Complex Numbers

Solve $\;$ $x^4 + 4 = 0$

Given: $\;$ $x^4 + 4 = 0$

i.e. $\;$ $\left(x^2 + 2i\right) \left(x^2 - 2i\right)= 0$

$\implies$ $\left(x^2 + 2i\right) = 0$ $\;$ OR $\;$ $\left(x^2 - 2i\right) = 0$

Now, $x^2 - 2i = 0$ $\implies$ $x = \left(2i\right)^{\frac{1}{2}}$

Let $2i = r \left(\cos \theta_1 + i \sin \theta_1\right)$

$\implies$ $r \cos \theta_1 = 0$, $\;\;$ $r \sin \theta_1 =2 $

$\therefore$ $\;$ $r = \sqrt{\left(0\right)^2 + \left(2\right)^2} = 2$

$\therefore$ $\;$ $\cos \theta_1 = 0$, $\;$ $\sin \theta_1 = 1$ $\implies$ $\theta_1 = \dfrac{\pi}{2}$

$\therefore$ $\;$ $2i = 2 \left[\cos \left(\dfrac{\pi}{2}\right) + i \sin \left(\dfrac{\pi}{2}\right)\right]$

$\begin{aligned} \therefore \; x = \left(2 i\right)^{\frac{1}{2}} & = 2^{\frac{1}{2}} \left[\cos \left(\dfrac{\pi}{2}\right) + i \sin \left(\dfrac{\pi}{2}\right)\right]^{\frac{1}{2}} \\\\ & = \sqrt{2} \left[\cos \left(2 k \pi + \dfrac{\pi}{2}\right) + i \sin \left(2 k \pi + \dfrac{\pi}{2}\right)\right]^{\frac{1}{2}} \\\\ & = \sqrt{2} \left[\cos \left(4 k + 1\right) \dfrac{\pi}{4} + i \sin \left(4 k + 1\right) \dfrac{\pi}{4}\right] \;\;\;\; k = 0,1 \\\\ & = \sqrt{2} \left(\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4}\right), \; \sqrt{2} \left(\cos \dfrac{5 \pi}{4} + i \sin \dfrac{5 \pi}{4}\right) \end{aligned}$

Now, $x^2 + 2i = 0$ $\implies$ $x = \left(-2i\right)^{\frac{1}{2}}$

Let $-2i = R \left(\cos \theta_2 + i \sin \theta_2\right)$

$\implies$ $R \cos \theta_2 = 0$, $\;\;$ $R \sin \theta_2 = - 2 $

$\therefore$ $\;$ $R = \sqrt{\left(0\right)^2 + \left(-2\right)^2} = 2$

$\therefore$ $\;$ $\cos \theta_2 = 0$, $\;$ $\sin \theta_2 = - 1$ $\implies$ $\theta_2 = 2 \pi - \dfrac{\pi}{2} = \dfrac{3 \pi}{2}$

$\begin{aligned} \therefore \; x = \left(-2 i\right)^{\frac{1}{2}} & = 2^{\frac{1}{2}} \left[\cos \left(\dfrac{3\pi}{2}\right) + i \sin \left(\dfrac{3\pi}{2}\right)\right]^{\frac{1}{2}} \\\\ & = \sqrt{2} \left[\cos \left(2 k \pi + \dfrac{3\pi}{2}\right) + i \sin \left(2 k \pi + \dfrac{3\pi}{2}\right)\right]^{\frac{1}{2}} \\\\ & = \sqrt{2} \left[\cos \left(4 k + 3\right) \dfrac{\pi}{4} + i \sin \left(4 k + 3\right) \dfrac{\pi}{4}\right] \;\;\;\; k = 0,1 \\\\ & = \sqrt{2} \left(\cos \dfrac{3\pi}{4} + i \sin \dfrac{3\pi}{4}\right), \; \sqrt{2} \left(\cos \dfrac{7 \pi}{4} + i \sin \dfrac{7 \pi}{4}\right) \end{aligned}$

$\therefore$ $\;$ $x = \sqrt{2} \; cis \left(\dfrac{\pi}{4}\right), \; \sqrt{2} \; cis \left(\dfrac{3\pi}{4}\right), \; \sqrt{2} \; cis \left(\dfrac{5\pi}{4}\right), \; \sqrt{2} \; cis \left(\dfrac{7\pi}{4}\right)$

Complex Numbers

Prove that if $\omega^3 = 1$, then $\dfrac{1}{1 + 2 \omega} - \dfrac{1}{1 + \omega} + \dfrac{1}{2 + \omega} = 0$

Given: $\;$ $\omega^3 = 1$

$\implies$ $\omega$ is the cube root of unity.

Then, $1 + \omega + \omega^2 = 0$

Now,

$\begin{aligned} \dfrac{1}{1 + 2 \omega} - \dfrac{1}{1 + \omega} & = \dfrac{1 + \omega - 1 - 2 \omega}{\left(1 + 2 \omega\right) \left(1 + \omega\right)} \\\\ & = \dfrac{- \omega}{1 + 3 \omega + 2 \omega^2} \\\\ & = \dfrac{-\omega}{\left(1 + \omega + \omega^2\right) + 2 \omega + \omega^2} \\\\ & = \dfrac{- \omega}{2 \omega + \omega^2} \\\\ & = \dfrac{-1}{2 + \omega} \end{aligned}$

$\begin{aligned} \therefore \; \dfrac{1}{1 + 2 \omega} - \dfrac{1}{1 + \omega} + \dfrac{1}{2 + \omega} & = \dfrac{-1}{2 + \omega} + \dfrac{1}{2 + \omega} \\\\ & = 0 \end{aligned}$

Hence proved.

Complex Numbers

Prove that if $\omega^3 = 1$, then $\left(\dfrac{-1 + i \sqrt{3}}{2}\right)^5 + \left(\dfrac{-1 - i \sqrt{3}}{2}\right)^5 = -1$

Given: $\;$ $\omega^3 = 1$

$\implies$ $\omega$ is the cube root of unity.

Then, $\omega = \dfrac{-1 + i \sqrt{3}}{2}$ $\;$ and $\;$ $\omega^2 = \dfrac{-1 - i \sqrt{3}}{2}$

Now,

$\begin{aligned} \left(\dfrac{-1 + i \sqrt{3}}{2}\right)^5 + \left(\dfrac{-1 - i \sqrt{3}}{2}\right)^5 & = \left(\omega\right)^5 + \left(\omega^2\right)^5 \\\\ & = \omega^3 \times \omega^2 + \left(\omega^3\right)^3 \times \omega \\\\ & = \omega^2 + \omega \;\;\; \left[\because \; \omega^3 = 1\right] \\\\ & = -1 \;\;\; \left[\because \; 1 + \omega + \omega^2 = 0\right] \end{aligned}$

Hence proved.

Complex Numbers

If $\;$ $x = a + b$, $\;$ $y = a \omega + b \omega^2$ $\;$ and $\;$ $z = a \omega^2 + b \omega$, $\;$ show that $\;$ $x^3 + y^3 + z^3 = 3 \left(a^3 + b^3\right)$ $\;$ where $\omega$ is the complex cube root of unity.

$x= a + b$

$\therefore$ $\;$ $x^3 = \left(a + b\right)^3 = a^3 + b^3 + 3 a^2 b + 3 a b^2$ $\;\;\; \cdots \; (1)$

$y = a\omega + b \omega^2$

$\begin{aligned} \therefore \; y^3 & = \left(a \omega + b \omega^2\right)^3 \\\\ & = a^3 \omega^3 + b^3 \left(\omega^2\right)^3 + 3 a^2 b \omega^4 + 3 a b^2 \omega^5 \\\\ & = a^3 \omega^3 + b^3 \left(\omega^3\right)^2 + 3 a^2 b \omega^3 \times \omega + 3 a b^2 \omega^3 \times \omega^2 \\\\ & = a^3 + b^3 + 3 a^2 b \omega + 3 ab^2 \omega^2 \;\;\; \cdots \; (2) \;\;\; [\text{Note: } \omega^3 = 1] \end{aligned}$

$z = a \omega^2 + b \omega$

$\begin{aligned} \therefore \; z^3 & = \left(a \omega^2 + b \omega\right)^3 \\\\ & = a^3 \left(\omega^2\right)^3 + b^3 \omega^3 + 3 a^2 b \omega^5 + 3 a b^2 \omega^4 \\\\ & = a^3 \left(\omega^3\right)^2 + b^3 \omega^3 + 3 a^2 b \omega^3 \times \omega^2 + 3 a b^2 \omega^3 \times \omega \\\\ & = a^3 + b^3 + 3 a^2 b \omega^2 + 3 a b^2 \omega \;\;\; \cdots \; (3) \end{aligned}$

$\therefore$ $\;$ We have from equations $(1)$, $(2)$ and $(3)$,

$\begin{aligned} x^3 + y^3 + z^3 & = a^3 + b^3 + 3 a^2 b + 3 ab^2 \\ & \hspace{1cm} a^3 + b^3 + 3 a^2 b \omega + 3 ab^2 \omega^2 \\ & \hspace{2cm} a^3 + b^3 + 3 a^2 b \omega^2 + 3 ab^2 \omega \\\\ & = 3 a^3 + 3 b^3 + 3 a^2 b \left(1 + \omega + \omega^2\right) + 3 ab^2 \left(1 + \omega^2 + \omega\right) \\\\ & = 3 \left(a^3 + b^3\right) \;\;\; [\because \; 1 + \omega + \omega^2 = 0] \end{aligned}$

Hence proved.

Complex Numbers

Find the value of $\;$ $\left(- \sqrt{3} - i\right)^{\frac{2}{3}}$

Let $\left(- \sqrt{3} - i\right) = r \left(\cos \theta + i \sin \theta\right)$

Then, $\;$ $r \cos \theta = - \sqrt{3}$; $\;\;$ $r \sin \theta = -1$

$\therefore$ $\;$ $r = \sqrt{\left(- \sqrt{3}\right)^2 + \left(-1\right)^2} = 2$

Now, $\;$ $\cos \theta = - \dfrac{\sqrt{3}}{2}$; $\;$ $\sin \theta = - \dfrac{1}{2}$ $\implies$ $\theta = -\pi + \dfrac{\pi}{6} = \dfrac{- 5 \pi}{6}$

$\begin{aligned} \therefore \; \left(- \sqrt{3} - i\right)^{\frac{2}{3}} & = 2^{\frac{2}{3}} \left[\cos \left(\dfrac{-5\pi}{6}\right) + i \sin \left(\dfrac{-5 \pi}{6}\right)\right]^{\frac{2}{3}} \\\\ & = 2 ^{\frac{2}{3}} \left\{\left[\cos \left(\dfrac{-5 \pi}{6}\right) + i \sin \left(\dfrac{-5 \pi}{6}\right)\right]^2\right\}^{\frac{1}{3}} \\\\ & = 2^{\frac{2}{3}} \left[\cos \left(\dfrac{-5 \pi}{3}\right) + i \sin \left(\dfrac{-5 \pi}{3}\right)\right]^{\frac{1}{3}} \\\\ & = 2^{\frac{2}{3}} \left[\cos \left(2 k \pi - \dfrac{5 \pi}{3}\right) + i \sin \left(2 k \pi - \dfrac{5 \pi}{3}\right)\right]^{\frac{1}{3}} \\\\ & = 2^{\frac{2}{3}} \left\{\cos \left[\left(6k - 5\right) \dfrac{\pi}{9}\right] + i \sin \left[\left(6k - 5\right) \dfrac{\pi}{9}\right] \right\} \;\;\; where \; k = 0, 1, 2 \end{aligned}$

$\therefore$ $\;$ The values of $\;$ $\left(- \sqrt{3} - i\right)^{\frac{2}{3}}$ $\;$ are

$2^{\frac{2}{3}} \left[\cos \left(\dfrac{-5 \pi}{9}\right) + i \sin \left(\dfrac{-5 \pi}{9}\right)\right]$, $\;$ $2^{\frac{2}{3}} \left[\cos \left(\dfrac{\pi}{9}\right) + i \sin \left(\dfrac{\pi}{9}\right)\right]$, $\;$ $2^{\frac{2}{3}} \left[\cos \left(\dfrac{7\pi}{9}\right) + i \sin \left(\dfrac{7\pi}{9}\right)\right]$

Complex Numbers

If $\;$ $a = \cos 2 \alpha + i \sin 2 \alpha$, $\;$ $b = \cos 2 \beta + i \sin 2 \beta$ $\;$ and $\;$ $c = \cos 2 \gamma + i \sin 2 \gamma$, $\;$ prove that $\;$ $\dfrac{a^2 b^2 + c^2}{abc} = 2 \cos 2 \left(\alpha + \beta - \gamma\right)$

Given: $\;$ $a = \cos 2 \alpha + i \sin 2 \alpha$, $\;$ $b = \cos 2 \beta + i \sin 2 \beta$, $\;$ $c = \cos 2 \gamma + i \sin 2 \gamma$

$\therefore$ $\;$ $a^2 = \left(\cos 2 \alpha + i \sin 2 \alpha\right)^2 = \cos 4 \alpha + i \sin 4 \alpha$

$b^2 = \left(\cos 2 \beta + i \sin 2 \beta\right)^2 = \cos 4 \beta + i \sin 4 \beta$

$c^2 = \left(\cos 2 \gamma + i \sin 2 \gamma\right)^2 = \cos 4 \gamma + i \sin 4 \gamma$

Now,

$\begin{aligned} a^2 b^2 & = \left(\cos 4 \alpha + i \sin 4 \alpha\right) \left(\cos 4 \beta + i \sin 4 \beta\right) \\\\ & = \left(\cos 4 \alpha \; \cos 4 \beta - \sin 4 \alpha \; \sin 4 \beta\right) + i \left(\sin 4 \alpha \; \cos 4 \beta + \cos 4 \alpha \; \sin 4 \beta\right) \\\\ & = \cos \left(4 \alpha + 4 \beta\right) + i \sin \left(4 \alpha + 4 \beta\right) \end{aligned}$

$\begin{aligned} \therefore \; a^2 b^2 + c^2 & = \left[\cos \left(4 \alpha + 4 \beta\right) + i \sin \left(4 \alpha + 4 \beta\right)\right] + \left[\cos 4 \gamma + i \sin 4 \gamma\right] \\\\ & = \left[\cos \left(4 \alpha + 4 \beta\right) + \cos 4 \gamma\right] + i \left[\sin \left(4 \alpha + 4 \beta\right) + \sin 4 \gamma\right] \\\\ & = 2 \cos \left(\dfrac{4 \alpha + 4 \beta + 4 \gamma}{2}\right) \cos \left(\dfrac{4 \alpha + 4 \beta - 4 \gamma}{2}\right) \\ & \hspace{1cm} + 2 i \sin \left(\dfrac{4 \alpha + 4 \beta + 4 \gamma}{2}\right) \cos \left(\dfrac{4 \alpha + 4 \beta - 4 \gamma}{2}\right) \\\\ & = 2 \cos \left(2 \alpha + 2 \beta - 2 \gamma\right)\left[\cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right)\right] \end{aligned}$

$\begin{aligned} abc & = \left(\cos 2 \alpha + i \sin 2 \alpha\right) \left(\cos 2 \beta + i \sin 2 \beta\right) \left(\cos 2 \gamma + i \sin 2 \gamma\right) \\\\ & = \left[\left(\cos 2 \alpha \; \cos 2 \beta - \sin 2 \alpha \; \sin 2 \beta \right) + i \left(\sin 2 \alpha \; \cos 2 \beta + \cos 2 \alpha \; \sin 2 \beta \right)\right] \\ & \hspace{9cm} \times \left(\cos 2 \gamma + i \sin 2 \gamma\right) \\\\ & = \left[\cos \left(2 \alpha + 2 \beta\right) + i \sin \left(2 \alpha + 2 \beta\right)\right] \times \left(\cos 2 \gamma + i \sin 2 \gamma\right) \\\\ & = \left[\cos \left(2 \alpha + 2 \beta\right) \; \cos 2 \gamma - \sin \left(2 \alpha + 2 \beta\right) \; \sin 2 \gamma\right] \\ & \hspace{1cm} + i \left[\sin \left(2 \alpha + 2 \beta\right) \; \cos 2 \gamma + \cos \left(2 \alpha + 2 \beta \right) \; \sin 2 \gamma\right] \\\\ & = \cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right) \end{aligned}$

$\begin{aligned} \therefore \; \dfrac{a^2 b^2 + c^2}{abc} & = \dfrac{2 \cos \left(2 \alpha + 2 \beta - 2 \gamma\right)\left[\cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right)\right]}{\cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right)} \\\\ & = 2 \cos \left(2 \alpha + 2 \beta - 2 \gamma\right) \\\\ & = 2 \cos 2 \left(\alpha + \beta - \gamma\right) \end{aligned}$

Hence proved.

Complex Numbers

If $x = \cos \alpha + i \sin \alpha$; $\;$ $y = \cos \beta + i \sin \beta$, $\;$ then prove that $\;$ $x^m y^n + \dfrac{1}{x^m y^n} = 2 \cos \left(m \alpha + n \beta\right)$

Given: $\;$ $x = \cos \alpha + i \sin \alpha$; $\;\;$ $y = \cos \beta + i \sin \beta$

$\implies$ $\dfrac{1}{x} = \cos \alpha - i \sin \alpha$; $\;$ $\dfrac{1}{y} = \cos \beta - i \sin \beta$

Now, $\;$ $x^m = \left(\cos \alpha + i \sin \alpha\right)^m = \cos \left(m \alpha\right) + i \sin \left(m \alpha \right)$

$y^n = \left(\cos \beta + i \sin \beta\right)^n = \cos \left(n \beta\right) + i \sin \left(n \beta\right)$

$\begin{aligned} \therefore \; x^m y^n & = \left[\cos \left(m \alpha\right) + i \sin \left(m \alpha\right)\right] \left[\cos \left(n \beta\right) + i \sin \left(n \beta\right)\right] \\\\ & = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) + i^2 \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\ & \hspace{2cm} + i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\ & = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) - \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\ & \hspace{2cm} + i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\ & = \cos \left(m \alpha + n \beta\right) + i \sin \left(m \alpha + n \beta\right) \end{aligned}$

$\dfrac{1}{x^m} = \left(\cos \alpha - i \sin \alpha\right)^m = \cos \left(m \alpha\right) - i \sin \left(m \alpha\right)$

$\dfrac{1}{y^n} = \left(\cos \beta - i \sin \beta\right)^n = \cos \left(n \beta\right) - i \sin \left(n \beta\right)$

$\begin{aligned} \therefore \; \dfrac{1}{x^m y^n} & = \left[\cos \left(m \alpha\right) - i \sin \left(m \alpha\right)\right] \left[\cos \left(n \beta\right) - i \sin \left(n \beta\right)\right] \\\\ & = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) + i^2 \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\ & \hspace{2cm} - i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\ & = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) - \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\ & \hspace{2cm} - i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\ & = \cos \left(m \alpha + n \beta\right) - i \sin \left(m \alpha + n \beta\right) \end{aligned}$

$\begin{aligned} \therefore \; x^m y^n + \dfrac{1}{x^m y^n} & = \cos \left(m \alpha + n \beta\right) + i \sin \left(m \alpha + n \beta\right) \\ & \hspace{2cm} + \cos \left(m \alpha + n \beta\right) - i \sin \left(m \alpha + n \beta\right) \\\\ & = 2 \cos \left(m \alpha + n \beta\right) \end{aligned}$

Hence proved.

Complex Numbers

If $\;$ $x + \dfrac{1}{x} = 2 \cos \theta$, $\;$ prove that $\;$ $x^n + \dfrac{1}{x^n} = 2 \cos \left(n \theta\right)$ $\;$ and $\;$ $x^n - \dfrac{1}{x^n} = 2 i \sin \left(n \theta\right)$

Let $x = \cos \theta + i \sin \theta$

$\begin{aligned} Then, \; \dfrac{1}{x} & = \dfrac{1}{\cos \theta + i \sin \theta} \\\\ & = \dfrac{\cos \theta - i \sin \theta}{\left(\cos \theta + i \sin \theta\right) \left(\cos \theta - i \sin \theta\right)} \\\\ & = \dfrac{\cos \theta - i \sin \theta}{\cos^2 \theta - i^2 \sin^2 \theta} \\\\ & = \cos \theta - i \sin \theta \end{aligned}$

so that $\;$ $x + \dfrac{1}{x} = 2 \cos \theta$

Now,

$x^n = \left(\cos \theta + i \sin \theta\right)^n = \cos \left(n \theta\right) + i \sin \left(n \theta\right)$

$\dfrac{1}{x^n} = \left(\cos \theta - i \sin \theta\right)^n = \cos \left(n \theta\right) - i \sin \left(n \theta\right)$

$\therefore \; x^n + \dfrac{1}{x^n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right) + \cos \left(n \theta\right) - i \sin \left(n \theta\right) = 2 \cos \left(n \theta\right)$

$x^n - \dfrac{1}{x^n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right) - \cos \left(n \theta\right) + i \sin \left(n \theta\right) = 2 i \sin \left(n \theta\right)$

Hence proved.

Complex Numbers

If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2px + \left(p^2 + q^2\right) = 0$ and $\tan \theta = \dfrac{q}{y + p}$, show that $\dfrac{\left(y + \alpha\right)^n - \left(y + \beta\right)^n}{\alpha - \beta} = q^{n - 1} \left(\dfrac{\sin n\theta}{\sin^n \theta}\right)$

The roots of the given quadratic equation $\;$ $x^2 - 2px + \left(p^2 + q^2\right) = 0$ $\;$ are

$\begin{aligned} x & = \dfrac{2p \pm \sqrt{4p^2 - 4p^2 - 4q^2}}{2} \\\\ & = \dfrac{2p \pm 2iq}{2} \\\\ & = p \pm iq \end{aligned}$

$\because$ $\;$ $\alpha$ and $\beta$ are the roots of the given quadratic equation, let

$\alpha = p + i q$ $\;$ and $\;$ $\beta = p - iq$

Given: $\;$ $\tan \theta = \dfrac{q}{y + p}$

$\implies$ $y = \dfrac{q}{\tan \theta} - p$

$\begin{aligned} \therefore \; \left(y + \alpha\right)^n & = \left(\dfrac{q}{\tan \theta} - p + p + iq\right)^n \\\\ & = \left[q \left(\dfrac{1}{\tan \theta} + i\right)\right]^n \\\\ & = q^n \left[\dfrac{\cos \theta + i \sin \theta}{\sin \theta}\right]^n \\\\ & = \dfrac{q^n \left[\cos \left(n\theta\right) + i \sin \left(n \theta\right)\right]}{\sin^n \theta} \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} \left(y + \beta\right)^n & = \left(\dfrac{q}{\tan \theta} - p + p - iq\right)^n \\\\ & = \left[q \left(\dfrac{1}{\tan \theta} - i\right)\right]^n \\\\ & = q^n \left[\dfrac{\cos \theta - i \sin \theta}{\sin \theta}\right]^n \\\\ & = \dfrac{q^n \left[\cos \left(n\theta\right) - i \sin \left(n \theta\right)\right]}{\sin^n \theta} \;\;\; \cdots \; (2) \end{aligned}$

$\alpha - \beta = p + iq - \left(p - iq\right) = 2 \; i \; q$ $\;\;\; \cdots \; (3)$

$\therefore$ $\;$ We have from equations $(1)$, $(2)$ and $(3)$,

$\begin{aligned} \dfrac{\left(y + \alpha\right)^n - \left(y + \beta\right)^n}{\alpha - \beta} & = \dfrac{\dfrac{q^n \left[\cos \left(n\theta\right) + i \sin \left(n \theta\right)\right]}{\sin^n \theta} - \dfrac{q^n \left[\cos \left(n\theta\right) - i \sin \left(n \theta\right)\right]}{\sin^n \theta}}{2 \;i \;q} \\\\ & = \dfrac{q^n \left[\cos \left(n \theta\right) + i \sin \left(n \theta\right) - \cos \left(n \theta\right) + i \sin \left(n \theta\right)\right]}{2 \;i \;q \; \sin^n \theta} \\\\ & = \dfrac{2 \; i \; q^n \; \sin \left(n \theta\right)}{2 \; i \; q \; \sin^n \theta} \\\\ & = q^{n - 1} \left[\dfrac{\sin \left(n \theta\right)}{\sin^n \theta}\right] \end{aligned}$

Hence proved.

Complex Numbers

Prove that $\left(1 + \cos \theta + i \sin \theta\right)^n + \left(1 + \cos \theta - i \sin \theta\right)^n = 2^{n + 1} \cos^{n} \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right)$, $\;$ $n \in N$

Let $\;$ $z = \cos \theta + i \sin \theta$

$\because$ $\;$ $\left|z\right| = 1$ $\implies$ $\overline{z} = \dfrac{1}{z}$

$\therefore$ $\;$ From equation $(1a)$, $\;$ $\overline{z} = \dfrac{1}{z} = \cos \theta - i \sin \theta$ $\;\;\; \cdots \; (1)$

$\begin{aligned} \therefore \; \left(1 + \cos \theta + i \sin \theta\right)^n + \left(1 + \cos \theta - i \sin \theta\right)^n & = \left(1 + z\right)^n + \left(1 + \dfrac{1}{z}\right)^n \\\\ & = \left(1 + z\right)^n + \dfrac{\left(1 + z\right)^n}{z^n} \\\\ & = \left(1 + z\right)^n \left(1 + \dfrac{1}{z^n}\right) \;\;\; \cdots \; (2) \end{aligned}$

From equation $(1)$,

$\begin{aligned} \dfrac{1}{z^n} & = \left(\cos \theta - i \sin \theta\right)^n \\\\ & = \cos \left(n \theta\right) - i \sin \left(n \theta\right) \;\; [\text{By De Moivre's theorem}] \end{aligned}$

$\begin{aligned} \therefore \; 1 + \dfrac{1}{z^n} & = 1 + \cos \left(n \theta\right) - i \sin \left(n \theta\right) \\\\ & = 2 \cos^2 \left(\dfrac{n \theta}{2}\right) - 2 i \sin \left(\dfrac{n \theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \\\\ & = 2 \cos \left(\dfrac{n \theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) - i \sin \left(\dfrac{n \theta}{2}\right)\right] \;\;\; \cdots \; (3a) \end{aligned}$

$\begin{aligned} Now, \; \left(1 + z\right) & = \left(1 + \cos \theta\right) + i \sin \theta \\\\ & = 2 \cos^2 \left(\dfrac{\theta}{2}\right) + 2 i \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right) \\\\ & = 2 \cos \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{\theta}{2}\right) + i \sin \left(\dfrac{\theta}{2}\right)\right] \end{aligned}$

$\begin{aligned} \therefore \; \left(1 + z\right)^n & = \left\{2 \cos \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{\theta}{2}\right) + i \sin \left(\dfrac{\theta}{2}\right)\right]\right\}^n \\\\ & = 2^n \cos^n \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) + i \sin \left(\dfrac{n \theta}{2}\right)\right] \;\;\; \cdots \; (3b) \end{aligned}$

$\therefore$ $\;$ We have from equations $(3a)$ and $(3b)$,

$\begin{aligned} \left(1 + z\right)^n \left(1 + \dfrac{1}{z^n}\right) & = 2^n \cos^n \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) + i \sin \left(\dfrac{n \theta}{2}\right)\right] \\ & \hspace{2.5cm} \times 2 \cos \left(\dfrac{n \theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) - i \sin \left(\dfrac{n \theta}{2}\right)\right] \\\\ & = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \\ & \hspace{1cm} \times \left[\cos \left(\dfrac{n \theta}{2}\right) + i \sin \left(\dfrac{n \theta}{2}\right)\right]\left[\cos \left(\dfrac{n \theta}{2}\right) - i \sin \left(\dfrac{n \theta}{2}\right)\right] \\\\ & = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \times \left[\cos^2 \left(\dfrac{n \theta}{2}\right) - i^2 \sin^2 \left(\dfrac{n \theta}{2}\right)\right] \\\\ & = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \times \left[\cos^2 \left(\dfrac{n \theta}{2}\right) + \sin^2 \left(\dfrac{n \theta}{2}\right)\right] \\\\ & = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \;\;\; \cdots \; (4) \end{aligned}$

$\therefore$ $\;$ From equations $(2)$ and $(4)$ we have,

$\left(1 + \cos \theta + i \sin \theta\right)^n + \left(1 + \cos \theta - i \sin \theta\right)^n = 2^{n + 1} \cos^{n} \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right)$, $\;$ $n \in N$

Hence proved.

Complex Numbers

Prove that $\left(1 + i\right)^n + \left(1 - i\right)^n = 2 ^{\frac{n + 2}{2}} \cos \left(\dfrac{n \pi}{4}\right)$, $\;$ $n \in N$

Let $\;$ $\left(1 + i\right) = r \left(\cos \theta + i \sin \theta\right)$ $\;\;\; \cdots \; (1)$

Equating the real and imaginary parts separately, we have

$r \; \cos \theta = 1$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $r \; \sin \theta = 1$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ $r = \sqrt{\left(1\right)^2 + \left(1\right)^2} = \sqrt{2}$

Substituting the value of $r$ in equations $(2a)$ and $(2b)$, we have

$\cos \theta = \dfrac{1}{\sqrt{2}}$, $\;\;$ $\sin \theta = \dfrac{1}{\sqrt{2}}$ $\implies$ $\theta = \dfrac{\pi}{4}$

Substituting the values of $r$ and $\theta$ in equation $(1)$ we have,

$\left(1 + i\right) = \sqrt{2} \left[\cos \left(\dfrac{\pi}{4}\right) + i \sin \left(\dfrac{\pi}{4}\right)\right]$

$\begin{aligned} \therefore \; \left(1 + i\right)^n & = \left\{\sqrt{2} \left[\cos \left(\dfrac{\pi}{4}\right) + i \sin \left(\dfrac{\pi}{4}\right)\right]\right\}^{n} \\\\ & = \left(\sqrt{2}\right)^n \left[\cos \left(\dfrac{\pi}{4}\right) + i \sin \left(\dfrac{\pi}{4}\right)\right]^n \\\\ & = 2^{\frac{n}{2}} \left[\cos \left(\dfrac{n \pi}{4}\right) + i \sin \left(\dfrac{n \pi}{4}\right)\right] \;\;\; \cdots \; (3a) \end{aligned}$

Replacing $+i$ with $-i$ in equation $(3a)$, we have

$\left(1 - i\right)^n = 2^{\frac{n}{2}} \left[\cos \left(\dfrac{n \pi}{4}\right) - i \sin \left(\dfrac{n \pi}{4}\right)\right]$ $\;\;\; \cdots \; (3b)$

Adding equations $(3a)$ and $(3b)$, we have

$\left(1 + i\right)^n + \left(1 - i\right)^n = 2^{\frac{n}{2}} \left[2 \cos \left(\dfrac{n \pi}{4}\right)\right]$

i.e. $\;$ $\left(1 + i\right)^n + \left(1 - i\right)^n = 2 ^{\frac{n + 2}{2}} \cos \left(\dfrac{n \pi}{4}\right)$

Hence proved.

Complex Numbers

If $\cos \alpha + \cos \beta + \cos \gamma = 0 = \sin \alpha + \sin \beta + \sin \gamma$, prove that

1. $\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma = 3 \cos \left(\alpha + \beta + \gamma\right)$


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

Given $\;\;$ $\cos \alpha + \cos \beta + \cos \gamma = 0 = \sin \alpha + \sin \beta + \sin \gamma$ $\;\;\; \cdots \; (1)$

Let

$a = \cos \alpha + i \sin \alpha = e^{i \alpha}$ $\;\;\; \cdots \; (2a)$

$b = \cos \beta + i \sin \beta = e^{i \beta}$ $\;\;\; \cdots \; (2b)$

$c = \cos \gamma + i \sin \gamma = e^{i \gamma}$ $\;\;\; \cdots \; (2c)$

$\therefore \; a + b + c = \left(\cos \alpha + \cos \beta + \cos \gamma\right) + i \left(\sin \alpha + \sin \beta + \sin \gamma\right)$

i.e. $\;$ $a + b + c = 0$ $\;\;$ [from equation $(1)$] $\;\;\; \cdots \; (3)$

$\therefore$ $\;$ We have from equation $(3)$, $\;$ $a^3 + b^3 + c^3 = 3 abc$ $\;\;\; \cdots \; (4)$

[Note:

$a + b + c = 0$ $\implies$ $a + b = -c$ $\;\;\; \cdots \; (A)$

$\therefore$ $\;$ $\left(a + b\right)^3 = \left(-c\right)^3$

i.e. $\;$ $a^3 + b^3 + 3ab \left(a + b\right) = - c^3$

i.e. $\;$ $a^3 + b^3 + c^3 = - 3 ab \left(a + b\right)$

i.e. $\;$ $a^3 + b^3 + c^3 = 3abc$ $\;$ (from equation $A$) ]

From equation $(2a)$,

$a^3 = \left(\cos \alpha + i \sin \alpha\right)^3 = \cos 3 \alpha + i \sin 3 \alpha$ $\;\;\; \cdots \; (5a)$

[By De Moivre's theorm: $\;$ $\left(\cos \theta + i \sin \theta \right)^n = \cos n \theta + i \sin n \theta$]

From equation $(2b)$,

$b^3 = \left(\cos \beta + i \sin \beta\right)^3 = \cos 3 \beta + i \sin 3 \beta$ $\;\;\; \cdots \; (5b)$

From equation $(2c)$,

$c^3 = \left(\cos \gamma + i \sin \gamma\right)^3 = \cos 3 \gamma + i \sin 3 \gamma$ $\;\;\; \cdots \; (5c)$

Adding equations $(5a)$, $(5b)$ and $(5c)$ we get,

$a^3 + b^3 + c^3 = \left(\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma\right)$

$\hspace{3cm}$ $+ i \left(\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma\right)$ $\;\;\; \cdots \; (6a)$

We have from equations $(2a)$, $(2b)$ and $(2c)$,

$\begin{aligned} 3abc & = 3 e^{i \alpha} \cdot e^{i \beta} \cdot e^{i \gamma} \\\\ & = 3 e^{i \left(\alpha + \beta + \gamma\right)} \\\\ & = 3 \left[\cos \left(\alpha + \beta + \gamma\right) + i \sin \left(\alpha + \beta + \gamma\right)\right] \\\\ & = 3 \cos \left(\alpha + \beta + \gamma\right) + 3 i \sin \left(\alpha + \beta + \gamma\right) \;\;\; \cdots \; (6b) \end{aligned}$

$\therefore$ $\;$ In view of equations $(6a)$ and $(6b)$, equation $(4)$ becomes

$\left(\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma\right) + i \left(\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma\right)$

$\hspace{4cm}$ $= 3 \cos \left(\alpha + \beta + \gamma\right) + 3 i \sin \left(\alpha + \beta + \gamma\right)$ $\;\;\; \cdots \; (7)$

1. Equating the real parts on either side of equation $(7)$ we have,
   
   $\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma = 3 \cos \left(\alpha + \beta + \gamma\right)$



2. Equating the imaginary parts on either side of equation $(7)$ we have,
   
   $\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma = 3 \sin \left(\alpha + \beta + \gamma\right)$

Complex Numbers

Simplify: $\;\;\;$ $\dfrac{\left(\cos 2 \theta - i \sin 2 \theta\right)^7 \left(\cos 3 \theta + i \sin 3 \theta\right)^{-5}}{\left(\cos 4 \theta + i \sin 4 \theta\right)^{12} \left(\cos 5 \theta - i \sin 5 \theta\right)^{-6}}$

$\begin{aligned} \dfrac{\left(\cos 2 \theta - i \sin 2 \theta\right)^7 \left(\cos 3 \theta + i \sin 3 \theta\right)^{-5}}{\left(\cos 4 \theta + i \sin 4 \theta\right)^{12} \left(\cos 5 \theta - i \sin 5 \theta\right)^{-6}} & = \dfrac{\left(e^{-i2 \theta}\right)^7 \cdot \left(e^{i 3 \theta}\right)^{-5}}{\left(e^{i 4 \theta}\right)^{12} \cdot \left(e^{-i 5 \theta}\right)^{-6}} \\\\ & = \dfrac{\left(e^{- i 14 \theta}\right) \cdot \left(e^{- i 15 \theta}\right)}{\left(e^{i 48 \theta}\right) \cdot \left(e^{i 30 \theta}\right)} \\\\ & = \dfrac{e^{- i 29 \theta}}{e^{i 78 \theta}} \\\\ & = e^{- i 107 \theta} \\\\ & = \cos \left(107 \theta\right) - i \sin \left(107 \theta\right) \end{aligned}$

Complex Numbers

Solve the equation $\;$ $x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = 0$ $\;$ if $\;$ $3 + i$ $\;$ is a root.

Given: $\;\;$ One root is $\left(3 + i\right)$

$\implies$ $\left(3 - i\right)$ is also a root.

Sum of roots $= 6$

Product of roots $= \left(3 + i\right) \left(3 - i\right) = 9 - i^2 = 10$

$\therefore$ $\;$ The corresponding factor is $\;$ $x^2 - 6x + 10$

$\therefore$ $\;$ $x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = \left(x^2 - 6x + 10\right) \left(x^2 + \lambda x + 2\right)$ $\;\;\; \cdots \; (1)$

$\lambda$ $\;$ is a real number (constant).

Equating the coefficients of the $x$ term in equation $(1)$ we have,

$- 32 = -12 + 10 \lambda$

i.e. $\;$ $10 \lambda = -20$ $\implies$ $\lambda = -2$

Substituting the value of $\lambda$ in equation $(1)$ we have,

$x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = \left(x^2 - 6x + 10\right) \left(x^2 - 2x + 2\right)$

$\therefore$ $\;$ $x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = 0$ $\implies$ $x^2 - 6x + 10 = 0$ $\;$ or $\;$ $x^2 - 2x + 2 = 0$

Now,

$\begin{aligned} x^2 - 2x + 2 = 0 \implies x & = \dfrac{2 \pm \sqrt{4 - 8}}{2} \\\\ & = \dfrac{2 \pm 2i}{2} \\\\ & = 1 \pm i \end{aligned}$

$\therefore$ $\;$ The roots are $\left(3 \pm i\right)$ and $\left(1 \pm i\right)$.

Complex Numbers

$P$ represents the variable complex number $z$. Find the locus of $P$, if $\;$ $arg \left(\dfrac{z - 1}{z + 3}\right) = \dfrac{\pi}{4}$

Let $z = x + iy$

Given: $\;\;\;$ $arg \left(\dfrac{z - 1}{z + 3}\right) = \dfrac{\pi}{4}$

i.e. $\;$ $arg \left(z - 1\right) - arg \left(z + 3\right) = \dfrac{\pi}{4}$

i.e. $\;$ $arg \left[\left(x - 1\right) + i y\right] - arg \left[\left(x + 3\right) + i y\right] = \dfrac{\pi}{4}$

i.e. $\;$ $\tan^{-1} \left(\dfrac{y}{x - 1}\right) - \tan^{-1} \left(\dfrac{y}{x + 3}\right) = \dfrac{\pi}{4}$

i.e. $\;$ $\tan^{-1} \left[\dfrac{\dfrac{y}{x - 1} - \dfrac{y}{x + 3}}{1 + \left(\dfrac{y}{x - 1}\right) \left(\dfrac{y}{x + 3}\right)} \right] = \dfrac{\pi}{4}$

i.e. $\;$ $\tan^{-1} \left[\dfrac{4y}{x^2 + 2x - 3 + y^2}\right] = \dfrac{\pi}{4}$

i.e. $\;$ $\dfrac{4y}{x^2 + y^2 + 2x - 3} = \tan \left(\dfrac{\pi}{4}\right) = 1$

i.e. $\;$ $x^2 + y^2 + 2x - 4y - 3 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the required equation of locus of $P$.

Complex Numbers

$P$ represents the variable complex number $z$. Find the locus of $P$, if $\;$ $\left|z - 5i\right| = \left|z + 5i\right|$

Let $z = x + iy$

Then, $z - 5i = x + i \left(y - 5\right)$; $\;\;\;$ $z + 5i = x + i \left(y + 5\right)$

$\left|z - 5i\right| = \sqrt{\left(x^2\right) + \left(y - 5\right)^2} = \sqrt{x^2 + y^2 - 10 y + 25}$

$\left|z - 5i\right| = \sqrt{\left(x^2\right) + \left(y + 5\right)^2} = \sqrt{x^2 + y^2 + 10 y + 25}$

$\therefore$ $\;$ $\left|z - 5i\right| = \left|z + 5i\right|$ $\implies$ $\sqrt{x^2 + y^2 - 10 y + 25} = \sqrt{x^2 + y^2 + 10 y + 25}$

i.e. $\;$ $x^2 + y^2 - 10 y + 25 = x^2 + y^2 + 10 y + 25$

i.e. $\;$ $y = 0$

$\therefore$ $\;$ The locus of point $P$ is $y = 0$ $\;$ i.e. $\;$ the $X$ axis.

Complex Numbers

$P$ represents the variable complex number $z$.Find the locus of $P$, if $\;$ $Im \left[\dfrac{2z + 1}{iz + 1}\right] = -2$

Let $P$ be the point $z = x + iy$

Then,

$\begin{aligned} \dfrac{2z + 1}{iz + 1} & = \dfrac{\left(2x + 1\right) + i \; 2y}{\left(1 - y\right) + i \; x} \\\\ & = \dfrac{\left[\left(2x + 1\right) + i \; 2y\right] \left[\left(1 - y\right) - i \;x\right]}{\left(1 -y\right)^2 - \left(i \; x\right)^2} \\\\ & = \dfrac{\left(2x + 1\right) \left(1 - y\right) - i \; x \left(2x + 1\right) + i \; 2y \left(1 - y\right) - i^2 \; 2xy}{1 -2y + y^2 - i^2 \; x^2} \\\\ & = \dfrac{\left(2x + 1\right) \left(1 - y\right) + 2xy}{x^2 + y^2 - 2y + 1} + i \;\dfrac{2y \left(1 - y\right) - x \left(2x + 1\right)}{x^2 + y^2 - 2y + 1} \end{aligned}$

$\therefore$ $\;$ $Im \left[\dfrac{2z + 1}{iz + 1}\right] = \dfrac{2y - 2 y^2 -2 x^2 - x}{x^2 + y^2 - 2y + 1}$

Given $\;\;$ $\dfrac{2y - 2 y^2 -2 x^2 - x}{x^2 + y^2 - 2y + 1} = -2$

i.e. $\;$ $2 y - 2 y^2 - 2 x^2 - x = - 2 x^2 - 2 y^2 + 4 y - 2$

i.e. $\;$ $x + 2y - 2 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the required equation of locus of $P$.

Complex Numbers

If $\;$ $arg \left(z -1\right) = \dfrac{\pi}{6}$ $\;$ and $\;$ $arg \left(z + 1\right) = \dfrac{2 \pi}{3}$, $\;$ then prove that $\;$ $\left|z\right| = 1$

Let $z = x + iy$

Then, $z -1 = \left(x - 1\right) + iy$ $\;$ and $\;$ $z + 1 = \left(x + 1\right) + iy$

Now, $\;$ $arg \left(z - 1\right) = \tan^{-1} \left(\dfrac{y}{x - 1}\right)$ $\;$ and $\;$ $arg \left(z + 1\right) = \tan^{-1} \left(\dfrac{y}{x + 1}\right)$

Given $\;\;$ $arg \left(z -1\right) = \dfrac{\pi}{6}$ $\;$ and $\;$ $arg \left(z + 1\right) = \dfrac{2 \pi}{3}$

$\implies$ $\tan^{-1} \left(\dfrac{y}{x - 1}\right) = \dfrac{\pi}{6}$; $\;$ $\tan^{-1} \left(\dfrac{y}{x + 1}\right) = \dfrac{2\pi}{3}$

$\implies$ $\dfrac{y}{x - 1} = \tan \left(\dfrac{\pi}{6}\right) = \dfrac{1}{\sqrt{3}}$; $\;$ $\dfrac{y}{x + 1} = \tan \left(\dfrac{2 \pi}{3}\right) = - \sqrt{3}$

i.e. $\;$ $x - \sqrt{3} y = 1$ $\;\;\; \cdots \; (1)$; $\;\;$ $\sqrt{3}x + y = -\sqrt{3}$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously we have,

$4x = -2$ $\implies$ $x = - \dfrac{1}{2}$

Substituting the value of $x$ in equation $(2)$ we have,

$y = - \sqrt{3} + \dfrac{\sqrt{3}}{2} = - \dfrac{\sqrt{3}}{2}$

$\therefore$ $\;$ $z = x + iy = \dfrac{-1}{2} - \dfrac{\sqrt{3}}{2}y$

$\therefore$ $\;$ $\left|z\right| = \sqrt{\left(\dfrac{-1}{2}\right)^2 + \left(\dfrac{-\sqrt{3}}{2}\right)^2}$

i.e. $\;$ $\left|z\right| = \sqrt{\dfrac{1}{4} + \dfrac{3}{4}} = 1$

Complex Numbers

Express the complex number $-1 + i \sqrt{3}$ in its polar form.

The given complex number is $\;$ $z = -1 + i \sqrt{3}$

$z$ $\;$ is of the form $\;$ $x + iy$ $\;$ where $\;$ $x = -1$, $\;$ $y = \sqrt{3}$

Modulus of $z = \left|z\right| = \sqrt{x^2 + y^2} = \sqrt{\left(-1\right)^2 + \left(\sqrt{3}\right)^2} = 2$

$\because$ $\;$ $x$ is negative and $y$ is positive, argument or amplitude $\theta$ of $z$ lies in the second quadrant in the complex plane.

Let $\;$ $\alpha = \tan^{-1} \left(\dfrac{\left|y\right|}{\left|x\right|}\right) = \tan^{-1} \left(\dfrac{\left|\sqrt{3}\right|}{\left|-1\right|}\right) = \tan^{-1} \left(\sqrt{3}\right) = \dfrac{\pi}{3}$

$\therefore$ $\;$ $\theta = \pi - \alpha = \pi - \dfrac{\pi}{3} = \dfrac{2 \pi}{3}$

$\therefore$ $\;$ Polar form of $z = r \left(\cos \theta + i \sin \theta\right) = r \; cis \; \theta = 2 \; cis \; \left(\dfrac{2\pi}{3}\right)$

Complex Numbers

Prove that the triangle formed by the points representing the complex numbers $\left(10 + 8i\right)$, $\left(-2 + 4i\right)$ and $\left(-11 + 31i\right)$ on the Argand plane is right angled.

Let $A$, $B$ and $C$ represent the complex numbers $\left(10 + 8i\right)$, $\left(-2 + 4i\right)$ and $\left(-11 + 31i\right)$ respectively on the Argand diagram.

Now,

$\begin{aligned} AB & = \left|\left(10 + 8i\right) - \left(-2 + 4i\right)\right| \\\\ & = \left|12 + 4i\right| \\\\ & = \sqrt{\left(12\right)^2 + \left(4\right)^2} = \sqrt{160} \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} BC & = \left|\left(-2 + 4i\right) - \left(-11 + 31i\right)\right| \\\\ & = \left|9 - 27i\right| \\\\ & = \sqrt{\left(9\right)^2 + \left(-27\right)^2} = \sqrt{810} \;\;\; \cdots \; (2) \end{aligned}$

$\begin{aligned} CA & = \left|\left(-11 + 31i\right) - \left(10 + 8i\right)\right| \\\\ & = \left|-21 + 23i\right| \\\\ & = \sqrt{\left(-21\right)^2 + \left(23\right)^2} = \sqrt{970} \;\;\; \cdots \; (3) \end{aligned}$

$\therefore$ $\;$ We have from equations $(1)$, $(2)$ and $(3)$

$CA^2 = AB^2 + BC^2$

$\implies$ $\angle ABC = 90^{\circ}$

i.e. $\;$ $\triangle ABC$ is a right angled triangle.