(1+ω)(1+ω²)(1+ω⁴)(1+ω⁸).........(1+ω²)¹¹ = 1
LHS = (1+ω)(1+ω²)(1+ω⁴)(1+ω⁸)...........2(1+ω²)¹¹
\((1+{ \omega }^{ 2^{ 0 } })^{ 6 }((1+{ \omega }^{ 2^{ 1 } })(1+{ \omega }^{ { 2 }^{ 2 } })((1+{ \omega }^{ { 2 }^{ 11 } })\)
[∵ ω⁴ = ω³.ωω⁸ = ω⁶.ω²]
There are 12 terms
= (1+ω)(1+ω²)(1+ω)(1+\({ \omega }^{ { 2 }^{ n } }\)).......... upto 12 terms
= (1+ω)⁶(1+ω²)⁶(-ω)⁶(-ω²)⁶
= (ω³)6 = 16 = 1 = RHS.

Consider \(\frac { 19-7i }{ 9+i } =\frac { 19-7i }{ 9+i } \times \frac { 9-i }{ 9-i } \)
= \(\frac { 171-19i-63i+7i^{ 2 } }{ (9)^{ 2 }-{ i }^{ 2 } } =\frac { 171-82i-7 }{ 81+1 } \)
= \(\frac { 164-82i }{ 82 } =\frac { 82(2-i) }{ 82 } \) = 2- i
Also \(\frac { 20-5i }{ 7-6i } =\frac { 20-5i }{ 7-6i } \times \frac { 7+6i }{ 7+6i } \)
= \(\frac { 140+120i-35i-30^{ 2 } }{ { 7 }^{ 2 }-(6i)^{ 2 } } \)
\(=\frac{140+85 i+30}{49+36}=\frac{170+85 i}{85}=\frac{\not 85(2+i)}{\not 85}=2+i\)
∴ \(\left( \frac { 19-7i }{ 9+i } \right) ^{ 12 }+\left( \frac { 20-5i }{ 7-6i } \right) ^{ 12 }\) = (2-i)12+(2+i)12
Let z = (2-i)¹²+(2+i)¹²
∴ \(\overline { z } \) = \(\overline { (2-i)^{ 12 }+(2+i)^{ 12 } } =\overline { (2-i)^{ 12 } } +\overline { (2+i)^{ 12 } } \)
[∵ \(\overline { { z }_{ 1 }+{ z }_{ 2 } } =\overline { { z }_{ 1 } } +\overline { { z }_{ 2 } } \)]
= (2 + i)¹²+(2 - i)¹²= z
∴ \(\overline { z } \) = z ⇒ z is purely real
∴ \(\left( \frac { 19-7i }{ 9+i } \right) ^{ 12 }+\left( \frac { 20-5i }{ 7-6i } \right) ^{ 12 }\) is real.

(1+i)¹⁸
Let 1+ i = \(r(cos\theta +isin\theta )\). Then , we get
\(r=\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 2 } ;\alpha ={ tan }^{ -1 }\left( \frac { 1 }{ 1 } \right) =\frac { \pi }{ 4 } \)
\(\theta =\alpha =\frac { \pi }{ 4 } \) (\(\because\) 1+i lies in the first Quadrant)
Therefore 1+ i = \(\sqrt { 2 } \left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \)
Raising the power 18 on both sides
\(\left( 1+i \right) ^{ 18 }=\left[ \sqrt { 2 } \left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \right] ^{ 18 }=\sqrt { 12 } ^{ 18 }\left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \)
By de Moivre’s theorem
\(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }\left( cos\frac { 18\pi }{ 4 } +isin\frac { 18\pi }{ 4 } \right) \)
= \({ 2 }^{ 9 }\left( cos\left( 4\pi +\frac { \pi }{ 2 } \right) +isin\left( 4\pi +\frac { \pi }{ 2 } \right) \right) ={ 2 }^{ 9 }\left( cos\frac { \pi }{ 2 } +isin\frac { \pi }{ 2 } \right) \)
\(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }\left( cos\frac { 18\pi }{ 4 } +isin\frac { 18\pi }{ 4 } \right) \)
= \(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }(i)=512i\)

\(\frac { 2\left( cos\frac { 9\pi }{ 4 } +isin\frac { 9\pi }{ 4 } \right) }{ 4\left( cos\left( \frac { -3\pi }{ 2 } + \right) sin\left( \frac { -3\pi }{ 2 } \right) \right) } \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { 9\pi }{ 4 } -\left( \frac { -3\pi }{ 2 } \right) \right) +isin\left( \frac { 9\pi }{ 4 } -\left( \frac { -3\pi }{ 2 } \right) \right) \right) \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { 9\pi }{ 4 } +\frac { 3\pi }{ 2 } \right) +isin\left( \frac { 9\pi }{ 4 } +\frac { 3\pi }{ 2 } \right) \right) \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { 15\pi }{ 4 } \right) +isin\left( \frac { 15\pi }{ 4 } \right) \right) =\frac { 1 }{ 2 } \left( cos\left( 4\pi -\frac { \pi }{ 4 } \right) +isin\left( 4\pi -\frac { \pi }{ 4 } \right) \right) \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { \pi }{ 4 } \right) -isin\left( \frac { \pi }{ 4 } \right) \right) =\frac { 1 }{ 2 } \left( \frac { 1 }{ \sqrt { 2 } } +i\frac { 1 }{ \sqrt { 2 } } \right) \)
\(\frac { 2\left( cos\frac { 9\pi }{ 4 } +isin\frac { 9\pi }{ 4 } \right) }{ 4\left( cos\left( \frac { -3\pi }{ 2 } + \right) sin\left( \frac { -3\pi }{ 2 } \right) \right) } \) = \(\frac { 1 }{ 2\sqrt { 2 } } -i\frac { 1 }{ 2\sqrt { 2 } } =\frac { \sqrt { 2 } }{ 4 } +i\frac { \sqrt { 2 } }{ 4 } \) Which is in rectangular form.

(1-ω+ω²)⁶ + (1+ω-ω²)⁶ = 128
LHS = (1-ω+ω²)6 + (1+ω-ω²)⁶
= (1+ω²-ω)⁶ +(-ω²+ω²)⁶
[∵ 1+ω+ω² = 0
⇒ 1+ω = -ω²
⇒ 1+ω² = -ω
= (-ω-ω)⁶ + (-2ω²)⁶
= (-2ω)⁶ + (-2ω²)⁶
= 2⁶.ω⁶ + 2⁶.ω¹²
= 2⁶[(ω³)² +(ω³)⁴]
= 2⁶[1+1]    [∵ ω³ = 1]
2⁶ \(\times\) 2¹ = 2⁷ = 128 = RHS

Let z = x + iy
Then \(\frac { 1+z }{ 1-z } \) = cos2θ + i sin 2θ
⇒ \(\frac { 1+x+iy }{ 1-x-iy } \) = cos 2θ + i sin 2θ ....(1)
Taking modulus,
\(\left| \frac { 1+x+iy }{ 1-x-iy } \right| \) = |cos 2θ+i sin 2θ| ⇒ \(\frac { |1+x+iy| }{ |1-x-iy| } \)
=\(\sqrt { cos^{ 2 }2\theta +sin^{ 2 }2\theta } \) = 1
⇒ |1 + x + iy| = |1 - x - iy|
⇒ \(\sqrt { (1+x)^{ 2 }+{ y }^{ 2 } } =\sqrt { (1+x)^{ 2 }+{ y }^{ 2 } } \)
⇒ (1 + x)²+ y² = (1-x)²+ y²

⇒ 4x = 0 ⇒ x = 0
From (1) \(\frac { (1+x)+iy }{ (1-x)-iy } \times \frac { (1-x)+iy }{ (1-x)+iy } \)
= cos2θ + isin 2θ
Choosing the imaginary part alone we get,
\(\frac { y(1+x)+y(1-x) }{ (1-x)^{ 2 }+{ y }^{ 2 } } \)= sin 2θ

\(\frac { 2y }{ 1+y^{ 2 } } \) = sin 2θ
⇒ \(\frac { 2tan\theta }{ 1+tan^{ 2 }\theta } \) = sin 2θ
∴ y must be equal to tan θ
⇒ y = tan θ
z = x+ iy
∴ z = 0 + i tan θ
⇒ z = tan θ

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

Let z =\(\left( \frac { 19+9i }{ 5-3i } \right) ^{ 15 }-\left( \frac { 8+i }{ I+2i } \right) ^{ 15 }\)
Here, \(\frac { 19+9i }{ 5-3i } =\frac { (19+9i)(5+3i) }{ (5-3i)(5+3i) } \)
= \(\frac { (95-27)+i(45+57) }{ { 5 }^{ 2 }+{ 3 }^{ 2 } } =\frac { 68+102i }{ 34 } \)
= 2 + 3i ................(1)
and \(\frac { 8+i }{ 1+2i } =\frac { (8+i)(1-2i) }{ (1+2i)(1-2i) } \)
= \(\frac { (8+2)+i(1-16) }{ { 1 }^{ 2 }+{ 2 }^{ 2 } } =\frac { 10-15i }{ 5 } \)
= 2 - 3i .............. (2)
Now z =\(\left( \frac { 19+9i }{ 5-3i } \right) ^{ 15 }-\left( \frac { 8+i }{ 1+2i } \right) ^{ 15 }\)
⇒ z = (2 + 3i)¹⁵ - (2 - 3i)¹⁵   (by (1) and (2))
Then by definition, \(\bar { z } =\left( \overline { (2+3i)^{ 15 }-(2-3i)^{ 15 } } \right) \)
= \(\left( \overline { 2+3i } \right) ^{ 15 }-\left( \overline { 2-3i } \right) ^{ 15 }\) (using properties of conjugates)
= (2 - 3i)¹⁵ - (2 + 3i)¹⁵ = -((2 + 3i)¹⁵ - (2-3i)¹⁵)
⇒ \(\\ \overline { z } \) = -z
Therefore, \(\left( \frac { 19+9i }{ 5-3i } \right) ^{ 15 }-\left( \frac { 8+i }{ 1+2i } \right) ^{ 15 }\) is purely imaginary.

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 = \(\left( 2+i\sqrt { 3 } \right) ^{ 10 }+\left( 2-i\sqrt { 3 } \right) ^{ 10 }\)
\(\bar { z } =\overline { \left( 2+i\sqrt { 3 } \right) +\left( 2-i\sqrt { 3 } \right) ^{ 10 } } \)
= \(\overline { \left( 2+i\sqrt { 3 } \right) ^{ 10 } } +\overline { \left( 2+i\sqrt { 3 } \right) ^{ 10 } } \)    \( \left(\because \overline{z_{1}+z_{2}}=\overline{z_{1}}+\overline{z_{2}}\right) \left(\because \overline{\left(z^{n}\right)}=(\bar{z})^{n}\right) \)
= \(\overline { \left( 2+i\sqrt { 3 } \right) ^{ 10 } } +\left( \overline { 2-i\sqrt { 3 } } \right) ^{ 10 }\)
= \(\left( 2-i\sqrt { 3 } \right) ^{ 10 }+\left( \overline { 2+i\sqrt { 3 } } \right) ^{ 10 }=z\)
\(\overline { z } \) = z \(\Rightarrow\) z is real

\(\overline { (2-i)^{ 12 }+(2+i)^{ 12 } } =\overline { (2-i)^{ 12 } } +\overline { (2+i)^{ 12 } } \)
\(\left( 2+i\sqrt { 3 } \right) ^{ 10 }-\left( 2-i\sqrt { 3 } \right) ^{ 10 }\)
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

 \(\left( -\sqrt { 3 } +3i \right) ^{ 31 }\)
Let \(-\sqrt { 3 } +3i=r\left( cos\theta +isin\theta \right) \). Then, we get
\(r=\sqrt { \left( -\sqrt { 3 } \right) ^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 12 } =2\sqrt { 3 } \)
\(\alpha ={ tan }^{ -1 }\left| \frac { 3 }{ -\sqrt { 3 } } \right| ={ tan }^{ -1 }\sqrt { 3 } =\frac { \pi }{ 3 } \)
\(\theta =\pi -\alpha =\pi -\frac { \pi }{ 3 } =\frac { 2\pi }{ 3 } \) (\(\because\)  \(\sqrt { 3 } +3i\) lies in II Quadrant)
Therefore,\(-\sqrt { 3 } +3i=2\sqrt { 3 } \left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) \)
Raising power 31 on both sides,
\(\left( -\sqrt { 3 } +3i \right) ^{ 31 }=\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) ^{ 31 }\)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\left( 20\pi +\frac { 2\pi }{ 3 } \right) +isin\left( 20\pi +\frac { 2\pi }{ 3 } \right) \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\left( \pi -\frac { \pi }{ 3 } \right) +isin\left( \pi -\frac { \pi }{ 3 } \right) \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( -cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } \right) =\left( 2\sqrt { 3 } \right) ^{ 31 }\left( -\frac { 1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) \).

We have to find \(1^{\frac{1}{3}}\). Let \(z=1^{\frac{1}{3}}\) then z³ =1 .
In polar form, the equation z³ = 1 can be written as
\(z^{3}=\cos (0+2 k \pi)+i \sin (0+2 k \pi)=e^{i 2 k \pi}\), k = 0, 1, 2, ....
Therefore \(z =\cos \left(\frac{2 k \pi}{3}\right)+i \sin \left(\frac{2 k \pi}{3}\right)=e^{i \frac{2 k \pi}{3}}\), k = 0, 1, 2.
Taking k = 0, 1, 2 , we get,
k = 0, z = cos 0 + isin 0 = 1.
k = 1, \(z=cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } =cos\left( \pi -\frac { \pi }{ 3 } \right) +isin\left( \pi -\frac { \pi }{ 3 } \right) \)
= \(-cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } =\frac { 1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \)
k = 2, \(z=cos\frac { 4\pi }{ 3 } +isin\frac { 4\pi }{ 3 } =cos\left( \pi +\frac { \pi }{ 3 } \right) +isin\left( \pi +\frac { \pi }{ 3 } \right) \)
= \(-cos\frac { \pi }{ 3 } -isin\frac { \pi }{ 3 } =-\frac { 1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \)
Therefore, the cube roots of unity are
\(1,\frac { -i+i\sqrt { 3 } }{ 2 } ,\frac { -1-i\sqrt { 3 } }{ 2 } \Rightarrow 1\) and \({ \omega }^{ 2 }\) where _{\({ e }^{ i\frac { 2\pi }{ 3 } }=\frac { -1+i\sqrt { 3 } }{ 2 } \)}

We have to find \(1^{\frac{1}{4}}\). Let z⁴ = \(1^{\frac{1}{4}}\). Then z⁴ = 1.
In polar form, the equation z = 1 can be written as
\(z^4=cos\left( 0+2k\pi \right) +isin\left( 0+2k\pi \right) ={ e }^{ i2k\pi }\), k = 0, 1, 2,...
Therefore,\({ \left( z \right) }^{ \frac { 1 }{ 4 } }=cos\left( \frac { 2k\pi }{ 4 } \right) +isin\left( \frac { 2k\pi }{ 4 } \right) ={ e }^{ i\frac { 2k\pi }{ 4 } }\), k=0,1,2,3.
Taking k = 0, 1, 2, 3, we get
k = 0, z = cos 0 + isin 0 = 1
k = 1, \(z=cos\left( \frac { \pi }{ 2 } \right) +isin\left( \frac { \pi }{ 2 } \right) =i\)
k = 2, \(z=cos\pi +isin\pi =-1\)
k = 3, \(z=cos\frac { 3\pi }{ 2 } +isin\frac { 3\pi }{ 2 } =-cos\frac { \pi }{ 2 } -isin\frac { \pi }{ 2 } =-i\)
Fourth roots of unity are 1, i, −1, −i \(\Rightarrow\) 1, \(\omega \), \({ \omega }^{ 2 }\) and \({ \omega }^{ 3 }\), where \(\omega ={ e }^{ i\frac { 2\pi }{ 4 } }=i\)

Let \({ z }^{ 3 }+8i=0\)
\(\Rightarrow\) z³ = -8i
= \(8(-i)=8\left( cos\left( -\frac { \pi }{ 2 } +2k\pi \right) isin\left( -\frac { \pi }{ 2 } +2k\pi \right) \right) \),k\(\in Z\)
\(z=\sqrt [ 3 ]{ 8 } \left( cos\left( \frac { -\pi +4k\pi }{ 6 } \right) +isin\left( \frac { -\pi +4k\pi }{ 6 } \right) \right) \)
Taking k = 0, 1, 2 we get,
k = 0, \(z=2\left( cos\left( -\frac { \pi }{ 6 } \right) +isin\left( -\frac { \pi }{ 6 } \right) \right) =2\left( -\frac { 1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \right) =2\left( \frac { \sqrt { 3 } }{ 2 } -i\frac { 1 }{ 2 } \right) \)
k = 1, \(z=2\left( cos\left( \frac { \pi }{ 2 } \right) +isin\left( \frac { \pi }{ 2 } \right) \right) =2=\left( 0+i \right) =0+2i=2i\)
k = 2,\(z=2\left( xcos\left( \frac { 7\pi }{ 6 } \right) +isim\left( \frac { 7\pi }{ 6 } \right) \right) =2\left( cos\left( \pi +\frac { \pi }{ 6 } \right) \right) +isin\left( \pi +\frac { \pi }{ 6 } \right) \)
= \(2\left( -cos\left( \frac { \pi }{ 6 } \right) -isin\left( \frac { \pi }{ 6 } \right) \right) =2\left( -\frac { \sqrt { 3 } }{ 2 } -i\frac { 1 }{ 2 } \right) =-\sqrt { 3 } -i\)
The values of z are \(\sqrt { 3 } -i,2i\) and \(-\sqrt { 3 } -i\)

|z| = 2 represents the circle with centre (0, 0) and radius 2.
Let A, B, and C be the vertices of the given triangle. Since the vertice z₁, z_2,and z₃  form an equilateral triangle inscribed in the circle|z| = 2, the sides of this triangle AB, BC, and CA subtend \(\frac { 2\pi }{ 3 } \) radians (120 degree) at the origin (circumcenter of the triangle).
(The complex number ze¹⁶ is a rotation of z by \(\theta\) radians in the counter clockwise direction about the origin.)
Therefore, we can obtain z₂ and z₃ and by the rotation of z₁ by \(\frac { 2\pi }{ 3 } and\ \frac { 4\pi }{ 3 } \) respectively.
Given that \(\vec { OA } ={ z }_{ 1 }=1+i\sqrt { 3 } \)
\(\vec { OB } ={ z }_{ 1 }e^{ i\frac { 2\pi }{ 3 } }=\left( 1+i\sqrt { 3 } \right) e^{ i\frac { 2\pi }{ 3 } }\)
= \(\left( 1+i\sqrt { 3 } \right) \left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) \)
= \(\left( 1+i\sqrt { 3 } \right) \left( -\frac { 1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) =-2;\)
\(\vec { OC } ={ z }_{ 1 }{ e }^{ i\frac { 4\pi }{ 3 } }={ z }_{ 2 }{ e }^{ i\frac { 2\pi }{ 3 } }=-2e3^{ i\frac { 2\pi }{ 3 } }\)
= \(-2\left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) \)
= \(-2\left( -\frac { 1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) =1-i\sqrt { 3 } \)
Therefore, z₂ = -2 and z₃ = 1-i\(\sqrt { 3 } \)

LHS = \(\left( \cfrac { 1+sin\frac { \pi }{ 10 } +icos\frac { \pi }{ 10 } }{ 1+sin\frac { \pi }{ 10 } -icos\frac { \pi }{ 10 } } \right) ^{ 10 }\)
Let z = \(sin\frac { \pi }{ 10 } +icos\frac { \pi }{ 10 } \)
∴ \(\frac { 1 }{ z } =sin\frac { \pi }{ 10 } -icos\frac { \pi }{ 10 } \)
∴ LHS =\(\left[ \frac { 1+z }{ 1+\frac { 1 }{ z } } \right] ^{ 10 }=\left[ \frac { 1+z }{ \frac { z+1 }{ z } } \right] ^{ 10 }\)= z¹⁰
= \(\left[ sin\frac { \pi }{ 10 } +icos\frac { \pi }{ 10 } \right] ^{ 10 }\)
= \({ i }^{ 10 }\left[ cos\frac { \pi }{ 10 } -isin\frac { \pi }{ 10 } \right] ^{ 10 }\)
= \({ i }^{ 10 }\left[ cos10\frac { \pi }{ 10 } -isin10\frac { \pi }{ 10 } \right] \) [By De Moivers theorum]
= i¹⁰[cos π- i sin π] = -1(-1-i(0)) = 1
Aliter: 
Let \(z=\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}\)
\(\frac{1}{z}=\sin \frac{\pi}{10}-i \cos \frac{\pi}{10}\)
\(\left[\frac{1+\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}}{1+\sin \frac{\pi}{10}-i \cos \frac{\pi}{10}}\right]^{10}=\left[\frac{1+z}{1+1 / z}\right]\)
\( =\left[\frac{(1+z)}{(z+1)} \cdot z\right]^{10} =z^{10} \)
\( =\left(\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}\right)^{10} \)
\( =\left[\cos \left(\frac{\pi}{2}-\frac{\pi}{10}\right)+i \sin \left(\frac{\pi}{2}-\frac{\pi}{10}\right)\right]^{10} \)
\( =\left[\cos \frac{4 \pi}{10}+i \sin \frac{4 \pi}{10}\right]^{10} \)
\( =\cos \frac{4 \pi}{10}(10)+i \sin \frac{4 \pi}{10}(10) \)
\( =\cos 4 \pi+i \sin 4 \pi \)   [\( \because\) By de Moiwe's theorem]  
= 1+ i(0) = 1

Given z = x + iy and arg\(\left( \frac { z-i }{ z+2 } \right) =\frac { \pi }{ 4 } \)
⇒ arg(z-i) - arg(z+2) = \(\frac { \pi }{ 4 } \)
⇒ arg(x + iy-i) - arg(x+iy+2) = \(\frac { \pi }{ 4 } \)
⇒ arg(x+i(y-1)-arg((x+2)+iy) = \(\frac { \pi }{ 4 } \)
⇒ \(tan^{ -1 }\left( \frac { y-1 }{ x } \right) -tan^{ -1 }\left( \frac { y }{ x+2 } \right) \) = \(\frac { \pi }{ 4 } \)
⇒ \(tan^{ -1 }\left( \frac { \frac { y-1 }{ x } -\frac { y }{ x+2 } }{ 1+\frac { y-1 }{ x } .\frac { y }{ x+2 } } \right) \)
= \(\frac { \pi }{ 4 } \)\(\left[ \because tan^{ -1 }x-tan^{ -1 }y=tan^{ -1 }\left( \frac { x-y }{ 1+xy } \right) \right] \)
\(\Rightarrow \frac{\left(\frac{(x+2)(y-1)- x y}{\not {x (\not x+\not2)}}\right)}{\left(\frac{x(x+2)+y(y-1)}{\not x(\not x+\not 2)}\right)}=\tan \frac{\pi}{4}=1\)
⇒ \(\frac { (x+2)(y-1)-xy }{ x(x+2)+y(y-1) } \) = 1
⇒ -x + 2y-2 = x²+ 2x + y²-y
⇒ x² + 2x + y²-y + x-2y + 2 = 0
⇒ x² + y²+3x-3y + 2 = 0
Hence proved.

Given cos α + cos β + cos \(\gamma\) = sin α + sin β + sin \(\gamma\)
∴ (cos α + cos β + cos \(\gamma\)) + i(sin α + sin β + sin \(\gamma\)) = 0
⇒ (cos α + i sin α) + (cos β + i sin β) + (cos \(\gamma\)+i sin \(\gamma\)) = 0
⇒ a + b + c = 0 where a = cos α + i sin α, b = cos β + i sin β, c = cos \(\gamma\) + i sin\(\gamma\)
If a + b + c = 0, then a³+b³+c³ = 3abc
∴ (cos α + i sin α)³ + (cos β + i sin β)³ + (cos \(\gamma\) + i sin \(\gamma\))³ = 3[ (cos α + i sin α) + (cos β + i sin β) + (cos \(\gamma\) + i sin \(\gamma\))
= 3[(cos(α + β + \(\gamma\)) + i sin(α + β + \(\gamma\))]
⇒ (cos 3α + cos β + cos \(\gamma\)) + i[sin 3α + sin 3β + sin 3\(\gamma\))]
= 3(cos(α + β + \(\gamma\)) + i sin(α + β + \(\gamma\))
Equating the real and imaginary parts, we get 
\( \cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=3 \cos (\alpha+\beta+\gamma) \)
\( \sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma=3 \sin (\alpha+\beta+\gamma) \)

Given z = x + iy
Im \(\left( \frac { 2z+1 }{ iz+1 } \right) \)= 0
⇒ Im\(\left( \frac { 2(x+iy)+1 }{ i(x+iy)+1 } \right) \)= 0
⇒ Im\(\left( \frac { (2x+1)+2iy }{ ix+i^{ 2 }y+1 } \right) \)
⇒ Im\(\left( \frac { (2x+1)+2iy }{ ix-y+1 } \right) \)
\(\left( \frac { (2x+1)+iy }{ (1-y)+ix } \right) \)
Multiply and divide by the conjugate of the denominator
We get Im\(\left( \frac { (2x+1)+2iy }{ (1-y)+ix } \times \frac { (1-y)-ix }{ (1-y)-ix } \right) \)=0
⇒ Im\(\left( \frac { (2x+1)+2iy\times (1-y)-ix }{ (1-y)^{ 2 }+{ x }^{ 2 } } \right) \)
Choosing the imaginably part we get,
\(\frac { (2x+1)(-x)+2y(1-y) }{ (1-y)^{ 2 }+{ x }^{ 2 } } \)
⇒ (2x+1)-x+2y(1-y) = 0
⇒ -2x²-x+2y-2y² = 0
⇒ 2x²+2y²+x-2y = 0
Hence, locus of z is 2x²+2y²+x-2y = 0

Given |z₁| = 1, |z₂|= 2, |z₃| = 3, |z₁ + z₂ +  z₃| = 1
|z₁|² = 1² ⇒ z₁ \(\overline { { z }_{ 1 } } \) = 1 ⇒ z₁ = \(\frac { 1 }{ { z }_{ 1 } } \)
|z₂|² = 4 ⇒ z₂ \(\overline { { z }_{ 2 } } \) = 1 ⇒ z₂ = \(\frac { 4 }{ { z }_{ 2 } } \)
|z₃|² = 9 ⇒ z₃ \(\overline { { z }_{ 3 } } \) = 1 ⇒ z₃ = \(\frac { 9 }{ { z }_{ 3 } } \)
∴ \(\left| 9,\frac { 1 }{ \overline { { z }_{ 1 } } } .\frac { 4 }{ \overline { { z }_{ 2 } } } +4.\frac { 1 }{ \overline { { z }_{ 1 } } } .\frac { 9 }{ \overline { { z }_{ 3 } } } +\frac { 4 }{ \overline { { z }_{ 2 } } } .\frac { 9 }{ \overline { z_{ 3 } } } \right| \)
\(\left| \frac { 36 }{ \overline { { z }_{ 1 } } \overline { { z }_{ 2 } } } +\frac { 36 }{ \overline { { z }_{ 1 } } \overline { { z }_{ 3 } } } +\frac { 36 }{ \overline { { z }_{ 2 } } \overline { { z }_{ 3 } } } \right| =\left| 36\left( \frac { \overline { { z }_{ 3 } } +\overline { { z }_{ 2 } } +\overline { { z }_{ 1 } } }{ \overline { { z }_{ 1 } } \overline { { z }_{ 2 } } \overline { { z }_{ 3 } } } \right) \right| \)
\(\\ \left[ \because |\overline { z_{ 1 } } +\overline { { z }_{ 2 } } +\overline { { z }_{ 3 } } |=|\overline { { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } } | \right] \)
=\(\frac { 36|\overline { { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } } | }{ |\overline { { z }_{ 1 } } ||\overline { { z }_{ 2 } } ||\overline { { z }_{ 3 } } | } =36\frac { |\overline { \overline { { z }_{ 1 } } +\overline { z_{ 2 } } +\overline { z_{ 3 } } | } }{ |\overline { { z }_{ 1 } } ||\overline { { z }_{ 2 } } ||\overline { { z }_{ 3 } } | } \)
\(\left[ \because |\overline { { z }_{ 1 } } |=|{ z }_{ 1 }|,|\overline { { z }_{ 2 } } =|{ z }_{ 21 }|,|\overline { { z }_{ 3 } } |=|\overline { { z }_{ 3 } } | \right] \)
\(=\frac{36(1)}{1(2)(3)}=\frac{\not 36}{\not 6}=6\)
∴ |9z₁ + z₂ + 4z₁z₃ + z₂z₃| = 6

Given that \(\left| { z }_{ 1 } \right| =\left| { z }_{ 2 } \right| =\left| { z }_{ 3 } \right| =r\Rightarrow { z }_{ 1 }\bar { { z }_{ 1 } } ={ z }_{ 2 }\bar { { z }_{ 2 } } ={ r }^{ 2 }\)
\(\Rightarrow { z }_{ 1 }=\frac { { r }^{ 2 } }{ \bar { { z }_{ 1 } } } ,{ z }_{ 2 }=\frac { { r }^{ 2 } }{ \bar { { z }_{ 2 } } } ,{ z }_{ 3 }=\frac { { r }^{ 2 } }{ { \bar { z } }_{ 3 } } \)
Therefore \({ z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 }=\frac { { r }^{ 2 } }{ { \bar { z } }_{ 1 } } +\frac { { r }^{ 2 } }{ \bar { { z }_{ 2 } } } +\frac { { r }^{ 2} }{ \bar { { z }_{ 3 } } } \)
= \({ r }^{ 2 }\left( \frac { \bar { { z }_{ 2 } } \bar { { z }_{ 3 } } +\bar { { z }_{ 1 } } \bar { { z }_{ 3 } } +\bar { { z }_{ 1 } } { \overline { z } }_{ 2 } }{ \overline { { z }_{ 1 } } \bar { { z }_{ 2 } } \bar { { z }_{ 3 } } } \right) \) 
\(\left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right| =\left| { r }^{ 2 } \right| \left| \frac { \overline { { z }_{ 2}{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } } }{ \overline { { z }_{ 1 }{ z }_{ 2 }{ z }_{ 3 } } } \right| \) \(\left(\because \bar{z}_{1}+\bar{z}_{2}=\overline{z_{1}+z_{2}}\right)\)
= \({ r }^{ 2 }\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ \left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \left| { z }_{ 3 } \right| } \)  \(\left( \because |z|=|\bar { z } |and\ \left| { z }_{ 1 }{ z }_{ 2 }{ z }_{ 3 } \right| =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \left| { z }_{ 3 } \right| \right) \)
= \(\left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right| ={ r }^{ 2 }\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ { r }^{ 3 } } =\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ r } \)
\(\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ \left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right| } \) = r   (given that \(z_{1}+z_{2}+z_{3} \neq 0\))
Thus,\(\left| \frac { { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } }{ { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } } \right| \) = r