Question 1: Find the modulus and argument of the following complex  numbers  and hence express each one of them in polar form:

i) $1+i$           ii) $\sqrt{3}+i$             iii) $1-i$            iv) $\frac{1-i}{1+i}$         v) $\frac{1}{1+i}$

vi) $\frac{1+2i}{1-3i}$          vii) $\sin 120^{\circ} - i \cos 120^{\circ}$           viii) $\frac{-16}{1+i \sqrt{3}}$

Answer:

i)       $z = 1+i$

$r = |z| = \sqrt{(1)^2+(1)^2} = \sqrt{2}$

Let $\tan \alpha = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\Rightarrow \tan \alpha = \Big|$ $\frac{1}{1}$ $\Big| \Rightarrow \alpha =$ $\frac{\pi}{4}$

Since the point $(1, 1)$ lies in the first quadrant, the $\text{arg}(z)$ is given by

$\theta = \alpha =$ $\frac{\pi}{4}$

Polar form of $1 + i$ is given by $= r ( \cos \theta + i \sin \theta)$

$= \sqrt{2} \big( \cos$ $\frac{\pi}{4}$ $+ i \sin$ $\frac{\pi}{4}$ $\big)$

ii)      $z = \sqrt{3}+i$

$r = |z| = \sqrt{(\sqrt{3})^2+(1)^2} = 2$

Let $\tan \alpha = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\Rightarrow \tan \alpha =$ $\Big|$ $\frac{1}{\sqrt{3}}$ $\Big|$ $\Rightarrow \alpha =$ $\frac{\pi}{6}$

Since the point $(\sqrt{3}, 1)$ lies in the first quadrant, the $\text{arg}(z)$ is given by

$\theta = \alpha =$ $\frac{\pi}{6}$

Polar form of $\sqrt{3}+i$ is given by $= r ( \cos \theta + i \sin \theta)$

$= 2 \big( \cos$ $\frac{\pi}{6}$ $+ i \sin$ $\frac{\pi}{6}$ $\big)$

iii)     $z = 1-i$

$r = |z| = \sqrt{(1)^2+(-1)^2} = \sqrt{2}$

Let $\tan \alpha = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\Rightarrow \tan \alpha =$ $\Big|$ $\frac{-1}{1}$ $\Big|$ $\Rightarrow \alpha =$ $\frac{\pi}{4}$

Since the point $(1, -1)$ lies in the fourth quadrant, the $\text{arg}(z)$ is given by

$\theta = -\alpha = -$ $\frac{\pi}{4}$

Polar form of $1 - i$ is given by $= r ( \cos \theta + i \sin \theta)$

$= \sqrt{2} \big( \cos$ $\frac{-\pi}{4}$ $+ i \sin$ $\frac{-\pi}{4}$ $\big)$

$= \sqrt{2} \big( \cos$ $\frac{\pi}{4}$ $- i \sin$ $\frac{\pi}{4}$ $\big)$

iv)     $z =$ $\frac{1-i}{1+i}$ $=$ $\frac{1-i}{1+i}$ $\times$ $\frac{1-i}{1-i}$ $=$ $\frac{-2i}{2}$ $=-i$

$z = -i$

$r = |z| = \sqrt{(0)^2+(1)^2} = 1$

Let $\tan \alpha = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\Rightarrow \tan \alpha =$ $\Big|$ $\frac{-1}{0}$ $\Big|$ $\Rightarrow \alpha =$ $\frac{-\pi}{2}$

Since the point $(0, -1)$ lies on the negative direction of the imaginary axis, the $\text{arg}(z)$ is given by

$\theta = \alpha =$ $\frac{3\pi}{2}$

Polar form of $\frac{1-i}{1+i}$ is given by $= r ( \cos \theta + i \sin \theta)$

$= 1 \big( \cos$ $\frac{3\pi}{2}$ $+ i \sin$ $\frac{3\pi}{2}$ $\big)$

$= \big( \cos$ $\frac{\pi}{2}$ $- i \sin$ $\frac{\pi}{2}$ $\big)$

v)      $z =$ $\frac{1}{1+i}$ $=$ $\frac{1}{1+i}$ $\times$ $\frac{1-i}{1-i}$ $=$ $\frac{1-i}{2}$ $=$ $\frac{1}{2}$ $- i$ $\frac{1}{2}$

$z =$ $\frac{1}{2}$ $- i$ $\frac{1}{2}$

$r = |z| =$ $\sqrt{(\frac{1}{2})^2+(\frac{-1}{2})^2}$ $=$ $\sqrt{\frac{1}{2}}$

Let $\tan \alpha = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\Rightarrow \tan \alpha =$ $\Big|$ $\frac{\frac{1}{2}}{-\frac{1}{2}}$ $\Big| = 1$ $\Rightarrow \alpha =$ $\frac{\pi}{4}$

Since the point $(\frac{1}{2}, -\frac{1}{2})$ lies in the fourth quadrant, the $\text{arg}(z)$ is given by

$\theta = -\alpha = -$ $\frac{\pi}{2}$

Polar form of $\frac{1}{1+i}$ is given by $= r ( \cos \theta + i \sin \theta)$

$= \frac{1}{\sqrt{2}} \big( \cos$ $\frac{-\pi}{4}$ $+ i \sin$ $\frac{-\pi}{4}$ $\big)$

$= \big( \cos$ $\frac{\pi}{4}$ $- i \sin$ $\frac{\pi}{4}$ $\big)$

vi)     $z =$ $\frac{1+2i}{1-3i}$ $=$ $\frac{1+2i}{1-3i}$ $\times$ $\frac{1+3i}{1+3i}$ $=$ $\frac{-5+5i}{10}$ $= -$ $\frac{1}{2}$ $+ i$ $\frac{1}{2}$

$z = -$ $\frac{1}{2}$ $+ i$ $\frac{1}{2}$

$r = |z| =$ $\sqrt{(\frac{-1}{2})^2+(\frac{1}{2})^2}$ $=$ $\sqrt{\frac{1}{2}}$

Let $\tan \alpha = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\Rightarrow \tan \alpha =$ $\Big|$ $\frac{\frac{1}{2}}{-\frac{1}{2}}$ $\Big| = 1$ $\Rightarrow \alpha =$ $\frac{\pi}{4}$

Since the point $\big(-$ $\frac{1}{2}$ $,$ $\frac{1}{2}$ $\big)$ lies in the second quadrant, the $\text{arg}(z)$ is given by

$\theta = \pi -\alpha = \pi - \frac{\pi}{4} =$ $\frac{3\pi}{4}$

Polar form of $\frac{1+2i}{1-3i}$ is given by $= r ( \cos \theta + i \sin \theta)$

$= \frac{1}{\sqrt{2}} \big( \cos$ $\frac{3\pi}{4}$ $+ i \sin$ $\frac{3\pi}{4}$ $\big)$

vii)    $\sin 120^{\circ} - i \cos 120^{\circ} =$ $\frac{\sqrt{3}}{2}$ $+ i$ $\frac{1}{2}$

$r = |z| =$ $\sqrt{(\frac{\sqrt{3}}{2})^2+(\frac{1}{2})^2}$ $= \sqrt{1} = 1$

Let $\tan \alpha = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\Rightarrow \tan \alpha = \Big|$ $\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$ $\Big| = \frac{1}{\sqrt{3}} \Rightarrow \alpha =$ $\frac{\pi}{6}$

Since the point $\big($ $\frac{\sqrt{3}}{2}$ $,$ $\frac{1}{2}$ $\big)$ lies in the first quadrant, the $\text{arg}(z)$ is given by

$\theta = \alpha =$ $\frac{\pi}{6}$

Polar form of $\sin 120^{\circ} - i \cos 120^{\circ}$ is given by $= r ( \cos \theta + i \sin \theta)$

$= \big( \cos$ $\frac{\pi}{6}$ $+ i \sin$ $\frac{\pi}{6}$ $\big)$

viii)    $z =$ $\frac{-16}{1+i \sqrt{3} }$ $=$ $\frac{-16}{1+i \sqrt{3} }$ $\times$ $\frac{1-i\sqrt{3}}{1-i \sqrt{3} }$ $=$ $\frac{-16+16\sqrt{3} i}{4}$ $= -4 + 4 \sqrt{3} i$

$z = -4 + 4 \sqrt{3} i$

$r = |z| = \sqrt{(-4)^2+(4 \sqrt{3})^2} = \sqrt{64} = 8$

Let $\tan \alpha = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\Rightarrow \tan \alpha = \Big|$ $\frac{4\sqrt{3}}{-4}$ $\Big| = \sqrt{3} \Rightarrow \alpha =$ $\frac{\pi}{3}$

Since the point $(-4 , 4 \sqrt{3})$ lies in the third quadrant, the $\text{arg}(z)$ is given by

$\theta = \pi - \alpha = \pi -$ $\frac{\pi}{3}$ $=$ $\frac{2\pi}{3}$

Polar form of $\frac{-16}{1+i \sqrt{3} }$ is given by $= r ( \cos \theta + i \sin \theta)$

$= 8 \big( \cos$ $\frac{2\pi}{3}$ $+ i \sin$ $\frac{2\pi}{3}$ $\big)$

$\\$

Question 2: Write $(i^{25})^3$  in polar form

Answer:

$z = (i^{25})^3 = i^{75} = i^{4 \times 18 + 3} = i^3 = -i$

$z = - i$

$r = |z| = \sqrt{(0)^2+(-1)^2} = 1$

Let $\tan \alpha = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\Rightarrow \tan \alpha = \Big|$ $\frac{1}{0}$ $\Big| \Rightarrow \alpha =$ $\frac{\pi}{2}$

Since the point $(0, -1)$ lies in the fourth quadrant, the $\text{arg}(z)$ is given by

$\theta = - \alpha = -$ $\frac{\pi}{2}$

Polar form of $(i^{25})^3$ is given by $= r ( \cos \theta + i \sin \theta)$

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

$= \big( \cos$ $\frac{\pi}{2}$ $- i \sin$ $\frac{\pi}{2}$ $\big)$

$\\$

Question 3: Express the following complex numbers in the form $r ( \cos \theta + i \sin \theta)$:

i) $1+ i \tan \alpha$           ii) $\tan \alpha - i$         iii) $1 - \sin \alpha + i \cos \alpha$            iv) $\frac{1-i}{\cos \frac{\pi}{3}+ i \sin \frac{\pi}{3} }$

Answer:

i)       Let $z = 1+ i \tan \alpha$

$\tan \alpha$  is a periodic function with period $\pi$.

Hence we take $\alpha \in \Big[ 0,$ $\frac{\pi}{2}$ $\Big) \cup \Big($ $\frac{\pi}{2}$ $, \pi \Big]$

$\underline{ \text{Case I:}}$ $\alpha \in \Big[ 0,$ $\frac{\pi}{2}$ $\Big)$

$z = 1 + i \tan \alpha$

$\Rightarrow |z| = \sqrt{1 + \tan^2 \alpha} = \sqrt{\sec^2 \alpha} = | \sec \alpha | = \sec \alpha$

Let $\beta$ be an acute angle given by $\tan \beta = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\tan \beta = | \tan \alpha | = \tan \alpha \Rightarrow \beta = \alpha$

We can see that $Re(z) > 0$ and $Im(z) > 0$, hence $z$ lies in first quadrant, $arg(z) = \beta = \alpha$.

Thus $z$ in the polar form is given by $z = \sec \alpha ( \cos \alpha + i \sin \alpha)$

$\underline{ \text{Case II:}}$ $\alpha \in \Big($ $\frac{\pi}{2}$ , $\pi \Big]$

$z = 1 + i \tan \alpha$

$\Rightarrow |z| = \sqrt{1 + \tan^2 \alpha} = \sqrt{\sec^2 \alpha} = | \sec \alpha | = -\sec \alpha$

Let $\beta$ be an acute angle given by $\tan \beta = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\tan \beta = | \tan \alpha | = - \tan \alpha \Rightarrow \tan \beta = \tan ( \pi - \alpha) \Rightarrow \beta = \pi - \alpha$

We can see that $Re(z) > 0$ and $Im(z) < 0$, hence $z$ lies in fourth quadrant, $arg(z) = -\beta = \alpha - \pi$.

Thus $z$ in the polar form is given by $z = -\sec \alpha \big( \cos (\alpha - \pi) + i \sin (\alpha - \pi) \big)$

ii)     Let $z = \tan \alpha - i$

$\tan \alpha$  is a periodic function with period $\pi$.

Hence we take $\alpha \in \Big[ 0,$ $\frac{\pi}{2}$ $\Big) \cup \Big($ $\frac{\pi}{2}$ $, \pi \Big]$

$\underline{ \text{Case I:}}$ $\alpha \in \Big[ 0,$ $\frac{\pi}{2}$ $\Big)$

$z = \tan \alpha - i$

$\Rightarrow |z| = \sqrt{\tan^2 \alpha + 1} = \sqrt{\sec^2 \alpha} = | \sec \alpha | = \sec \alpha$

Let $\beta$ be an acute angle given by $\tan \beta = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\tan \beta = \Big|$ $\frac{1}{\tan \alpha}$ $\Big| = | \cot \alpha | = \cot \alpha$

$\Rightarrow \tan \beta = \tan \big($ $\frac{\pi}{2}$ $- \alpha \big) \Rightarrow \beta =$ $\frac{\pi}{2}$ $- \alpha$

We can see that $Re(z) > 0$ and $Im(z) < 0$, hence $z$ lies in fourth quadrant, therefore $arg(z) = -\beta = \alpha -$ $\frac{\pi}{2}$.

Thus $z$ in the polar form is given by

$z = \sec \alpha \Big\{ \cos (\alpha -$ $\frac{\pi}{2}$ $\big) + i \sin \big( \alpha -$ $\frac{\pi}{2}$ $) \Big\}$

$\underline{ \text{Case II:}}$ $\alpha \in \Big($ $\frac{\pi}{2}$ , $\pi \Big]$

$z = \tan \alpha - i$

$\Rightarrow |z| = \sqrt{\tan^2 \alpha+1 } = \sqrt{\sec^2 \alpha} = | \sec \alpha | = -\sec \alpha$

Let $\beta$ be an acute angle given by $\tan \beta = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\tan \beta = \Big|$ $\frac{1}{\tan \alpha}$ $\Big| = | \cot \alpha | = - \cot \alpha$

$\Rightarrow \tan \beta = \tan \big( \alpha -$ $\frac{\pi}{2}$ $\big) \Rightarrow \beta = \alpha -$ $\frac{\pi}{2}$

We can see that $Re(z) < 0$ and $Im(z) < 0$, hence $z$ lies in third quadrant, therefore $arg(z) = \pi + \beta =$ $\frac{\pi}{2}$ $+ \alpha$.

Thus $z$ in the polar form is given by

$z = -\sec \alpha \Big\{ \cos ($ $\frac{\pi}{2}$ $+ \alpha \big) + i \sin ($ $\frac{\pi}{2}$ $+ \alpha \big) \Big\}$

iii)     Let $z = (1 - \sin \alpha) + i \cos \alpha$

Both Sine and Cosine functions are periodic function with period $2\pi$

Hence let us take $\alpha \in [ 0, 2\pi ]$

$\Rightarrow |z| = \sqrt{ (1- \sin \alpha)^2 + \cos^2 \alpha } = \sqrt{2- 2 \sin \alpha} = \sqrt{2} \sqrt{1- \sin \alpha}$

$\Rightarrow |z| =$ $\sqrt{2}$ $\sqrt{ (\cos \frac{\alpha}{2} -\sin \frac{\alpha}{2})^2 }$ $= \sqrt{2} \Big| \cos$ $\frac{\alpha}{2}$ $-\sin$ $\frac{\alpha}{2}$ $\Big|$

Let $\beta$ be an acute angle given by $\tan \beta = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\tan \beta =$ $\frac{| \cos \alpha |}{|1- \sin \alpha |}$ $= \Big|$ $\frac{\cos^2 \frac{\alpha}{2} - \sin^2 \frac{\alpha}{2}}{( \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2})^2}$ $\Big|$ $= \Big|$ $\frac{\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2}}{ \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}}$ $\Big|$ $= \Big|$ $\frac{1 + \tan \frac{\alpha}{2}}{ 1 - \tan \frac{\alpha}{2}}$ $\Big|$

$\Rightarrow \tan \beta = \Big| \tan ($ $\frac{\pi}{4}$ $+$ $\frac{\alpha}{2}$ $) \Big|$

$\underline{ \text{Case I:}}$ $0 \leq \alpha <$ $\frac{\pi}{2}$

In this case $\cos$ $\frac{\alpha }{2}$ $> \sin$ $\frac{\alpha }{2}$ and $\frac{\pi}{4}$ $+$ $\frac{\alpha}{2}$ $\in \Big[$ $\frac{\pi }{4}$ $,$ $\frac{\pi}{2}$ $\Big)$

$\Rightarrow |z| = \sqrt{2} \Big( \cos$ $\frac{\alpha }{2}$ $- \sin$ $\frac{\alpha }{2}$ $\Big)$

and $\tan \beta = \Big| \tan ($ $\frac{\pi}{4}$ $+$ $\frac{\alpha}{2}$ $) \Big| = \tan ($ $\frac{\pi}{4}$ $+$ $\frac{\alpha}{2}$ $)$

$\Rightarrow \beta =$ $\frac{\pi}{4}$ $+$ $\frac{\alpha}{2}$

Clearly, $z$ lies in the first quadrant, therefore $arg( z) =$ $\frac{\pi}{4}$ $+$ $\frac{\alpha}{2}$

Hence the polar form of $z$ is

$\sqrt{2} \Big( \cos$ $\frac{\alpha}{2}$ $- \sin$ $\frac{\alpha}{2}$ $\Big) \Big\{ \cos ($ $\frac{\pi}{4}$ $+$ $\frac{\alpha}{2}$ $) + i \sin ($ $\frac{\pi}{4}$ $+$ $\frac{\alpha}{2}$ $) \Big\}$

$\underline{ \text{Case II:}}$ $\frac{\pi}{2}$ $< \alpha <$ $\frac{3\pi}{2}$

In this case $\cos$ $\frac{\alpha }{2}$ $< \sin$ $\frac{\alpha }{2}$ and $\frac{\pi}{4}$ $+$ $\frac{\alpha}{2}$ $\in \Big($ $\frac{\pi }{2}$ $,$ $\pi$ $\Big)$

$|z| = \Big|\sqrt{2} \Big( \cos$ $\frac{\alpha }{2}$ $- \sin$ $\frac{\alpha }{2}$ $\Big) \Big| = - \sqrt{2} \Big( \cos$ $\frac{\alpha }{2}$ $- \sin$ $\frac{\alpha }{2}$ $\Big)$

and $\tan \beta = \Big| \tan ($ $\frac{\pi}{4}$ $+$ $\frac{\alpha }{2}$ $) \Big| = - \tan ($ $\frac{\pi}{4}$ $+$ $\frac{\alpha }{2}$ $) = \tan \{ \pi - ($ $\frac{\pi}{4}$ $+$ $\frac{\alpha }{2}$ $) \}$

$\tan \beta = \tan ($ $\frac{3\pi}{4}$ $-$ $\frac{\alpha}{2}$ $)$

$\Rightarrow \beta = \Big($ $\frac{3\pi}{4}$ $-$ $\frac{\alpha}{2}$ $\Big)$

Clearly, $z$ lies in the fourth quadrant, therefore

$arg( z) = - \beta = - \Big($ $\frac{3\pi}{4}$ $-$ $\frac{\alpha}{2}$ $\Big)$ $=$ $\frac{\alpha}{2}$ $-$ $\frac{3\pi}{4}$

Hence the polar form of $z$ is

$-\sqrt{2} \Big( \cos$ $\frac{\alpha}{2}$ $- \sin$ $\frac{\alpha}{2}$ $\Big) \Big\{ \cos ($ $\frac{\alpha}{2}$ $-$ $\frac{3\pi}{2}$ $) + i \sin ($ $\frac{\alpha}{2}$ $-$ $\frac{3\pi}{4}$ $) \Big\}$

$\underline{ \text{Case III:}}$ $\frac{3\pi}{2}$ $< \alpha <$ $2\pi$

In this case $\cos$ $\frac{\alpha }{2}$ $< \sin$ $\frac{\alpha }{2}$ and $\frac{\pi}{4}$ $+$ $\frac{\alpha}{2}$ $\in \Big($ $\pi ,$ $\frac{5\pi}{4}$ $\Big)$

$|z| = \Big|\sqrt{2} \Big( \cos$ $\frac{\alpha }{2}$ $- \sin$ $\frac{\alpha }{2}$ $\Big) \Big| = - \sqrt{2} \Big( \cos$ $\frac{\alpha }{2}$ $- \sin$ $\frac{\alpha }{2}$ $\Big)$

and $\tan \beta = \Big| \tan ($ $\frac{\pi}{4}$ $+$ $\frac{\alpha }{2}$ $) \Big| = \tan ($ $\frac{\pi}{4}$ $+$ $\frac{\alpha }{2}$ $) = - \tan \{ \pi - ($ $\frac{\pi}{4}$ $+$ $\frac{\alpha }{2}$ $) \}$

$\tan \beta = \tan ($ $\frac{\alpha}{2}$ $-$ $\frac{3\pi}{4}$ $)$

$\Rightarrow \beta = \Big($ $\frac{\alpha}{2}$ $-$ $\frac{3 \pi}{4}$ $\Big)$

Clearly, $z$ lies in the first quadrant, therefore

$arg( z) = \beta = \Big($ $\frac{\alpha}{2}$ $-$ $\frac{3\pi}{4}$ $\Big)$

Hence the polar form of $z$ is

$-\sqrt{2} \Big( \cos$ $\frac{\alpha}{2}$ $- \sin$ $\frac{\alpha}{2}$ $\Big) \Big\{ \cos ($ $\frac{\alpha}{2}$ $-$ $\frac{3\pi}{4}$ $) + i \sin ($ $\frac{\alpha}{2}$ $-$ $\frac{3\pi}{4}$ $) \Big\}$

iv)     Let $z =$ $\frac{1-i}{\cos \frac{\pi}{3}+ i \sin \frac{\pi}{3} }$

$=$ $\frac{1 - i }{\frac{1}{2} + i \frac{\sqrt{3}}{2} }$ $=$ $\frac{2-2i}{1 + i \sqrt{3}}$ $\times$ $\frac{1 - i \sqrt{3}}{1 - i \sqrt{3}}$ $=$ $\frac{2 - 2i - i 2 \sqrt{3} + 2 \sqrt{3} i^2}{1+3}$

$=$ $\frac{2 -2\sqrt{3}-2i ( 1 + \sqrt{3})}{4}$ $=$ $\frac{(1-\sqrt{3}) - i ( 1 + \sqrt{3})}{2}$ $=$ $\frac{(1-\sqrt{3})}{2}$ $+ i$ $\frac{( -1 - \sqrt{3})}{2}$

$\therefore z =$ $\frac{(1-\sqrt{3})}{2}$ $+ i$ $\frac{( -1 - \sqrt{3})}{2}$

$|z| =$ $\sqrt{ \Big( \frac{1-\sqrt{3}}{2} \Big)^2 + \Big( \frac{-1-\sqrt{3}}{2} \Big)^2 }$ $=$ $\sqrt{\frac{1+3-2\sqrt{3}}{4}+ \frac{1+3 + 2\sqrt{3}}{4}}$ $=$ $\sqrt{\frac{8}{4}}$ $= \sqrt{2}$

Let $\beta$ be an acute angle given by $\tan \beta = \Big|$ $\frac{Im(z)}{Re(z)}$ $\Big|$

$\tan \beta = \Big|$ $\frac{\frac{1+\sqrt{3}}{2}}{\frac{1-\sqrt{3}}{2}}$ $\Big| = \Big|$ $\frac{1+\sqrt{3}}{1-\sqrt{3}}$ $\Big|$ $= \Big|$ $\frac{\tan \frac{\pi}{4} + \tan \frac{\pi}{3} }{1 - \tan \frac{\pi}{4} \tan \frac{\pi}{3}}$ $\Big|$ $= \tan ($ $\frac{\pi}{4}$ $+$ $\frac{\pi}{3}$ $) = \tan$ $\frac{7\pi}{12}$

$\Rightarrow \beta =$ $\frac{7\pi}{12}$

Clearly, $z$ lies in the fourth quadrant, therefore $arg( z) = \beta = -$ $\frac{7\pi}{12}$

Hence the polar form of $z$ is $\sqrt{2} \Big( \cos$ $\frac{7\pi}{12}$ $- \sin$ $\frac{7\pi}{12}$ $\Big)$

$\\$

Question 4: If $z_1$ and $z_2$ are two complex numbers such that $|z_1|= |z_2|$ and $arg(z_1)+arg(z_2)= \pi$ , then show that $z_1 = - \overline{z_2}$

Answer:

Let $\theta_1$ by $arg(z_1)$ and $\theta_2$ by $arg(z_2)$

Given $|z_1| = |z_2|$;  and $arg(z_1)+arg(z_2)= \pi$

Since $z_1$ is a complex number

$z_1 = |z_1| ( \cos \theta_1 + i \sin \theta_1)$

$= |z_2| [ \cos (\pi - \theta_2) + i \sin (\pi - \theta_2) ]$

$= |z_2| [ - \cos \theta_2 + i \sin \theta_2 ]$

$= - |z_2|[\cos \theta_2 - i \sin \theta_2 ]$

$= - \overline{z_2}$

Hence $z_1 = - \overline{z_2}$

$\\$

Question 5: If $z_1, z_2$ and $z_3, z_4$ are two pairs of conjugate complex numbers, prove that $arg \Big($ $\frac{z_1}{z_4}$ $\Big) + arg \Big($ $\frac{z_2}{z_3}$ $\Big) = 0$

Answer:

$z_1, z_2$ and $z_3, z_4$ are two pairs of conjugate complex numbers.

$\therefore z_1 = r_1 e^{i \theta_1}, z_2 = r_1 e^{-i \theta_1}, z_3 = r_2 e^{i \theta_2} , z_4 = r_2 e^{-i \theta_2}$

$\therefore$ $\frac{z_1}{z_4}$ $=$ $\frac{r_1 e^{i \theta_1}}{r_2 e^{-i \theta_2}}$ $= \Big($ $\frac{r_1}{r_2}$ $\Big)$ $e^{i( \theta_1 - \theta_2)}$

$\Rightarrow arg \Big($ $\frac{z_1}{z_4}$ $\Big) = \theta_1 - \theta_2$   … … … … … i)

$\therefore$ $\frac{z_2}{z_3}$ $=$ $\frac{r_1 e^{-i \theta_1}}{r_2 e^{i \theta_2}}$ $= \Big($ $\frac{r_1}{r_2}$ $\Big)$ $e^{i( -\theta_1 + \theta_2)}$

$\Rightarrow arg \Big($ $\frac{z_2}{z_3}$ $\Big) = \theta_2 - \theta_1$   … … … … … ii)

$arg \Big($ $\frac{z_1}{z_4}$ $\Big) + arg \Big($ $\frac{z_2}{z_3}$ $\Big) = \theta_1 - \theta_2 + \theta_2 - \theta_1$

$\Rightarrow arg \Big($ $\frac{z_1}{z_4}$ $\Big) + arg \Big($ $\frac{z_2}{z_3}$ $\Big) = 0$. Hence proved.

$\\$

Question 6: Express $\sin$ $\frac{\pi}{5}$ $+ i \Big( 1 - \cos$ $\frac{\pi}{5}$ $\Big)$ in polar form.

Answer:

Let $z = \sin$ $\frac{\pi}{5}$ $+ i \Big( 1 - \cos$ $\frac{\pi}{5}$ $\Big)$

$|z| =$ $\sqrt{ \big( \sin \frac{\pi}{5} \big)^2 + \big(1 - \cos \frac{\pi}{5} \big)^2 }$

$=$ $\sqrt{ \sin^2 \frac{\pi}{5} + 1 +\cos^2 \frac{\pi}{5} - 2 \cos \frac{\pi}{5} }$

$= \sqrt{2}$ $\Big( \sqrt{1 -\cos \frac{\pi}{5} } \Big)$

$= \sqrt{2}$ $\sqrt{ 2 \sin^2 \frac{\pi}{10}}$

$= 2 \sin$ $\frac{\pi}{10}$

$\tan \beta =$ $\frac{|1 - \cos \frac{\pi}{5} |}{|\sin \frac{\pi}{5} |}$ $= \Big|$ $\frac{2 \sin^2 \frac{\pi}{10}}{ 1 \sin \frac{\pi}{10} \cos \frac{\pi}{10}}$ $\Big| = | \tan$ $\frac{\pi}{10}$ $|$

$\Rightarrow \beta =$ $\frac{\pi}{10}$

Clearly, $z$ lies in the first quadrant. Therefore $arg(z) =$ $\frac{\pi}{10}$. Hence the polar form of $z$ is  $2 \sin$ $\frac{\pi}{10}$ $\Big( \cos$ $\frac{\pi}{10}$ $+ i \sin$ $\frac{\pi}{10}$ $\Big)$