + \frac{1}{\cos \theta + i \sin \theta} = 2\cos\theta = 2x$

But

$\cos\left(\frac{13\pi}{7}\right) = cos\left(2\pi - \frac{\pi}{7}\right) = \cos\frac{\pi}{7}, \cos\frac{11\pi}{7} = \cos\frac{3\pi}{7}, \cos\frac{9\pi}{7} = \cos\frac{5\pi}{7}$

Dividing (1) by $$z^3$$, we get

$z^3 - z^2 + z - 1 + \frac{1}{z} - \frac{1}{z^2} + \frac{1}{z^3} = 0$
$\left(z^3 + \frac{1}{z^3}\right) - \left(z^2 + \frac{1}{z^2}\right) + \left(z + \frac{1}{z}\right) - 1 = 0$
$\left(z + \frac{1}{z}\right)^3 - 3z.\frac{1}{z}\left(z + \frac{1}{z}\right) - \left[\left(z + \frac{1}{z}\right)^2 - 2z.\frac{1}{z}\right] + z + \frac{1}{z} - 1 = 0 \Rightarrow 8x^3 - 4x^2 -4x + 1 = 0$

Roots of this equation are $$\cos \frac{\pi}{7}, \cos \frac{3\pi}{7}$$ and $$\cos \frac{5\pi}{7}$$.

10. Given, $$z^{10} - 1 = 0 \Rightarrow z^{10} = 1 = \cos 0 + i \sin 0$$

$$\therefore z = (\cos 0 + i \sin 0)\frac{1}{10} = \cos\frac{2r\pi}{10} + i \sin\frac{2r\pi}{10}$$

$= \pm 1, \cos\frac{\pi}{5} \pm i\sin\frac{\pi}{5}, \cos\frac{2\pi}{5} \pm i\sin\frac{2\pi}{5}, \cos\frac{3\pi}{5} \pm i\sin\frac{3\pi}{5}, \cos\frac{4\pi}{5} \pm i\sin\frac{4\pi}{5}$

Quadratic equation whose roots are $$\pm 1$$ is $$z^2 - 1 = 0$$.

And quadratic equation whose roots are $$\cos\frac{\pi}{5} \pm \sin\frac{\pi}{5}$$ is $$z^2 - 2\cos\frac{\pi}{5}z + 1 = 0$$

Thus,

$z^{10} - 1 = (z^2 - 1)\left(z^2 - 2\cos\frac{\pi}{5}z + 1\right)\left(z^2 - 2\cos\frac{2\pi}{5}z + 1\right)\left(z^2 - 2\cos\frac{3\pi}{5}z + 1\right)\left(z^2 - 2\cos\frac{4\pi}{5}z + 1\right)$

Dividing both sides by $$z^5$$, we get

$z^5 - \frac{1}{z^5} = \left(z - \frac{1}{z}\right)\left(z + \frac{1}{z} - 2\cos\frac{\pi}{5}\right)\left(z + \frac{1}{z} - 2\cos\frac{2\pi}{5}\right)\left(z + \frac{1}{z} - 2\cos\frac{3\pi}{5}\right)\left(z + \frac{1}{z} - 2\cos\frac{4\pi}{5}\right)$

Putting $$z = \cos \theta + i \sin \theta$$ in the above equation, so that $$z^5 - \frac{1}{z^5} = 2i\sin 5\theta$$, we get

$2i\sin 5\theta = 2i\sin\theta.2\left(\cos \theta - \cos\frac{\pi}{5}\right)2\left(\cos \theta - \cos\frac{2\pi}{5}\right)2\left(\cos \theta - \cos\frac{3\pi}{5}\right)2\left(\cos \theta - \cos\frac{4\pi}{5}\right)$
$\therefore \sin 5\theta = 16 \sin \theta\left(\cos \theta - \cos\frac{\pi}{5}\right)\left(\cos \theta - \cos\frac{2\pi}{5}\right)\left(\cos \theta - \cos\frac{3\pi}{5}\right)\left(\cos \theta - \cos\frac{4\pi}{5}\right)$
$= 16\sin \theta\left(\cos\theta -\cos\frac{\pi}{5}\right)\left(\cos\theta +\cos\frac{\pi}{5}\right)\left(\cos\theta -\cos\frac{2\pi}{5}\right)\left(\cos\theta +\cos\frac{2\pi}{5}\right)$
$= 16\sin \theta\left(\cos^2\theta - \cos^2\frac{\pi}{5}\right)\left(\cos^2\theta - \cos^2\frac{2\pi}{5}\right)$
$= 16\sin \theta\left(\sin^2\frac{\pi}{5} - \sin^2\theta\right)\left(\sin^2\frac{2\pi}{5} - \sin^2\theta\right)$
$= 16\sin\theta\sin^2\frac{\pi}{5}\sin^2\frac{2\pi}{5}\left(1 - \frac{\sin^2\theta}{\sin^2\frac{\pi}{5}}\right)\left(1 - \frac{\sin^2\theta}{\sin^2\frac{2\pi}{5}}\right)$
$= 16\sin\theta\sin^2{36^\circ}\sin^2{72^\circ}\left(1 - \frac{\sin^2\theta}{\sin^2\frac{\pi}{5}}\right)\left(1 - \frac{\sin^2\theta}{\sin^2\frac{2\pi}{5}}\right)$
$= 16\sin\theta\left(\frac{\sqrt{10 - 2\sqrt{5}}}{2}\right)^2\left(\frac{\sqrt{10 + 2\sqrt{5}}}{2}\right)^2\left(1 - \frac{\sin^2\theta}{\sin^2\frac{\pi}{5}}\right)\left(1 - \frac{\sin^2\theta}{\sin^2\frac{2\pi}{5}}\right)$

Thus,

$\sin5\theta = 5\sin\theta\left(1 - \frac{\sin\theta}{\sin^2\frac{\pi}{5}}\right)\left(1 - \frac{\sin\theta}{\sin^2\frac{2\pi}{5}}\right)$
11. Given, $$x^7 + 1 = 0$$ or $$x^7 = -1 = \cos\pi + i \sin\pi$$

$\therefore x = \left(\cos\pi + i \sin\pi\right)\frac{1}{7} = \cos\frac{2r\pi + \pi}{7} + i \sin \frac{2r\pi + \pi}{7}, r = 0, 1, 2, ..., 6$
$= \cos \frac{\pi}{7} \pm i \sin\frac{\pi}{7}, \cos \frac{2\pi}{7} \pm i \sin\frac{2\pi}{7}, \cos \frac{3\pi}{7} \pm i \sin\frac{3\pi}{7}, \cos \pi + i \sin \pi(= -1)$
$x^7 + 1 = (x + 1)\left(x^2 - 2\cos\frac{\pi}{7}x + 1\right)\left(x^2 - 2\cos\frac{2\pi}{7}x + 1\right)\left(x^2 - 2\cos\frac{3\pi}{7}x + 1\right)$

Putting $$x = i$$, we get

$i^7 + 1 = (1 + i)\left(-2i\cos\frac{\pi}{7}\right)\left(-2i\cos\frac{2\pi}{7}\right)\left(-2i\cos\frac{3\pi}{7}\right)$
$1 - i = 8(1 + i)\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7} = -8(1 - i)\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7}$
$\therefore \cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7} = -\frac{1}{8}$
12. $(\cos\alpha + i \sin\alpha)^n = \cos^n\alpha + i{^nC_1}\cos^{n-1}\alpha \sin\alpha + i^2{^nC_2}\cos^{n-2}\alpha \sin^2\alpha + ... + i^n{^nCn}\sin^n\alpha$
$\Rightarrow \cos n\alpha + i \sin n\alpha = (\cos^n\alpha - {^nC_2}\cos^{n-2}\alpha \sin^2\alpha) + i({^nC_1}\cos^{n-1}\alpha \sin\alpha)$

Equating imaginary parts, we get

$\therefore \sin n\alpha = {^nC_1}\cos^{n-1}\alpha \sin\alpha - {^nC_3}\cos^{n-3}\alpha \sin^3\alpha + ...$
$\therefore \sin (2n+1)\alpha = {^{2n+1}C_1}\cos^{2n}\alpha \sin\alpha - {^{2n+1}C_3}\cos^{2n-2}\alpha \sin^3\alpha + ...$
(1)$\Rightarrow \sin (2n+1)\alpha = = \sin^{2n+1}\alpha[{^{2n+1}\cot^{2n}}\alpha - {^{2n+1}C_3}cot^{2n-2}\alpha + ...]$
$\text{when} \alpha = \frac{\pi}{2n+1}, \frac{2\pi}{2n+1}, ..., \frac{n\pi}{2n+1}, \sin(2n+1)\alpha = 0$
$\therefore \cot^2\frac{\pi}{2n+1}, \cot^2\frac{2\pi}{2n+1}, ..., \cot^2\frac{n\pi}{2n+1}$

are the roots of the equation. From the second term coefficient we get sum of roots in a polynomial.

$\therefore \cot^2\frac{\pi}{2n+1}+ \cot^2\frac{2\pi}{2n+1}+ ...+ \cot^2\frac{n\pi}{2n+1} = \frac{{^{2n+1}C_3}}{^{2n+1}C_1}$
13. Let

(2)$C = \cos\theta \cos\theta + \cos^2\theta \cos 2\theta + ... + cos^n\theta \cos n\theta$

and

(3)$C = \cos\theta \sin\theta + \cos^2\theta \sin 2\theta + ... + cos^n\theta \sin n\theta$

Now,

$C + iS = \cos\theta(cos\theta + i \sin\theta) + \cos^2\theta(\cos 2\theta + i \sin 2\theta) + ... + \cos^n\theta(\cos n\theta + i \sin n\theta)$
$= \cos\theta.e^{i\theta} + \cos^2\theta.e^{2i\theta} + ... + \cos^n\theta.e^{ni\theta}$

$$= x + x^2 + ... + x^n$$, where $$x = \cos\theta e^{i\theta}$$

$= \frac{x(x^n - 1)}{x - 1} = \frac{\cos\theta e^{i\theta}(\cos^n\theta e^{in\theta - 1})}{\cos\theta e^{i\theta} - 1}$
$=\frac{\cos\theta[\cos^n\theta(\cos n\theta + i \sin n\theta) - 1]}{\cos\theta - e^{-i\theta}}$
$\frac{\cos\theta[(\cos^n]theta\cos n\theta -1) + i cos^n\theta\sin n\theta]}{\cos\theta - (\cos\theta - i \sin\theta)}$
$= -i \cot\theta(\cos^n\theta \cos n\theta - 1) + i \cos^n\theta \sin] n\theta$

Equating imaginary parts, we get

$S = -\cot\theta(\cos^n\theta \cos n\theta - 1) = \cot\theta(1 - \cos^n\theta\cos n]theta)$
14. $$L.H.S. = -3 -4i$$

$= 5\left(-\frac{3}{5} - i\frac{4}{5}\right) = 5\left(\cos\left(\pi + tan^{-1}\frac{4}{5}\right) + i \sin\left(\pi + tan^{-1}\frac{4}{5}\right)\right)$
$= 5e^{i\left(\pi + tan^{-1}\frac{4}{5}\right)}$
15. Putting

$x^4 = \frac{\sqrt{3} - 1}{2\sqrt{2}} + \frac{\sqrt{3} + 1}{2\sqrt{2}}$

in polar form we get

$x^4 = \cos\frac{5\pi}{12} + i \sin\frac{5\pi}{12}$
$\therefore x = \cos\frac{(24r + 5)\pi}{48} + i \sin\frac{(24r + 5)\pi}{48}, r = 0, 1, 2, 4$
16. $L.H.S. = \left(\frac{1 + \cos \phi + i \sin\phi}{1 + \cos \phi - i \sin\phi}\right)^n$
$= \left(\frac{(1 + \cos\phi + i \sin\phi)(1 + \cos\phi + i \sin\phi)}{(1 + \cos\phi)^2 + \sin^2\phi}\right)^n$
$= \left(\frac{1+ 2\cos\phi + \cos^2\phi - \sin^2\phi + 2i \sin\phi(1 + \cos\phi)}{1 + 2\cos\phi + \cos^2\phi + \sin^2\phi}\right)^n$
$= \left(\frac{2\cos\phi(1 + \cos\phi) + 2i \sin\phi(1 + \cos\phi)}{2\cos\phi(1 + \cos\phi)}\right)^n$
$= (\cos\phi + i \sin\phi^n) = \cos n\phi + i \sin n\phi$
17. This problem has been left as an exercise for the reader.

18. This problem has been left as an exercise for the reader.

19. This problem has been left as an exercise for the reader.

20. This problem has been left as an exercise for the reader.

21. Given, $$Ae^{2i\theta} + Be^{-2i\theta} = 5cos2\theta - 7sin2\theta$$

Let, $$A = a + ib$$ and $$B = c + id$$. Substituting and equating real and imaginary parts we have $$a + c = 5, d - b = 7$$ and $$b + a + d - c = 0$$. Since we have four unknowns and three equations it is impossible to find their values by simple calculation. However, by trial and error we can arrive at multiple solutions of these equation. For example, for $$a, b, c$$ and $$d$$ values of $$1, -2, 4$$ and $$5$$ satisfy the equations. Similarly, $$2, -3, 3$$ and $$4$$ also satisfy these equations. Similarly, more values for these can be found and there exists infinite solutions for these unknowns as is evident from theory of equations.

22. Given $$\sqrt{1 - c^2} = nc - 1,$$ squaring and solving we have $$1 - c^2 = n^2c^2 - 2nc + 1$$ i.e. $$(1 + n^2)c^2 - 2nc = 0$$ which has two solutions $$c = 0$$ and $$c = \frac{2n}{1 + n^2}$$. Clearly, the solution $$c = 0$$ does not satisfy our desired equation as substituting $$c = 0$$ makes it $$1 = 0$$ which is false. Thus, substituting the other value and taking R.H.S.

$R.H.S. = \frac{c}{2n}(1 + nx)\left(1 + \frac{n}{x}\right)$
$\Rightarrow \frac{1}{1 + n^2}(1 + n(\cos\theta + i \sin\theta)) + \left(1 + n(\cos\theta + i \sin\theta)\right)$
$\Rightarrow \frac{1}{1 + n^2}(1 + n^2 + 2n \cos\theta)$
$\Rightarrow 1 + c \cos\theta$
23. From the given equality, we have

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

Let $$\frac{2r\pi}{n} = \theta$$

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

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.

24. $L.H.S. = \frac{xy(x + y) - (x + y)}{xy(x - y)+(x - y)}$

Dividing numerator and denominator by $$xy$$

$= \frac{x + y - \frac{1}{x} - \frac{1}{y}}{x - y + \frac{1}{y} - \frac{1}{x}}$
$= \frac{\cos\alpha + i \sin\alpha + \cos\beta + i \sin\beta - \cos\alpha + i\sin\alpha - \cos\beta - i \sin\beta}{\cos\alpha + i \sin\alpha - \cos\beta - i \sin\beta - \cos\alpha - i\sin\alpha + \cos\beta - i \sin\beta}$
$= \frac{\sin\alpha + \sin\beta}{sin\alpha - \sin\beta}$
25. $(1 + x)^n = {^nC_0} + {^nC_1}x + {^nC_3}x^2 + {^nC_3}x^3 + ...$

Substituting $$x = 1, \omega$$ and $$\omega^{2}$$ and adding we get(from $$1 + \omega + \omega^2 = 0$$)

$2^n = {^nC_0} + {^nC_1} + {^nC_3} + {^nC_3} + ...$
$(-w)^2n = {^nC_0} + {^nC_1}\omega + {^nC_3}\omega^2 + {^nC_3}\omega^3 + ...$
$(-w)^4n = {^nC_0} + {^nC_1}\omega^2 + {^nC_3}\omega^4 + {^nC_3}\omega^6 + ...$

Now there can be three cases as n is a multiple of 3 and n is not multiple of 3. Adding all those we get the desired result.

26. Hint: Put $$x = 1, \omega, \omega^2$$ and multiply the three equations thus obtained with $$1, \omega^2$$ and $$\omega$$ respectively and add.

27. Hint: Put $$x = 1, \omega, \omega^2$$ and multiply the three equations thus obtained with $$1, \omega$$ and $$\omega^2$$ and respectively add.

28. Hint: Put $$1, \omega, \omega^2$$ and add.

29. Hint: Put $$1, \omega, \omega^2$$ and add.

30. Solving the system of linear equations we get

$x^{\prime\prime} = \frac{AA^{\prime} + BB^{\prime} + CC^{\prime}}{3} y^{\prime\prime} = \frac{AA^{\prime} + BB^{\prime}\omega^2 + CC^{\prime}\omega}{3} z^{\prime\prime} = \frac{AA^{\prime} + BB^{\prime}\omega + CC^{\prime}\omega^2}{3}$

Further $$AA^{\prime} + BB^{\prime} + CC^{\prime} = 3(xx^{\prime} + zy^{\prime} + yz^{\prime})$$

And so $$x^{\prime\prime = xx^{\prime}} + zy^{\prime} + yz^{\prime}$$. Analogously $$y^{\prime\prime = yy^{\prime} + xz^{\prime} + zx^{\prime}}, z^{\prime\prime} = zz^{\prime} + yx^{\prime} + xy^{\prime}$$ (the last two expressions emerge from the first one as a result of circular permutation).

31. We have the identity

$(\alpha\delta - \beta\gamma)(\alpha^{\prime}\delta^{\prime} - \beta^{\prime}\gamma^{\prime}) = (\alpha\alpha^{\prime} + \beta\gamma^{\prime}) + (\gamma\beta^{\prime} + \delta\delta^{\prime}) - (\alpha\beta^{\prime} + \beta\delta^{\prime})(\gamma\alpha^{\prime} + \delta\gamma^{\prime})$

Let us put there

$\alpha = x + yi, \beta = z + ti, \gamma = -(z - ti), \delta = x - yi$
$\alpha^{prime} = a + bi, \beta^{\prime} = c + di, \gamma^{\prime} = -(c - di), delta^{\prime} = a - bi$

Then,

$\alpha\delta - \beta\gamma = x^2 + y^2 + z^2 + t^2$
$\alpha^{\prime}\delta^{\prime} = beta^{\prime}\gamma^{\prime} = a^2 + b^2 + c^2 + d^2$
$\alpha\alpha^{\prime} + \beta\gamma^{\prime} = (ax - by - cz - dt) + i(bx + ay + dz - ct)$
$\gamma\beta^{\prime} + \delta\delta^{\prime} = \overline{\alpha\alpha^{\prime} + \beta\gamma^{\prime}}$
$\therefore (\alpha\alpha^{\prime} + \beta\gamma^{\prime})(\gamma\beta^{\prime} + \delta\delta^{\prime}) = (ax - by - cz - dt)^2 + (bx + ay + dz - ct)^2$

Further

$\alpha\beta^{\prime} + \beta\delta^{\prime} = (cx - dy + az + bt) + i(dx + cy -bz + at)$
$\gamma\alpha^{\prime} + \delta\gamma^{\prime} = -(cx - dy + az + by) + i(dz + cy - bz + at)$

i.e.

$-(\alpha\beta^{\prime} + \beta\delta^{\prime})(\gamma\alpha^{\prime} + \delta\gamma^{\prime}) = (cx - dy + az + bt)^2 + (dx + cy - bz + at)^2$

Substituting the obtained expressions into the original identity, we find

$(a^2 + b^2 + c^2 + d^2)(x^2 + y^2 + z^2 + t^2) =$
$(ax - by -cz -dt)^2 + (bx + ay -dz + ct)^2 + (cx + dy + az -bt)^2 + (dx - cy + bz + at)^2$
32. Expanding the expression $$(\cos\theta + i \sin\theta)^n$$ using binomial expansion, using De Moivre’s formula and comparing real and imaginary parts we have

$\cos n\theta = \cos^n\theta - {^nC_2}\cos^{n-1}\theta\sin^2\theta + ...$
$\sin n\theta = n\cos^{n-1}\theta\sin\theta - (^nC_3)\cos^{n-3}\theta\sin^3\theta + ...$

Taking into account the parity of n and dividing both members of these equalities by $$\cos^n\theta$$, we get the desired result.

33. We have,

$\cos x = \frac{(\cos x + i \sin x)+(\cos x - i \sin x)}{2}$

Put $$z = \cos x + i \sin x$$. Then $$\cos x - i \sin x = z^{-1}$$

$cos^{2m}x = \left(\frac{z + z^{-1}}{2}\right)^{2m} = \frac{1}{2^{2m}}\sum_{k=0}^{2m} C^k_{2m}z^{-k}z^{2m - k}$

Further

$2^{2m}\cos^{2m}x = \sum_{k=0}^{m-1}C^k_{2m} z^{2(m - k)} + C^m_{2m} + \sum_{k = m + 1}^{2m}C^k_{2m}z^{2(m - k)}$

In the second term we put $$m-k =-(m-k^{\prime})$$. Then this sum is rewritten in the following manner

$\sum_{k^{\prime}=m-1}^0C_{2m}^{2m-k^{\prime}}z^{-2(m-k^{\prime})} = \sum_{k=0}^{m-1}C_{2m}^kz^{-2(m-k)}$

And so

$2^{2m}\cos^{2m}\theta = \sum_{k=0}^{m-1}C^k_2m \left(z^{2(m-k)}+z^{-2(m-k)}\right)+C^m_{2m}$

However,

$z^{2(m-k)} + z^{-2(m-k)} = 2\cos 2(m-k)$
$\therefore 2^{2m}cos^{2m}x = \sum_{k = 0}^{k = m - 1} 2 {2m \choose k} cos2(m - k)x + {2m \choose m}$
34. Replacing $$x$$ by $$\frac{\pi}{2} -x$$ we can solve this one.

35. This can be done 233.

36. This can be done as 234.

37. From the expression

$u_n + iv_n = (\cos\alpha + i \sin\alpha) + r[\cos(\alpha + \theta) + i \sin(\alpha + \theta)] + ... + r^n[\cos(\alpha + b\theta) + i \sin(\alpha + b\theta)]$

Putting $$\cos \theta + i \sin\theta = z$$, then

$u_n + iv_n = (\cos\alpha + i\sin\alpha{1 + rz + ... + (rz)^n})$
$= (\cos\alpha + i \sin\alpha)\frac{(rz)^{n+1} - 1}{rz - 1}$

Let us transform the fraction $$\frac{(rz)^{n+1} - 1}{rz - 1}$$, separating the real part from the imaginary one.

We have

$\frac{(rz)^{n+1} - 1}{rz - 1} = \frac{[(rz)^{n+1}][r\overline{z} - 1]}{(rz -1)r\overline{z} - 1}$
$= \frac{r^{n+1}[\cos n\theta + i \sin n\theta] - r[\cos\theta - i \sin\theta]}{1 -2 r\cos\theta + r^2} + \frac{-r^{n+1}[\cos(n+1)\theta + i\sin(n+1)\theta] + 1}{1 -2r \cos\theta + r^2}$

Multiplying the last fraction by $$\cos\alpha + i \sin\alpha$$ and separating the real and imaginary parts, we get the required result

$u_n + iv_n = \frac{r^{n+1}[\cos (n\theta + \alpha) + i \sin (n\theta + \alpha)] - r[\cos(\alpha -\theta) - i \sin(\alpha - \theta)]}{1 -2 r\cos\theta + r^2} + \frac{-r^{n+1}\{\cos[(n+1)\theta + \alpha] + i\sin[(n+1)\theta + \alpha]\} + \cos\alpha + i \sin\alpha}{1 -2r \cos\theta + r^2}$

Hence,

$u_n = \frac{\cos\alpha - r\cos(\alpha - \theta) - r^{n+1}\cos[(n+1)\theta + \alpha] + r^{n+2}\cos(n\theta + \alpha)}{1 - 2r \cos\theta + r^2}$
$v_n = \frac{\sin\alpha - r\sin(\alpha - \theta) - r^{n+1}\sin[(n+1)\theta + \alpha] + r^{n+2}\sin(n\theta + \alpha)}{1 - 2r \cos\theta + r^2}$

Putting in these formulas $$\alpha = 0, r =1$$, we find

$1 + \cos\theta + \cos 2\theta + ... + \cos n\theta = \frac{\sin\frac{n+1}{2}\theta\cos\frac{n}{2}\theta}{\sin\frac{\theta}{2}}$
$\sin\theta + \sin 2\theta + ... + \sin n\theta = \frac{sin\frac{(n+1)\theta}{2}\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}$
38. We have,

$S + S^{\prime}i = \sum_{k=0}^nC^k_n(\cos k\theta + i \sin k\theta) = \sum_{k=0}^n(\cos\theta + i \sin\theta)^k$
$= (1 + \cos\theta + i \sin\theta)^n = \left[2\cos^2\frac{\theta}{2} + 2i \sin\frac{\theta}{2}\cos\frac{\theta}{2}\right]^n$
$= 2^n\cos^n\frac{\theta}{2}\left(\cos\frac{\theta}{2} + i \sin\frac{\theta}{2}\right)^n$
$= 2^n\cos^n\frac{\theta}{2}\left(\cos\frac{n\theta}{2} + i \sin\frac{n\theta}{2}\right)$

Equating real and imaginary parts, we have

$S = 2^n\cos^n\frac{\theta}{2}\cos\frac{n\theta}{2},~~~ S^{\prime} = 2^n\cos^n\frac{\theta}{2}\sin\frac{n\theta}{2}$
39. This problem is solved in previous one where $$S$$ of this problem has been considered as $$S^{\prime}$$.

40. Putting

$S = \sin^{2p}\alpha + \sin^{2p}2\alpha + \sin^{2p}n\alpha = \sum_{l=1}^n \sin^{2p}l\alpha$

But from problem 234

$\sin^{2p}l\alpha = \frac{1}{2^{2p -1}}(-1)^p\sum_{k=0}^{p-1}(-1)^k\cos 2(p-k)l\alpha + \frac{1}{2^{2p}}C^p_{2p}$
$\therefore s = \frac{(-1)^p}{2^{2p - 1}}\sum_{k = 0}^{p - 1}(-1)^kC^k_2p\sum_{l=1}^n\cos2(p - k)l\alpha + \frac{n}{2^{2p}}C^p_{2p}$

Putting $$2(p-k)\alpha = \xi$$. Then

$\sum_{l = 1}^n \cos 2(p - k)l\alpha = \cos\xi + ... + \cos n\xi = \frac{sin\frac{n\xi}{2}\cos\frac{n+1}{2}\xi}{sin\frac{\xi}{2}}$

(from 237). Let us denote

$\frac{sin\frac{n\xi}{2}\cos\frac{n+1}{2}\xi}{sin\frac{\xi}{2}} = \sigma_k$

Then we can prove that $$\sigma_k = 0$$ if $$k$$ is of the same parity as $$p \{k\equiv p(mod 2)\}$$ and $$\sigma_k = -1$$ if $$k$$ and $$p$$ are of different parity $$\{k \equiv p + 1(mode 2)\}$$, and we get

$S = \frac{(-1)^p+1}{2^{2p - 1}}\sum_{\substack{k=0\\k\equiv p + 1(mod 2)}}^{p-1} (-1)^kC^k_{2p} + n\frac{1}{2^{2p}}C^p_{2p}.$

Hence,

$S = \frac{1}{2^{2p - 1}}\sum_{\substack{k=0\\k\equiv p + 1(mod 2)}}^{p-1} C^k_{2p} + n \frac{1}{2^{2p}}C^p{2p}$

But we can prove that

$\sum_{\substack{k=0\\k\equiv p + 1(mod 2)}}^{p-1} C^k_{2p} = 2^{2p -2}$

Hence, out formula is deduced.

41. Rewriting the polynomial as

$x^n - a^n - nxa^{n-1} + na^n = (x^ - a^n) -na^{n-1}(x-a) = (x - a)(x^{n-1} + ax^{n-2} + ... + a^{n-1} - na^{n-1})$

At $$x = a$$ the second factor of the last product vanishes and, consequently, is dividiable by $$x - a$$; therefore the given polynomial is divisible by $$(x - a)^2$$.

42. Considering the given expression as a polynomial in $$y$$, let us put $$y = 0$$. We see that at $$y = 0$$ the polynomial vanishes (for any $$x$$). Therefore out polynomial is divisible by $$y$$. Since it is symmetrical both with respect to $$x$$ and $$y$$ (remains unchanged on permutation of these letters), it is divisible by $$x$$ as well. Thus, the polynomial is divisible by $$xy$$. To prove that it is divisible by $$x + y$$ let us put in it $$y = -x$$. It is evident that for odd $$n$$ we have

$(x - x)^n - x^n -(-x)^n = 0.$

Consequently, out polynomial is divisible by $$x + y$$. It only remains to prove the divisibility of the polynomial by

$x^2 + xy + y^2 = (y - x\omega)(y - \omega^2)$

where $$1 + \omega + \omega^2 = 0$$.

For this purposed it only remains to replace $$y$$ first by $$x\omega$$ and then by $$x\omega^2$$ and to make sure that with these substitutions the polynomial vanishes. Since, by hypothesis, $$n$$ is not divisible by three, it follows that $$n = 3l + 1$$ or $$n = 3l + 2$$. At $$y = x\omega$$ the polynomial attains the following value:

$(x + x\omega)^n - x^n - (x\omega)^n = x^n(\omega^{2n} + 1 + \omega^n) = 0$

Likewise we prove that at $$y = x\omega^2$$ the polynomial vanishes as well and consequently, its divisibility by $$xy(x+y)(x^2+xy+y^2)$$ is proved.

43. Put $$f(x) = x^{4a} + x^{4b+1} + x^{4c+2} + x^{4d+3}$$. On the other hand $$x^3 + x^2 + x + 1 = (x+1)(x^2+1) = (x+1)(x+i)(x-i)$$

It only remains to show that $$f(-1) = f(i) = f(-i) = 0$$.

44. $$x^2 + 1 = 0 => x = \pm i$$ but $$f(i) = (\cos\theta + i \sin]\theta)^n - \cos n\theta - i \sin n\theta = 0$$ (by de Moivre’s Formula). Similarly, it can be proven that $$f(-i) = 0$$ thus our divisibility requirement is satisfied.

45. Resolving the polynomial $$x^2 -2kx\cos\theta + k^2$$ into linear factor of $$x$$. For this factor we find the roots of this quadratic equation. We get,

$$x = k(\cos\theta \pm i \sin\theta)$$

Let us denote $$x^n \sin\theta - k^{n - 1}x \sin n\theta + k^n \sin(n - 1)\theta = f(x)$$, then it can be proven that $$f(k(\cos\theta \pm i \sin\theta)) = 0$$.

46. We have

$s = \sum_{k=0}^{n-1}x^p_k = \sum_{k=0}^{n-1}z^{kp}$

where $$z = \cos\frac{2\pi}{n} + i \sin\frac{2\pi}{n}$$. Thus,

$\sum_{k = 0}^{n - 1}x^p_k = 1 + z^p + z^{2p} + ... + z^{(n-1)p}.$

But

$z^p = \cos \frac{2p\pi}{n} + i \sin \frac{2p\pi}{n}.$

It is obvious that $$z^p = 1$$ if and only if $$p$$ is divisible by $$n$$. In this case $$s = n$$.

And if $$z^p \neq 1$$, then $$s = 1 + z^p + z^{2p} + ... + z^{(n-1)p} = \frac{z^{np - 1}}{z^p - 1}$$, since $$z^{np} = 1$$.

47. We have

$\sum_{k=0}^{n-1}\left|A_k\right|^2 = \sum_{k=0}^{n-1}A_k\overline{A_k}.$

But

$A_k\overline{A_k} = \left(x + y\epsilon^k + z\epsilon^{2k} + ... +w\epsilon^{(n - 1)}k) \times (\overline{x} + \overline{y}\epsilon^{-k} + \overline{z}^{-2k} + ... + \overline{w}\epsilon^{-(n-1)k}\right)$
$= x\overline{x} + y\overline{y} + ... + w\overline{w} + x\left(\overline{y}\epsilon^{-k} + \overline{z}\epsilon^{-2k} + ... + \overline{w}\epsilon^{-(n-1)k}\right) + y\epsilon^k\left(\overline{x} + \overline{z}\epsilon{-2k} + ... + \overline{w}\epsilon^{-(n-1)k}\right) + w\epsilon^{(n-1)k}\left(\overline{x} + \overline{y}^{-k} + ... + \overline{u}\epsilon^{-(n-2)k}\right)$
$A_k\overline{A_k} = n(\left|x\right|^2 + \left|y\right|^2 + ... + \left|w\right|^2) + x\sum_{k=0}^{n-1}\left(\overline{y}\epsilon^{-k} + \overline{z}\epsilon^{-2k + ... + \overline{w}\epsilon^{-(n-1)k}}\right) + y\sum_{k=0}^{n-1}\left(\overline{x}\epsilon^{k} + \overline{z}\epsilon^{-k} + ... + \overline{w}\epsilon^{-n(n-2)k}\right) + ... + + w\sum_{k=0}^{n-1}\left(\overline{x}\epsilon^{(n-1)k + \overline{y}\epsilon^{(n-2)k}} + ... + \overline{u}\epsilon^k\right)$

But $$\sum_{k=0}^{n-1}\epsilon^lk = 0$$ if $$l$$ is not divisible by $$n$$ (see previous problem).

Therefore all the sums in the right member vanish and we get

$\left|A_0\right|^2 + \left|A_1\right|^2 + ... + \left|A_{n-1}\right|^2 = n\left(\left|x\right|^2 + \left|y\right|^2 + ... + \left|w\right|^2\right)$
48. Denoting the roots of index $$2n$$ from unity by $$x_8$$ so that

$x_s = \cos\frac{2s\pi}{n} + i \sin\frac{2s\pi}{n}~~~~~(s = 1, 2, ..., 2n)$
$\therefore x^{2n} - 1 = \prod_{s=1}^{2n}(x - x_s) = \prod_{s=1}^{n-1}(x-x_s)\prod_{s=n+1}^{2n - 1}(x - x_s).(x^2 - 1),$

$$\because x_n = -1, x_{2n} = 1$$. But $$x_{2n-s} = \overline{x_s}$$, thus,

$x^{2n} - 1 = (x^2 - 1)\prod_{s=1}^{n-1}(x - x_s)(x - \overline{x_s}) = (x^2 - 1)\prod_{s=1}^{n-1}\left(x^2 - 2x\cos\frac{s\pi}{n} + 1\right)$
49. This is solved as 248.

50. This is solved as 248 and 249.