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

$\displaystyle \text{i) } 1+i \hspace{1.0cm} \text{ii) } \sqrt{3}+i \hspace{1.0cm} \text{iii) } 1-i \hspace{1.0cm} \text{iv) } \frac{1-i}{1+i} \hspace{1.0cm} \text{v) } \frac{1}{1+i}$

$\displaystyle \text{vi) } \frac{1+2i}{1-3i} \hspace{1.0cm} \text{vii) } \sin 120^{\circ} - i \cos 120^{\circ} \hspace{1.0cm} \text{viii) } \frac{-16}{1+i \sqrt{3}}$

$\displaystyle \text{i) } z = 1+i$

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

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

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

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

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

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

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

$\displaystyle \text{ii) } z = \sqrt{3}+i$

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

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

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

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

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

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

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

$\displaystyle \text{iii) } z = 1-i$

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

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

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

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

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

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

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

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

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

$\displaystyle z = -i$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

$\displaystyle z = - i$

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{i) } 1+ i \tan \alpha \hspace{1.0cm} \text{ii) } \tan \alpha - i \hspace{1.0cm} \text{iii) } 1 - \sin \alpha + i \cos \alpha \hspace{1.0cm} \text{iv) } \frac{1-i}{\cos \frac{\pi}{3}+ i \sin \frac{\pi}{3} }$

$\displaystyle \text{i) } \text{Let } z = 1+ i \tan \alpha$

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

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

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

$\displaystyle z = 1 + i \tan \alpha$

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

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

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

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

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

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

$\displaystyle z = 1 + i \tan \alpha$

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

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

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

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

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

$\displaystyle \text{ii) } \text{Let } z = \tan \alpha - i$

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

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

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

$\displaystyle z = \tan \alpha - i$

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

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

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

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

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

Thus $\displaystyle z$ in the polar form is given by

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

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

$\displaystyle z = \tan \alpha - i$

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

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

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

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

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

Thus $\displaystyle z$ in the polar form is given by

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

$\displaystyle \text{iii) } \text{Let } z = (1 - \sin \alpha) + i \cos \alpha$

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

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

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

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

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

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

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

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

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

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

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

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

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

Hence the $\displaystyle \text{Polar form of } z$ is

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

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

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

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

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

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

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

$\displaystyle \text{Clearly, } z$ lies in the fourth quadrant, therefore

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

Hence the $\displaystyle \text{Polar form of } z$ is

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

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

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

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

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

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

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

$\displaystyle \text{Clearly, } z$ lies in the first quadrant, therefore

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

Hence the $\displaystyle \text{Polar form of } z$ is

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{Let } \theta_1$ by $\displaystyle arg(z_1) \text{ and } \theta_2$ by $\displaystyle arg(z_2)$

$\displaystyle \text{Given } |z_1| = |z_2|$; $\displaystyle \text{and } arg(z_1)+arg(z_2)= \pi$

$\displaystyle \text{Since } z_1$ is a complex number

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

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

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

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

$\displaystyle = - \overline{z_2}$

$\displaystyle \text{Hence } z_1 = - \overline{z_2}$

$\displaystyle \\$

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

$\displaystyle z_1, z_2 \text{ and } z_3, z_4$ are two pairs of conjugate complex numbers.

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

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