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)