\(\left( -\sqrt { 3 } +3i \right) ^{ 31 }\)
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,
\(\left( -\sqrt { 3 } +3i \right) ^{ 31 }=\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) ^{ 31 }\)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\left( 20\pi +\frac { 2\pi }{ 3 } \right) +isin\left( 20\pi +\frac { 2\pi }{ 3 } \right) \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\left( \pi -\frac { \pi }{ 3 } \right) +isin\left( \pi -\frac { \pi }{ 3 } \right) \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( -cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } \right) =\left( 2\sqrt { 3 } \right) ^{ 31 }\left( -\frac { 1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) \).

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 \(-\sqrt { 3 } -i\)

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, \(z={ 2 }^{ \frac { 1 }{ 3 } }\left( cos\frac { 25\pi }{ 18 } +sin\frac { 25\pi }{ 18 } \right) ={ 2 }^{ \frac { 1 }{ 3 } }\left( -cos\frac { 7\pi }{ 18 } -sin\frac { 7\pi }{ 18 } \right) \)

|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\(\sqrt { 3 } \)

(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
\(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }\left( cos\frac { 18\pi }{ 4 } +isin\frac { 18\pi }{ 4 } \right) \)
= \({ 2 }^{ 9 }\left( cos\left( 4\pi +\frac { \pi }{ 2 } \right) +isin\left( 4\pi +\frac { \pi }{ 2 } \right) \right) ={ 2 }^{ 9 }\left( cos\frac { \pi }{ 2 } +isin\frac { \pi }{ 2 } \right) \)
\(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }\left( cos\frac { 18\pi }{ 4 } +isin\frac { 18\pi }{ 4 } \right) \)
= \(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }(i)=512i\)