Question

The polar form of complex number z=$\dfrac{{i - 1}}{{\cos \dfrac{\pi }{3} + \sin \dfrac{\pi }{3}}}$ is.
A.$\dfrac{1}{{\sqrt 2 }}\left( {\cos \dfrac{{3\pi }}{{12}} + i\sin \dfrac{{3\pi }}{{12}}} \right)$
B.$\sqrt 2 \left( {\cos \dfrac{{5\pi }}{{12}} + i\sin \dfrac{{5\pi }}{{12}}} \right)$
C.$\sqrt 2 \left( {\cos \dfrac{{7\pi }}{{12}} + i\sin \dfrac{{7\pi }}{{12}}} \right)$
D.$\dfrac{1}{{\sqrt 2 }}\left( {\cos \dfrac{{5\pi }}{{12}} + i\sin \dfrac{{5\pi }}{{12}}} \right)$

Answer 1

Hint: Now in this question we have to find the polar form of a complex number now before all that we should know that what is a complex number it is a number in the form a+bi, where a and b are real numbers , and i is a solution of the equation ${x^2} = - 1$.Because no real number satisfies this equation, i is called an imaginary number. Now one should also know what is a polar form of a complex number. The polar form of a complex number is another way to represent a complex number. The form z=a+bi is called the rectangular coordinate form of a complex number. The horizontal axis and vertical axis is the imaginary axis.

Complete step-by-step answer:
Let z=$\dfrac{{i - 1}}{{\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}}}$
=$\dfrac{{i - 1}}{{\cos \left( {\dfrac{{180}}{3}} \right) + i\sin \left( {\dfrac{{180}}{3}} \right)}}$
=$\dfrac{{i - 1}}{{\cos {{60}^ \circ } + i\sin {{60}^ \circ }}}$
=$\dfrac{{i - 1}}{{\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i}}$
=$\dfrac{{i - 1}}{{\dfrac{{1 + \sqrt 3 i}}{2}}}$
=$\dfrac{{2\left( {i - 1} \right)}}{{1 + \sqrt 3 i}}$
Rationalizing:
=$\dfrac{{2\left( {i - 1} \right)}}{{1 + \sqrt 3 i}} \times \dfrac{{1 - \sqrt 3 i}}{{1 - \sqrt 3 i}}$
Now using the rationalising we are able to find that:
$
   = \dfrac{{2\left( {1 - i} \right)\left( {1 - \sqrt 3 i} \right)}}{{\left( {1 + \sqrt 3 i} \right)\left( {1 - \sqrt 3 i} \right)}} \\
   = \dfrac{{2\left[ {i\left( {1 - \sqrt 3 i} \right) - 1\left( {1 - \sqrt 3 i} \right)} \right]}}{{\left( {1 - \sqrt 3 i} \right)\left( {1 - \sqrt 3 i} \right)}} \\
   = \dfrac{{2\left[ {i - \sqrt 3 {i^2} - 1 + \sqrt 3 i} \right]}}{{\left( {1 - \sqrt 3 i} \right)\left( {1 - \sqrt 3 i} \right)}} \\
 $
 Using $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$
 $
   = \dfrac{{2\left[ { - 1 + i + \sqrt 3 i - \sqrt 3 {i^2}} \right]}}{{{1^2} - {{\left( {\sqrt 3 i} \right)}^2}}} \\
   = \dfrac{{2\left[ { - 1 + i + \sqrt 3 i - \sqrt 3 i\left( {{i^2}} \right)} \right]}}{{1 - 3{i^2}}} \\
    \\
 $
Now, we know that:
$
   \Rightarrow {i^2} = - 1 \\
   = \dfrac{{2\left[ { - 1 + i + \sqrt 3 i - \sqrt 3 i\left( { - 1} \right)} \right]}}{{1 - 3\left( { - 1} \right)}} \\
   = \dfrac{{2\left[ { - 1 + i + \sqrt 3 i + \sqrt 3 } \right]}}{{1 + 3}} \\
 $
$
   = \dfrac{{2\left[ {\left( { - 1 + \sqrt 3 } \right) + \left( {i + \sqrt 3 i} \right)} \right]}}{4} \\
   = \dfrac{{\left( { - 1 + \sqrt 3 } \right) + i\left( {1 + \sqrt 3 } \right)}}{2} \\
   = \dfrac{{\sqrt 3 - 1}}{2} + i\dfrac{{\sqrt 3 + 1}}{2} \\
    \\
 $
Now, z=$\dfrac{{\sqrt 3 - 1}}{2} + i\dfrac{{\sqrt 3 + 1}}{2}$ ………(1)
Let polar form be
z=r$\left( {\cos \theta + i\sin \theta } \right)$ ………..(2)

From (1) and (2)
$
  \dfrac{{\sqrt 3 - 1}}{2} + i\dfrac{{\sqrt 3 + 1}}{2} = r\left( {\cos \theta + \sin \theta } \right) = r\cos \theta + ir\sin \theta \\
  {\text{ }} \downarrow {\text{ }} \downarrow {\text{ }} \downarrow {\text{ }} \downarrow {\text{ }} \\
 $
(real part) (imaginary part) (real part) (imaginary part)


Comparing real part
$\dfrac{{\sqrt 3 - 1}}{2} = r\cos \theta $

Comparing Imaginary parts
$\dfrac{{\sqrt 3 + 1}}{2} = r\sin \theta $

Squaring both sides
 $
  {\left( {\dfrac{{\sqrt 3 - 1}}{2}} \right)^2} = {\left( {r\cos \theta } \right)^2} \\
  \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{{2^2}}} = {r^2}{\cos ^2}\theta \\
 $
Using ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
$
  \dfrac{{{{\left( {\sqrt 3 } \right)}^2} + {{\left( 1 \right)}^2} - 2\sqrt 3 \times 1}}{4} = {r^2}{\cos ^2}\theta \\
  \dfrac{{3 + 1 - 2\sqrt 3 }}{4} = {r^2}{\cos ^2}\theta \\
  \dfrac{{4 + 2\sqrt 3 }}{4} = {r^2}{\cos ^2}\theta {\text{ }}.......{\text{(3)}} \\
 $
Squaring both sides
$
  {\left( {\dfrac{{\sqrt 3 + 1}}{2}} \right)^2} = {\left( {r\sin \theta } \right)^2} \\
  \dfrac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{{2^2}}} = {r^2}{\sin ^2}\theta \\
 $
Using ${\left( {a - b} \right)^2} = {a^2} - {b^2} - 2ab$
  $
  \dfrac{{{{\left( {\sqrt 3 } \right)}^2} + {{\left( 1 \right)}^2} + 2\sqrt 3 \times 1}}{4} = {r^2}{\sin ^2}\theta \\
  \dfrac{{3 + 1 + 2\sqrt 3 }}{4} = {r^2}{\sin ^2}\theta \\
  \dfrac{{4 + 2\sqrt 3 }}{4} = {r^2}{\sin ^2}\theta {\text{ }}.......{\text{(4)}} \\
 $
Adding (3) and (4)
$
  \dfrac{{4 - 2\sqrt 3 }}{4} + \dfrac{{4 + 2\sqrt 3 }}{4} = {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta \\
  \dfrac{1}{4}\left( {4 - 2\sqrt 3 + 4 + 2\sqrt 3 } \right) = {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) \\
 $
$
  \dfrac{1}{4}\left( {4 + 4 - 2\sqrt 3 + 2\sqrt 3 } \right) = {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) \\
  \dfrac{1}{4}\left( {8 - 0} \right) + = {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) \\
 $
$
  \dfrac{8}{4} = {r^2} \times 1{\text{ }}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta = 1} \right) \\
  2 = {r^2} \\
  \sqrt 2 = r \\
  r = \sqrt 2 \\
 $
$
  \dfrac{{\sqrt 3 - 1}}{2} + \dfrac{{\sqrt 3 + 1}}{2} = r\cos \theta + ir\sin \theta \\
  {\text{ }} \downarrow {\text{ }} \downarrow {\text{ }} \downarrow {\text{ }} \downarrow \\
 $
(Real part) (Imaginary (Real (Imaginary
                      part) part) part)


Comparing real part Comparing imaginary part
$\dfrac{{\sqrt 3 - 1}}{2} = {\text{ r cos }}\theta $ $\dfrac{{\sqrt 3 + 1}}{2} = {\text{ r sin }}\theta $
Put r = $\sqrt 2 $ Put r =$\sqrt 2 $
$
  \dfrac{{\sqrt 3 - 1}}{2} = \sqrt 2 \cos \theta \\
  \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} = \cos \theta \\
$
$
  \dfrac{{\sqrt 3 + 1}}{2} = \sqrt 2 \sin \theta \\
  \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} = \sin \theta \\
 $
Hence,$\cos \theta = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\& \sin \theta = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
Rough
$
  \sin {75^ \circ } = \sin \left( {{{45}^ \circ } + {{30}^ \circ }} \right) \\
    \\
 $
Using $\sin \left( {x + y} \right) = \cos y\sin x + \cos x\sin y$
$
   = \left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right) + \left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( {\dfrac{1}{2}} \right) \\
   = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} + \dfrac{1}{{2\sqrt 2 }} \\
 $
$\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
So,$\sin {75^ \circ } = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
Since $\sin \theta $ and $\cos \theta $ both positive
Hence $\theta $ lies in the 1st Quadrant
Hence, argument= ${75^ \circ }$
                               $ = {75^ \circ } \times \dfrac{\pi }{{180}}$
                                $ = \dfrac{{5\pi }}{{12}}$
Hence, $r = \sqrt 2 ,\& {\text{ }}\theta = \dfrac{{5\pi }}{{12}}$
Thus, Polar form of z:
$
   = r\left( {\cos \theta + i\sin \theta } \right) \\
   = \sqrt 2 \left( {\cos \dfrac{{5\pi }}{{12}} + i\sin \theta \dfrac{{5\pi }}{{12}}} \right) \\
 $
Hence, option B is the correct option

Note:In this question it should be noted that we use various concepts of complex numbers and the polar form of complex numbers. Briefly it means complex number is a number that can be expressed in the form of a+bi, where a and b are real numbers and I is the solution of equation ${x^2} = - 1$ . The polar form of a complex number whereas is another way of representing a complex number. In this question we deal with various trigonometry too like finding the value of $\sin {75^ \circ }$ and basic concepts of representing the complex number in polar form.