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Question: The polar form of complex number z=\(\dfrac{{i - 1}}{{\cos \dfrac{\pi }{3} + \sin \dfrac{\pi }{3}}}\...

The polar form of complex number z=i−1cos⁡π3+sin⁡π3\dfrac{{i - 1}}{{\cos \dfrac{\pi }{3} + \sin \dfrac{\pi }{3}}} is.
A.12(cos⁡3π12+isin⁡3π12)\dfrac{1}{{\sqrt 2 }}\left( {\cos \dfrac{{3\pi }}{{12}} + i\sin \dfrac{{3\pi }}{{12}}} \right)
B.2(cos⁡5π12+isin⁡5π12)\sqrt 2 \left( {\cos \dfrac{{5\pi }}{{12}} + i\sin \dfrac{{5\pi }}{{12}}} \right)
C.2(cos⁡7π12+isin⁡7π12)\sqrt 2 \left( {\cos \dfrac{{7\pi }}{{12}} + i\sin \dfrac{{7\pi }}{{12}}} \right)
D.12(cos⁡5π12+isin⁡5π12)\dfrac{1}{{\sqrt 2 }}\left( {\cos \dfrac{{5\pi }}{{12}} + i\sin \dfrac{{5\pi }}{{12}}} \right)

Explanation

Solution

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 x2=−1{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=i−1cos⁡π3+isin⁡π3\dfrac{{i - 1}}{{\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}}}
=i−1cos⁡(1803)+isin⁡(1803)\dfrac{{i - 1}}{{\cos \left( {\dfrac{{180}}{3}} \right) + i\sin \left( {\dfrac{{180}}{3}} \right)}}
=i−1cos⁡60∘+isin⁡60∘\dfrac{{i - 1}}{{\cos {{60}^ \circ } + i\sin {{60}^ \circ }}}
=i−112+32i\dfrac{{i - 1}}{{\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i}}
=i−11+3i2\dfrac{{i - 1}}{{\dfrac{{1 + \sqrt 3 i}}{2}}}
=2(i−1)1+3i\dfrac{{2\left( {i - 1} \right)}}{{1 + \sqrt 3 i}}
Rationalizing:
=2(i−1)1+3i×1−3i1−3i\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:
=2(1−i)(1−3i)(1+3i)(1−3i) =2[i(1−3i)−1(1−3i)](1−3i)(1−3i) =2[i−3i2−1+3i](1−3i)(1−3i)  = \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 (a−b)(a+b)=a2−b2\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}
=2[−1+i+3i−3i2]12−(3i)2 =2[−1+i+3i−3i(i2)]1−3i2   = \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:
⇒i2=−1 =2[−1+i+3i−3i(−1)]1−3(−1) =2[−1+i+3i+3]1+3  \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}} \\\
=2[(−1+3)+(i+3i)]4 =(−1+3)+i(1+3)2 =3−12+i3+12   = \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=3−12+i3+12\dfrac{{\sqrt 3 - 1}}{2} + i\dfrac{{\sqrt 3 + 1}}{2} ………(1)
Let polar form be
z=r(cos⁡θ+isin⁡θ)\left( {\cos \theta + i\sin \theta } \right) ………..(2)

From (1) and (2)
3−12+i3+12=r(cos⁡θ+sin⁡θ)=rcos⁡θ+irsin⁡θ  ↓ ↓ ↓ ↓   \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
3−12=rcos⁡θ\dfrac{{\sqrt 3 - 1}}{2} = r\cos \theta

Comparing Imaginary parts
3+12=rsin⁡θ\dfrac{{\sqrt 3 + 1}}{2} = r\sin \theta

Squaring both sides
(3−12)2=(rcos⁡θ)2 (3−1)222=r2cos⁡2θ  {\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 (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab
(3)2+(1)2−23×14=r2cos⁡2θ 3+1−234=r2cos⁡2θ 4+234=r2cos⁡2θ .......(3)  \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
(3+12)2=(rsin⁡θ)2 (3+1)222=r2sin⁡2θ  {\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 (a−b)2=a2−b2−2ab{\left( {a - b} \right)^2} = {a^2} - {b^2} - 2ab
(3)2+(1)2+23×14=r2sin⁡2θ 3+1+234=r2sin⁡2θ 4+234=r2sin⁡2θ .......(4)  \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)
4−234+4+234=r2cos⁡2θ+r2sin⁡2θ 14(4−23+4+23)=r2(cos⁡2θ+sin⁡2θ)  \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) \\\
14(4+4−23+23)=r2(cos⁡2θ+sin⁡2θ) 14(8−0)+=r2(cos⁡2θ+sin⁡2θ)  \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) \\\
84=r2×1 (cos⁡2θ+sin⁡2θ=1) 2=r2 2=r r=2  \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 \\\
3−12+3+12=rcos⁡θ+irsin⁡θ  ↓ ↓ ↓ ↓  \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
3−12= r cos θ\dfrac{{\sqrt 3 - 1}}{2} = {\text{ r cos }}\theta 3+12= r sin θ\dfrac{{\sqrt 3 + 1}}{2} = {\text{ r sin }}\theta
Put r = 2\sqrt 2 Put r =2\sqrt 2
3−12=2cos⁡θ 3−122=cos⁡θ  \dfrac{{\sqrt 3 - 1}}{2} = \sqrt 2 \cos \theta \\\ \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} = \cos \theta \\\
3+12=2sin⁡θ 3+122=sin⁡θ  \dfrac{{\sqrt 3 + 1}}{2} = \sqrt 2 \sin \theta \\\ \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} = \sin \theta \\\
Hence,cos⁡θ=3−122&sin⁡θ=3+122\cos \theta = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\& \sin \theta = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}
Rough
sin⁡75∘=sin⁡(45∘+30∘)   \sin {75^ \circ } = \sin \left( {{{45}^ \circ } + {{30}^ \circ }} \right) \\\ \\\
Using sin⁡(x+y)=cos⁡ysin⁡x+cos⁡xsin⁡y\sin \left( {x + y} \right) = \cos y\sin x + \cos x\sin y
=(12)(32)+(12)(12) =322+122  = \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 }} \\\
3+122\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}
So,sin⁡75∘=3+122\sin {75^ \circ } = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}
Since sin⁡θ\sin \theta and cos⁡θ\cos \theta both positive
Hence θ\theta lies in the 1st Quadrant
Hence, argument= 75∘{75^ \circ }
=75∘×π180= {75^ \circ } \times \dfrac{\pi }{{180}}
=5π12= \dfrac{{5\pi }}{{12}}
Hence, r=2,& θ=5π12r = \sqrt 2 ,\& {\text{ }}\theta = \dfrac{{5\pi }}{{12}}
Thus, Polar form of z:
=r(cos⁡θ+isin⁡θ) =2(cos⁡5π12+isin⁡θ5π12)  = 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 x2=−1{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∘\sin {75^ \circ } and basic concepts of representing the complex number in polar form.