i¹⁷²⁹ = i¹⁷²⁸ i¹ = i

\(\theta =\frac { 3\pi }{ 2 } \)
When θ = \(\frac { 3\pi }{ 2 } \)
Rotation of z is \(ze^{ i\theta }=ze^{ i\frac { 3\pi }{ 2 } }\)
= \(2\sqrt { 2 } e^{ -\frac { \pi }{ 4 } }.e^{ i\frac { 3\pi }{ 2 } }=2\sqrt { 2 } e^{ i\left( i\frac { \pi }{ 2 } -\frac { \pi }{ 4 } \right) }\)
= \(2\sqrt { 2 } e^{ i5^{ \frac { \pi }{ 4 } } }\)

(z+w)²
= [(5-2i)+(-1+ 3i)]² = (4+ i)²
= 16 + i²+ 8i =16 -1+8i
= 15+ 8i

Represent z, iz and z + iz in the Argand diagram.
z = 2 + 3i can be represented as (2,3)
iz = i(2 + 3i)
= 2i + 3i²
= 2i-3
= -3 + 2i can be represented as (-3, 2)
z + iz = 2 + 3i - 3 + 2i = -1 + 5i can be represented as (-1, 5) in the argand diagram.

\(\frac { cos\frac { \pi }{ 6 } -isin\frac { \pi }{ 6 } }{ 2\left( cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } \right) } \)
=\(\frac { 1 }{ 2 } \left[ \frac { cos\left( \frac { -\pi }{ 6 } \right) +isin\left( \frac { -\pi }{ 6 } \right) }{ cos\left( \frac { \pi }{ 3 } \right) +isin\left( \frac { \pi }{ 3 } \right) } \right] \)
[∵ cos(-θ)=cosθ and sin(-θ)=-sinθ
\(=\frac { 1 }{ 2 } \left[ cos\left( \frac { -\pi }{ 6 } -\frac { \pi }{ 3 } \right) +isin\left( \frac { -\pi }{ 6 } -\frac { \pi }{ 3 } \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ cos\left( \frac { -\pi -2\pi }{ 6 } \right) +isin\left( \frac { -\pi -2\pi }{ 6 } \right) \right] \)
\(\frac { 1 }{ 2 } \left[ cos\left( \frac { -3\pi }{ 6 } \right) +isin\left( -\frac { 3\pi }{ 6 } \right) \right] \)
\(\\ \frac { 1 }{ 2 } \left[ cos\left( -\frac { \pi }{ 2 } \right) +isin\left( -\frac { \pi }{ 2 } \right) \right] \\ \)
\(\frac { 1 }{ 2 } \left[ cos\left( \frac { \pi }{ 2 } \right) -isin\left( \frac { \pi }{ 2 } \right) \right] \)
\(\frac { 1 }{ 2 } [0-i(1)]=\frac { -i }{ 2 } \).

|z+i| = |z-1|
⇒ |x + iy +i| = |x + iy-1|
⇒ |x + i(y + 1)| = |(x - 1) + iy|
⇒ \(\sqrt { { x }^{ 2 }+(y+1)^{ 2 } } =\sqrt { (x-1)^{ 2 }+y^{ 2 } } \)
⇒ x² + (y + 1)² =(x- 1)² + y²
[ squaring both sides]
\(\Rightarrow \not x^{2}+ \not y^{2}+2 y+ \not1= \not x^{2}-2x+ \not 1+ \not y^2\)
⇒ 2y + 2x = 0
⇒ x + y = 0
Hence, the Cartesian  equation is x + y = 0

Since z is purely imaginary
z = -\(\overline { z } \)
∴ 2n\(\left[ cos\frac { n\pi }{ 6 } +isin\frac { n\pi }{ 6 } \right] \)
= -2n\(\left[ cos\frac { n\pi }{ 6 } +isin\frac { n\pi }{ 6 } \right] \) [From (1) & (2)]
\(\Rightarrow \cos \frac{n \pi}{6}+ i{\not \sin \frac{m \pi}{6}}=-\cos \frac{n \pi}{6}+i {\not\sin \frac{m \pi}{6}}\)
⇒ 2cos\(\frac { n\pi }{ 6 } \) = 0
⇒ \(cosn\frac { \pi }{ 6 } =0=cos\frac { \pi }{ 2 } \)
[∴ cos \(\frac { \pi }{ 2 } \) = 0]

⇒ n = \(\frac{6}{2}\)
⇒ n = 3
Hence z is purely imaginary

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

\(\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \left( cos\frac { \pi }{ 12 } +isin\frac { \pi }{ 12 } \right) \)
[∵ (cosθ₁+isinθ₁)(cosθ₂+isonθ₂)
= cos(θ₁+θ₂)+isin(θ₁+θ₂)
= \(cos\left( \frac { 2\pi +\pi }{ 12 } \right) +isin\left( \frac { 2\pi +\pi }{ 12 } \right) \)
\(cos\left( \frac { 3\pi }{ 12 } \right) +isin\left( \frac { 3\pi }{ 12 } \right) \)
\(cos\left( \frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \)
\(\frac { 1 }{ \sqrt { 2 } } +i\frac { 1 }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \)(1+i)
Aliter:
\( \left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right) \)
\( =\cos \left(\frac{\pi}{6}+\frac{\pi}{12}\right)+i \sin \left(\frac{\pi}{6}+\frac{\pi}{12}\right) \)
\( =\left(\cos \frac{3 \pi}{12}+i \sin \frac{3 \pi}{12}\right) \)
\( =\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \)
\( =\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}=\frac{1+i}{\sqrt{2}} \)

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