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 \(1^{\frac{1}{3}}\). 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 _{\({ e }^{ i\frac { 2\pi }{ 3 } }=\frac { -1+i\sqrt { 3 } }{ 2 } \)}

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 \(-\sqrt { 3 } -i\)

Now, \(\frac { z-1 }{ z+1 } =\frac { x+iy-1 }{ x+iy+1 } =\frac { \left( x-1 \right) +iy }{ \left( x+1 \right) +iy } =\frac { \left[ \left( x-1 \right) +iy \right] \left[ \left( x+1 \right) -iy \right] }{ \left[ \left( x+1 \right) +iy \right] \left[ \left( x+1 \right) -iy \right] } \)
\(\Rightarrow \frac { z-1 }{ z+1 } =\frac { \left( { x }^{ 2 }+{ y }^{ 2 }-1 \right) +i\left( 2y \right) }{ \left( x+1 \right) ^{ 2 }+{ y }^{ 2 } } \)
Since, arg \(\left( \frac { z-1 }{ z+2 } \right) =\frac { \pi }{ 2 } \Rightarrow { tan }^{ -1 }\left( \frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 }-1 } \right) \)= \(\frac { \pi }{ 2 } \)
\(\Rightarrow \frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 }-1 } =tan\frac { \pi }{ 2 } \) ⇒ x²+ y² − 1 = 0
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }=1\)

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 
\( \cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=3 \cos (\alpha+\beta+\gamma) \)
\( \sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma=3 \sin (\alpha+\beta+\gamma) \)