|(1+3i)³| = |1+3i|³ = \(\left[ \sqrt { { 1 }^{ 2 }+{ 3 }^{ 2 } } \right] ^{ 3 }\) =\(\left( \sqrt { 10 } \right) ^{ 3 }\)
=\((\sqrt { 10 } )^{ 3 }=\sqrt { 10 } \times \sqrt { 10 } \times \sqrt { 10 } \times \sqrt { 10 } =10\sqrt { 10 } \).

LHS = (1+ω²)³ - (1+ω)³
= (-ω)³ - (-ω²)³
[∴ 1 + ω + ω² = 0]
= -ω³ + ω⁶  [∴ ω³ = 1]
= -1 + 1 = 0

Z = \(\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 107 }+\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 107 }\)
\(\bar { z } =\left( \overline { \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } } \right) ^{ 107 }+\left( \overline { \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } } \right) ^{ 107 }\)
= \(\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 107 }+\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 107 }\) = z
Since z = \(\bar { z } \), Im(z) = 0

(cos θ + i sin θ )² = cos 2θ + isin 2θ
[By De moivre's theorem]
⇒ cos 2θ + isin 2θ = x + iy
Equating the real and imaginary parts we get,
x = cos 2θ, y = sin 2θ
∴ x² + y² = cos²2θ + sin²2θ = 1
Hence proved

Let z₁ = a+ib and z₂ = c+id
Given |z₁| = Iz₂|
⇒ \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } } =\sqrt { { c }^{ 2 }+{ d }^{ 2 } } \)
Squaring both sides we get, a² + b² = c² + d²
This cannot imply that a = c and b = d
∴ z₁ and z₂ need not be equal

\(-\sqrt { 3 } +i\)

Modulus = 2 and
\(a={ tan }^{ -1 }\left| \frac { y }{ x } \right| ={ tan }^{ -1 }\frac { 1 }{ \sqrt { 3 } } =\frac { \pi }{ 6 } \)
Since the complex number  \(-\sqrt { 3 } +i\) lies in the second quadrant has the principal value
\(\theta =\pi -\alpha =\pi -\frac { \pi }{ 6 } =\frac { 5\pi }{ 6 } \)
Therefore the modulus and principal argument of \(-\sqrt { 3 } +i\) are 2 and \(\frac { 5\pi }{ 6 } \) respectively.

 i¹⁷²⁹ = i¹⁷²⁸ i¹ = i

\({ i }^{ 59 }+\frac { 1 }{ { i }^{ 59 } } \)
i^{4 \(\times\) 14 + 3} + i^{-(4 \(\times\) 14 + 3)}
= (i⁴)¹⁴.i³ + (i⁴)^-14.i^-3
= 1.i³+1.i^-3  [∵ i⁴ = 1]
= -i + i  [∴ i³ = -i and i^-3= i]
= 0

\(\sqrt { 3 } +i\)

Modulus = \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { \left( \sqrt { 3 } \right) ^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 3+1 } =2\)
\(\alpha =tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }\frac { 1 }{ \sqrt { 3 } } =\frac { \pi }{ 6 } \)
Since the complex number \(\sqrt { 3+ } i\) lies in the first quadrant, has the principal value
\(\theta =\alpha =\frac { \pi }{ 6 } \)
Therefore, the modulus and principal argument of \(\sqrt { 3+ } i\) are 2 and \(\frac { \pi }{ 6 } \) respectively.

\(\left| \frac { 2+i }{ -1+2i } \right| =\frac { \left| 2+i \right| }{ \left| -1+2i \right| } =\frac { \sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 } } }{ \sqrt { \left( -1 \right) ^{ 2 }+{ 2 }^{ 2 } } } =1\) \(\left( \because \left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| =\left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| ,{ z }_{ 2 }\neq 0 \right) \)

\(\overline { (2-i)^{ 12 }+(2+i)^{ 12 } } =\overline { (2-i)^{ 12 } } +\overline { (2+i)^{ 12 } } \)
\(\left( 2+i\sqrt { 3 } \right) ^{ 10 }-\left( 2-i\sqrt { 3 } \right) ^{ 10 }\)
Now \(\overline { z } \) = \(\overline { (2+\sqrt { 3 } )^{ 10 }-(2-i\sqrt { 3 } )^{ 10 } } \)
\(\overline { z } \) = \(\overline { (2+i\sqrt { 3 } )^{ 10 } } -(2-i\sqrt { 3 } )^{ 10 }\)
[∵ \(\overline { { z }_{ 1 }-{ z }_{ 2 } } =\overline { { z }_{ 1 } } -\overline { { z }_{ 2 } } \)]
= \((2-i\sqrt { 3 } )^{ 10 }-(2+i\sqrt { 3 } )^{ 10 }\)
= -\(\left[ (2-i\sqrt { 3 } )^{ 10 }-(2+i\sqrt { 3 } )^{ 10 } \right] \)
∴ \(\overline { z } \) = -\(\ { z } \) ⇒ z is purely imaginary
Hence \((2-i\sqrt { 3 } )^{ 10 }-(2+i\sqrt { 3 } )^{ 10 }\)is purely imaginary

Given z₁= 2 - i and z₂= -4+3i
z₁z₂ = (2-i)(-4+3i)
= -8 + 6i + 4i - 3i²
= -8 +10i - 3(-1)
= -8 +10i + 3 = -5 +10i
Inverse of z₁z₁ is \(\frac { 1 }{ { z }_{ 1 }{ z }_{ 2 } } \)
= \(\frac { 1 }{ -5+10i } \times \frac { -5-10i }{ -5-10i } \)
= \(\frac { -5-10i }{ (-5)^{ 2 }-(10i)^{ 2 } } \)
= \(\frac { -5-10i }{ 25-100i^{ 2 } } \)
= \(\frac { -5-10i }{ 25+100 } \)  [∵ i² = -1]
\(=\frac{\not{5}(-1-2 i)}{\not{5}\langle(2 5)}=\frac{-1-2 i}{25}\)
∴ Inverse of z₁z₂ is \(\frac { 1 }{ 25 } \) (-1-2i)
Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } \) is \(\frac { 1 }{ \frac { { z }_{ 1 } }{ { z }_{ 2 } } } =\frac { { z }_{ 2 } }{ { z }_{ 2 } } \)
∴ Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } =\frac { { z }_{ 2 } }{ { z }_{ 1 } } =\frac { -4+3i }{ 2-i } \times \frac { 2+i }{ 2+i } \)
= \(\frac { -4+3i }{ 2-i } \times \frac { 2+i }{ 2+i } \)
= \(\frac { -8-4i+6i+3i^{ 2 } }{ 2^{ 2 }-({ i }^{ 2 }) } \)
= \(\frac { -8+2i-3 }{ 4+1 } =\frac { -11+2i }{ 5 }\)
\( =\frac { 1 }{ 5 } \)(-11 + 2i)
∴ Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } \) is \(\frac { -11+2i }{ 5 }\)or \(\frac { 1 }{ 5 } \)(-11 + 2i)

We have z = (2+3i)(1−i) = (2+3)+(3−2)i = 5+i
\(\Rightarrow\) \({ z }^{ -1 }=\frac { 1 }{ z } =\frac { 1 }{ 5+i } \)
Multiplying the numerator and denominator by the conjugate of the denominator, we get
\({ z }^{ -1 }=\frac { \left( 5-i \right) }{ \left( 5+i \right) \left( 5-i \right) } =\frac { 5-i }{ { 5 }^{ 2 }+{ I }^{ 2 } } =\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)
\(\Rightarrow\)\({ z }^{ -1 }=\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)

i¹⁹⁴⁷+ i¹⁹⁵⁰
i¹⁹⁴⁷+i¹⁹⁵⁰ = i¹⁹⁴⁴.i³+i¹⁹⁴⁸.i²
[∴ 1944 is a multiple of 4, or 1948 is also a multiple of 4]
= (i⁴)486.i².i¹+(i⁴)487.i² [i⁴ = 1]
= (1⁴⁸⁶)(-1) + (1)⁴⁸⁷(-1) [i²= -1]
= -i-1
= -(1- i)

(i)⁷= (i)⁴⁺³ = (i)³ = -i

Given p+iq = \(\frac { (a+i)^{ 2 } }{ 2a-i } \) ...........(1)
Taking conjugate both sides we get,
p-iq =\(\frac { (a+i)^{ 2 } }{ 2a+i } \) ..........(2)
Multiplying (1) and (2) we get,
(p+iq)(p-iq) = \(\frac { (a+i)^{ 2 } }{ 2a-i } \times \frac { (a-i)^{ 2 } }{ 2a+i } \)
p²+q² = \(\frac { [(a+i)(a-i)^{ 2 }] }{ 4a^{ 2 }+1 } =\frac { ({ a }^{ 2 }+1)^{ 2 } }{ 4a^{ 2 }+1 } \).
Hence proved

\(\frac { z+1 }{ z-i } =\frac { x+iy+1 }{ x+iy-i } =\frac { (x+1)+iy }{ x+i(y-i) } \)
=\(\frac { (x+1)+iy }{ x+i(y-1) } \times \frac { x-i(y-1) }{ x-i(y-1) } \)
Choosing the real part alone ,we get
\(\frac { (x+1)x-{ i }^{ 2 }y(y-1) }{ { x }^{ 2 }+(y-1)^{ 2 } } \)= 0
⇒ x (x + 1) +y (y - 1) = 0
x² + x + y² - y = 0 when is the locus of z.

Let z = x + iy be a complex number
∴ \(\left| \frac { x+iy-3 }{ x+iy+3 } \right| \)  = 2
⇒ \(\left| \frac { x+iy-3 }{ x+iy+3 } \right| \) = 2
⇒ |(x-3) + iy| = 2|(x+3)+iy|
⇒ \(\sqrt { (x-3)^{ 2 }+{ y }^{ 2 } } =2\sqrt { (x+3)^{ 2 }+{ y }^{ 2 } } \)
Squaring both sides we get,
(x - 3)² + y² = 4[(x + 3)² + y²]
⇒ x² + 9 - 6x + y² = 4 [x²+ 9 + 6x + y²]
x² + 9 - 6x + 1 = 4x² + 36 + 24x + 4y²
⇒ 3x² + 3y² + 30x + 27 = which represent a circle.

Let x = \((-27)^{ \frac { 1 }{ 3 } }=3^{ 3\times \frac { 1 }{ 3 } }(-1)^{ \frac { 1 }{ 3 } }\)
= 3\((cos\pi +isin\pi )^{ \frac { 1 }{ 3 } }\)
= 3\(\left[ cos\frac { 1 }{ 3 } (2k\pi +\pi )+isin\frac { 1 }{ 3 } (2k\pi +\pi ) \right] \), k = 0, 1, 2..
∴ The roots of -27 are
When k = 0, 3\(\left[ cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } \right] =3\ c\ is\ \frac { \pi }{ 3 } \)
When k = 1, -3
When k = 2, 3 c is\(\frac { 5\pi }{ 3 } \).

-1 = i² = i \(\times\) i =\(\sqrt { -1 } \times \sqrt { -1 } =\sqrt { (-1) } \times \sqrt { (-1) } \)
= \(\sqrt { 1 } \)
⇒ -1 = i
In the above proof we have used \(\sqrt { -1 } \times \sqrt { -1 } \)
= \(\sqrt { (-1)(-1) } \) which is wrong
Since \(\sqrt { ab } =\sqrt { a } .\sqrt { b } \) is true only at least one of a and b is non-negative.

Consider the equation |z+2−i| = 2.
This can be written as |z−(−2+i) | = 2.
The above equation represents the circle with centre z₀ = -2+i and radius r = 2. Therefore z + 2−i<2 represents all points inside the circle with centre at −2+i and radius 2 as shown in figure.

Area of the triangle formed by the vertices z, iz and Z+ iz is 50 sq. units
Let z = x + iy
Then iz = i(x + iy) = ix + i²y = -y + ix
z + iz = x + iy-y + ix
= (x - y) + i(x + y)
If A denotes the area of the triangle formed by z, iz and z + iz, then 
A = \(\frac { 1 }{ 2 }\ \left| \begin{matrix} x & y & 1 \\ x-y & x+y & 1 \\ -y & x & 1 \end{matrix} \right| \)
R₂ ⟶ R₂-R₁-R₃, we get
A = \(\frac { 1 }{ 2 }\ \left| \begin{matrix} x & y & 1 \\ 0 & 0 & -1 \\ -y & x & 1 \end{matrix} \right| \)
Expanding along R₂ we get
A = \(\frac { 1 }{ 2 }\ \left[ +1\left| \begin{matrix} x & y \\ -y & x \end{matrix} \right| \right] =\frac { 1 }{ 2 } \)(x²+y²)
Given A = 50 sq units
∴ 50 = \(\frac{1}{2}\)(x²+y²) ⇒ 100 = x²+ y²
Then \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { 100 } \) = 10
∴ |z| = 10  [∵ |z| = \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)]
AIiter :
Given area of triangle = 50 sq. unit
\(\frac{1}{2}\left|\begin{array}{ccc} x & y & 1 \\ -x-y & x+y & 1 \\ -y & x & 1 \end{array}\right|=50\)
\(\stackrel{R_{2} \rightarrow R_{2}-R_{3}}{\rightarrow} \frac{1}{2}\left|\begin{array}{ccc} x & y & 1 \\ 0 & 0 & -1 \\ -y & x & 1 \end{array}\right|=50\)
\(\left.\frac{1}{2}\left[\begin{array}{cc} x & y \\ -y & x \end{array}\right]\right]\) = 50
\(\frac{1}{2}\left[x^{2}+y^{2}\right]=50\)
\(x^{2}+y^{2}=100\)
\(|z|^{2}=100\)
|z| = 10

|z²-3| ≤ |z²|+|-3| [Triangle law of inequality]
≤ |z|²+3≤1+3  [∴ |z| = 1]
|z²-3| ≤ 4 ..............(1)
Also, |z²- 3| ≥ ||z²|-|-3||
≥ ||z|²-3|   [∵ |-3| = 3]
≥ |1²-3|  [∵ |z| = 1]
≥ |-2| .
|z²-3| ≥ 2.............(2)
From (1) and (2) we get 2 ≤ |z²-3| ≤ 4
Hence proved.

It is enough to prove that the sides of the triangle are equal.
Let z₁ = 1, \({ z }_{ 2 }=\frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \) and \({ z }_{ 3 }=\frac { -1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \)
The length of the sides of the triangles are
\(\left| { z }_{ 1 }-{ z }_{ 2 } \right| =\left| 1-\left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) \right| =\left| \cfrac { 3 }{ 2 } -\cfrac { \sqrt { 3 } }{ 2 } i \right| =\sqrt { \frac { 9 }{ 4 } +\frac { 3 }{ 4 } } =\frac { 2\sqrt { 3 } }{ 2 } =\sqrt { 3 } \)
\(\left\lfloor { z }_{ 2 }-{ z }_{ 3 } \right\rfloor =\left| \left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) -\left( \frac { -1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \right) \right| =\sqrt { \left( \sqrt { 3 } \right) ^{ 2 } } =\sqrt { 3 } \)
\(\left| { z }_{ 3 }-{ z }_{ 1 } \right| =\left| \left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) -1 \right| =\left| \frac { -3 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } i \right| =\sqrt { \frac { 9 }{ 4 } +\frac { 3 }{ 4 } } =\sqrt { 3 } \)
Since the sides are equal, the given points form an equilateral triangle

We consider \(\frac { 1+i }{ 1-i } =\frac { \left( 1+i \right) \left( 1+i \right) }{ \left( 1-i \right) \left( 1+i \right) } =\frac { 1+2i }{ 1+1 } =\frac { 2i }{ 2 } =i\)
and \(\frac { 1-i }{ 1+i } =\left( \frac { 1+{ i } }{ 1-i } \right) ^{ -1 }=\frac { 1 }{ i } =-i\)
Therefore,\(\left( \frac { 1+i }{ 1-i } \right) ^{ 3 }-\left( \frac { 1-i }{ 1-i } \right) ^{ 2 }\)= i³-(-i)³ = - i - i = -2i

Given z₁ = 2 + 5i, z₂ = -3 - 4i and z₃ = 1 + i
Additive inverse of z₁ is
-z₁ = -(2 + 5i)
= -2 - 5i
Multiplicative inverse of z₁ is
\(\frac { 1 }{ { z }_{ 1 } } =\frac { 1 }{ 2+5i } \times \frac { 2-5i }{ 2-5i } \)
[Multiply and divide by the conjugate of denominator]
= \(\frac { 2-5i }{ { 2 }^{ 2 }-(5i)^{ 2 } } =\frac { 2-5i }{ 4-25^{ 2 } } =\frac { 2-5i }{ 4+25 } \)
(z₁)^-1 = \(\frac { 1 }{ 29 } \)(2- 5i)   [∴ i² = -1]
Additive inverse of z₂ is
-z₂ = -(3 - 4i)
= 3 + 4i
Multiplicative inverse of z₂ is
\(\frac { 1 }{ z_{ 2 } } =\frac { 1 }{ -3-4i } \times \frac { -3+4i }{ -3+4i } \)
= \(\frac { -3+4i }{ (-3)^{ 2 }-(4i)^{ 2 } } \)
= \(\frac { -3+4i }{ 9-16i^{ 2 } } =\frac { -3+4i }{ 9+16 } \)
(z₂)^-1 = \(\frac { 1 }{ 25 } \)(-3 + 4i)
Additive inverse of z₃ is
-z₃ = -(1 + i)
= -1- i
Multiplicative inverse of z₃ is
\(\frac { 1 }{ { z }_{ 3 } } =\frac { 1 }{ 1+i } \times \frac { 1-i }{ 1-i } =\frac { 1-i }{ { 1 }^{ 2 }-({ i }^{ 2 }) } \)
= \(\frac { 1-i }{ 1+i } \)
(z₃)^-1 \(=\frac { 1 }{ 2 } \)(1 - i)

Given (3 -i) x - (2 - i) y + 2i + 5
= 2x + (-1 + 2i) y + 3 + 2i
⇒ 3x - ix - 2y + iy + 2i + 5 = 2x - y + 2iy + 3 + 2i
choosing the real and imaginary parts
(3x-2y + 5) + i (-x + y + 2) = 2x - y + 3 + i (2y+ 2)
Equating the real and imaginary parts both sides, we get
3x- 2y+ 5 = 2x-y+3
⇒ 3x - 2y + 5 - 2x +y - 3 = 0
⇒ x-y = -2... (1)
-x+y+2 = 2y+2
⇒ -x+y+2-2y-2 = 0
⇒ -x-y = 0 ⇒ x+y = 0.. (2)
(1)-(2) we get,

y = 1
Substituting y = 1 in (2) we get.
x+1 = 0 ⇒ x = -1
∴ x = -1 and y = 1

Let z₁ = (2+i)x + (1−i)y + 2i−3 = (2x+y−3) + i(x−y+ 2)and
z₂ = x+(−1+2i)y+1+i = (x−y+1) + i(2y+1)
Given that z₁ = z₂ 
Therefore (2x+y−3) + i(x−y+2) = (x−y+1) + i(2y+1).
Equating real and imaginary parts separately, gives
2x+y−3 = x−y+1 \(\Rightarrow\) x+2y = 4
x−y+2 = 2y +1 \(\Rightarrow\) x−3y = −1
Solving the above equations, gives
x = 2 and y = 1.

Comparing the two given lines with
\(\vec { r } =\vec { a } +t\vec { b } ,\vec { r } =\vec { c } +s\vec { d } \)
we have, \(\vec { a } =-\hat { -1 } -3\hat { j } -5\hat { k } ,\vec { b } =3\hat { i } +5\hat { j } +7\hat { k } ,\vec { c } =2\hat { i } +4\hat { j } +6\hat { k } \) and \(\vec { d } =\hat { i } +4\hat { j } +7\hat { k } \)
We know that the two given lines are coplar, if \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )\)=0
Here, \(\vec { b } \times \vec { d } \left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{matrix} \right| =7\hat { i } -14\hat { j } +7\hat { k } \) and \(\vec { c } -\vec { a } =3\hat { i } +7\hat { j } +11\hat { k } \)
Then, \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )=(3\hat { i } +7\hat { j } +11\hat { k } )(7\hat { i } -14\hat { j } +7\hat { k } )\)
Therefore the two given lines are coplanar.Then we find the non parametric form of vector equation of the plane containing the two given coplanar lines. We know that the plane containing the two given coplanar lines is
\((\vec { r } -\vec { a } ).(\vec { b } \times \vec { d } )\)=0
which implies that \((\vec { r } -(-\hat { i } -3\hat { j } -5\hat { k } )).(7\hat { i } -14\hat { j } +7\hat { k } )\)=0. Thus, the required non-parametric vector equation of the plane containing the two given coplanar lines is \(\vec { r } .(\hat { i } -2\hat { j } +\hat { k } )\)=0.

Let 2-2i = r(cosθ + isinθ)
r =\(\sqrt { { 2 }^{ 2 }+(-2)^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 } \)
The principal value α =tan^-1\(\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { -z }{ z } \right| |\)
= tan^-1(1) = \(\frac { \pi }{ 4 } \)
Since the complex number 2 - 2i lies in the quadrant
θ = -α = -\(\frac { \pi }{ 4 } \)
∴ 2-2i = \(2\sqrt { 2 } \left[ cos\left( -\frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \right] ^{ \frac { 1 }{ 3 } }\)
∴ \((2\sqrt { 2 } )^{ \frac { 1 }{ 3 } }\left[ cos\left( -\frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \right] ^{ \frac { 1 }{ 3 } }\)
= \(8^{ \frac { 1 }{ 6 } }\left[ cos\frac { 1 }{ 3 } \left( 2k\pi -\frac { \pi }{ 4 } \right) +isin\frac { 1 }{ 3 } \left( 2k\pi -\frac { \pi }{ 4 } \right) \right] \)
k = 0, 1, 2
The roots are
∴ When k = 0, \(8^{ \frac { 1 }{ 6 } }cis\left( -\frac { \pi }{ 12 } \right) \)
when k = 1, \(8^{ \frac { 1 }{ 6 } }cis\left( \frac { 7\pi }{ 12 } \right) \)
when k = 2, \(8^{ \frac { 1 }{ 6 } }cis\left( \frac { 15\pi }{ 12 } \right) \)
∴ The product of the root
= \(8^{ \frac { 1 }{ 6 } }cis\left( -\frac { \pi }{ 12 } +\frac { 7\pi }{ 12 } +\frac { 15\pi }{ 12 } \right) \)
= \(8^{ \frac { 1 }{ 6 } }cis\left( \frac { 21\pi }{ 12 } \right) =8^{ \frac { 1 }{ 6 } }cis\left( \frac { 7\pi }{ 12 } \right) \)
= \(8^{ \frac { 1 }{ 6 } }cis\left( 2\pi -\frac { \pi }{ 4 } \right) =8^{ \frac { 1 }{ 6 } }cis\left( -\frac { \pi }{ 4 } \right) \)
= \(8^{ \frac { 1 }{ 6 } }\left[ cos\left( -\frac { \pi }{ 4 } \right) +isin\left( -\frac { \pi }{ 4 } \right) \right] \)
= \(8^{ \frac { 1 }{ 6 } }\left[ cos\left( \frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \right] \)
= \(8^{ \frac { 1 }{ 6 } }\left[ \frac { 1 }{ \sqrt { 2 } } -\frac { i }{ \sqrt { 2 } } \right] =2^{ 3\times \frac { 1 }{ 6 } }\left( \frac { 1-i }{ \sqrt { 2 } } \right) =2^{ 1/2 }\left( \frac { 1-i }{ \sqrt { 2 } } \right) \)
= 1-i

LHS = \(\left( \frac { i+\sqrt { 3 } }{ -i+\sqrt { 3 } } \right) ^{ 2\omega }+\left( \frac { i-\sqrt { 3 } }{ i+\sqrt { 3 } } \right) ^{ 2\omega }\)
= \(\left( \frac { \sqrt { 3 } +i }{ \sqrt { 3 } -i } \times \frac { \sqrt { 3 } +i }{ \sqrt { 3 } +i } \right) ^{ 2\omega }+\left( \frac { -\sqrt { 3 } +i }{ \sqrt { 3 } +i } \times \frac { \sqrt { 3 } -i }{ \sqrt { 3 } -1 } \right) ^{ 2\omega }\)
= \(\left( \frac { 3-1+2\sqrt { 3 } i }{ 3+1 } \right) ^{ 2\omega }+\left( \frac { -3+1+2\sqrt { 3 } i }{ 3+1 } \right) ^{ 2\omega }\)
= \(\left( \frac { 2+2\sqrt { 3 } i }{ 4 } \right) ^{ 2\omega }+\left( \frac { -2+2\sqrt { 3 } i }{ 4 } \right) ^{ 2\omega }\)
= \(\left( \frac { 1+\sqrt { 3 } i }{ 2 } \right) ^{ 2\omega }+\left( \frac { -2+2\sqrt { 3 } i }{ 2 } \right) ^{ 2\omega }\)
=\(\left[ -\left( \frac { -1-\sqrt { 3 } i }{ 2 } \right) \right] ^{ 2\omega }+\left[ \frac { -1+\sqrt { 3 } i }{ 2 } \right] ^{ 2\omega }\)
= (-ω²)^2ω+(ω)^2ω
[∴ ω = \(\frac { -1+i\sqrt { 3 } }{ 2 } \), ω² = \(\frac { -1-i\sqrt { 3 } }{ 2 } \)]
= ω^4ω+ω^2ω
= (ω³)¹³³. ω¹ + (ω³)⁶⁶.ω²
= 1.ω+1.ω²  [∴ 1+ω+ω² = 0 & ω³ = 1]
= ω + ω²
= -1 = RHS

(1+5ω²+ω⁴)(1+5ω+ω²)(5+ω+ω²)
= (1+5ω²+ω)(1+5ω+ω²)(5+ω+ω²)
[∴ ω⁴ = ω³.ω¹= ω]
= (1+ω+5ω²)(1+ω²+5ω)(5+ω+ω²)
= (-ω²+5ω²)(-ω+5ω)(5-1)
= (4ω²)(4ω)(4) = 64 ω³
= 64(1) = 64   [∴ ω³ = 1]
RHS
Hence proved

Let \(z=cos2\theta +isin2\theta \)
As |z| = |z|² = z\(\bar { z } \) = 1, we get \(\bar { z } =\frac { 1 }{ z } =cos2\theta -isin2\theta \)
Therefore, \(\frac { 1+cos2\theta +isin2\theta }{ 1+cos2\theta -isin2\theta } =\frac { 1+z }{ 1+\frac { 1 }{ z } } =\frac { \left( 1+z \right) z }{ z+1 } =z\)
Therefore, \(\left( \frac { 1+cos2\theta +isin2\theta }{ 1+cos2\theta -isin2\theta } \right) ^{ 30 }={ z }^{ 30 }=\left( cos2\theta +isin2\theta \right) ^{ 30 }\)
= \(cos60\ \theta +isin60\ \theta \)

We have to find \((\sqrt{3}+1)^{\frac{1}{3}}\). Let \(z=(\sqrt{3}+i)^{\frac{1}{3}}\). Then \({ z }^{ 3 }=\sqrt { 3 } +i=r\left( cos\theta +isin\theta \right) \)
Then, \(r=\sqrt { 3+1 } =2\) and \(\alpha =\theta =\frac { \pi }{ 6 } \) (\(\because \sqrt{3}+i\) lies in the first quadrant)
Therefore, \({ z }^{ 3 }=\sqrt { 3 } +i=2\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \)
\(\Rightarrow z=\sqrt [ 3 ]{ 2 } \left( cos\left( \frac { \pi +12k\pi }{ 18 } \right) +isin\left( \frac { \pi +12k\pi }{ 18 } \right) \right) \), k = 0, 1, 2.
Taking k = 0, 1, 2, we get
k = 0, z \(={ 2 }^{ \frac { 1 }{ 3 } }\left( cos\frac { \pi }{ 18 } +sin\frac { \pi }{ 18 } \right) \)
k = 1, \(z={ z }^{ \frac { 1 }{ 3 } }\left( cos\frac { \pi }{ 18 } +sin\frac { \pi }{ 18 } \right) \)
k = 2, \(z={ 2 }^{ \frac { 1 }{ 3 } }\left( cos\frac { 25\pi }{ 18 } +sin\frac { 25\pi }{ 18 } \right) ={ 2 }^{ \frac { 1 }{ 3 } }\left( -cos\frac { 7\pi }{ 18 } -sin\frac { 7\pi }{ 18 } \right) \)