⇒ 3x + (2x - 3y)i = 6 + 3i⁹
⇒ 3x + (2x - 3y)i = 6 + 3 . i⁴ . i⁴ . i¹
⇒ 3x + (2x - 3y)i = 6 + 3i
Equating the real and imaginary parts we get,
3x = 6 ⇒ x = 2
2x - 3y = 3 ⇒ 2(2) - 3y = 3
⇒ 4 - 3y = 3
⇒ 4 - 3 = 3y
⇒ 3y = 1 ⇒ y = \(\frac{1}{3}\)
∴ x = 2, y = \(\frac{1}{3}\).

Let z = -2
z = 2(-1) = 2(cos π + i sin π)
∴ arg(z) = π

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

Given z = 5i¹¹ + 7i³
= 5i⁴ . i⁴ . i² . i¹ + 7. i² . i¹
= 5(1)(1)(-1)(i) + 7(-1)(i)
= -5i - 7i = -12i
∴ Re(z) = 0 and In(z) = -12

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

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.

i²i³...i⁴⁰ = i^1+2+3...+40 = \(i^{\frac{40 \times 41}{2}}=i^{820}=i^{0}=1 .\)

|3z-6+12i| = 8
⇒ 3|z-2+4i| = 8
⇒ |z-(2 - 4i) = \(\frac{8}{3}\).
It is of the form |z - z₀| = r and so it represents a circle.
Its centre is (2 - 4i) and radius is \(\frac{8}{3}\).

Let z = -6+8i
|z| =\(\sqrt { (-6)^{ 2 }+8^{ 2 } } \)
= \(\sqrt { 36+64 } =\sqrt { 100 } \) = 10
\(\sqrt { a+ib } =\pm \left( \sqrt { \frac { |z|+a }{ 2 } } +i\frac { b }{ |b| } \sqrt { \frac { |z|-a }{ 2 } } \right) \)
[Here |z| = 10, a = -6, b = 8]
\(\sqrt { -6+8i } \pm \left( \sqrt { \frac { 10-6 }{ 2 } } +i\frac { 8 }{ |8| } \sqrt { \frac { 10+6 }{ 2 } } \right) \)
= \(\pm \left( \sqrt { \frac { 4 }{ 2 } } +i\sqrt { \frac { 16 }{ 2 } } \right) \)
= \(\pm (\sqrt { 2 } +i\sqrt { 8 } )\)
= \(\\ \pm (\sqrt { 2 } +i2\sqrt { 2 } )\)
Aliter :
Square root of -6 + 8i
Let a + ib = - 6 + 8i
a = -6, b = 8
\(|z|=\sqrt{6^{2}+8^{2}}=\sqrt{100}=10\)
\(\sqrt{a+i b}=\pm\left[\sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}+i \frac{b}{|b|} \sqrt{\frac{\sqrt{a^{2}+b^{2}}-a}{2}}\right]\)
\(=\pm\left[\sqrt{\frac{10-6}{2}}+i \sqrt{\frac{10+6}{2}}\right]\)
\(=\pm[\sqrt{2}+i \quad 2 \sqrt{2}]\)

\(\cfrac { 10-5i }{ 6+2i } \)
\(\frac { 10-5i }{ 6+2i } \times \frac { 6-2i }{ 6-2i } \)
[Multiply and divided by the conjugate of the denominator]
= \(\frac { 60-20i-30i+10{ i }^{ 2 } }{ { 6 }^{ 2 }-(2i)^{ 2 } } =\frac { 60-50i-10 }{ 36+4 } \)
= \(\frac { 50-50i }{ 40 } =\frac { 50(1-i) }{ 40 } \)
\(=\frac { 5 }{ 4 } \)(1 - i)

\(Re\left( \frac { 1 }{ z } \right) \)
= \(Re\left( \frac { 1 }{ x+iy } \right) =Re\left( \frac { 1 }{ x+iy } \times \frac { x-iy }{ x-iy } \right) \)
= \(Re(\left( \frac { x-iy }{ { x }^{ 2 }({ i }^{ 2 }{ y }^{ 2 }) } \right) =Re\left( \frac { x-iy }{ { x }^{ 2 }+{ y }^{ 2 } } \right) \)
∴ Real part is \(\frac { x }{ { x }^{ 2 }+{ y }^{ 2 } } \)

\(1+i\sqrt { 3 } \)
\(r=||z|=\sqrt { { 1 }^{ 2 }+\left( \sqrt { 3 } \right) ^{ 2 } } \)
|\(\theta ={ tan }^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right) =\frac { \pi }{ 3 } \)
Hence \(ang(z)=\frac { \pi }{ 3 } \)
Therefore, the polar form of \(1+i\sqrt { 3 } \) can be written as
\(1+i\sqrt { 3 } =2\left( cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } \right) \)
\(=2\left( cos\left( \frac { \pi }{ 3 } +2k\pi \right) +isin\left( \frac { \pi }{ 3 } +2k\pi \right) \right) ,k\varepsilon z\).

\(3-i\sqrt { 3 } \)
Let x + iy = \(3-i\sqrt { 3 } \)
= r(cos θ + i sin θ)
r = \(\\ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { 3^{ 2 }+(\sqrt { 3 } )^{ 2 } } =\sqrt { 9+3 } \)
= \(\sqrt { 12 } =2\sqrt { 3 } \)
α = \(tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { -\sqrt { 3 } }{ 3 } \right| =tan^{ -1 }\left| \frac { 1 }{ \sqrt { 3 } } \right| =\frac { \pi }{ 6 } \)
Since the complex number \(3-i\sqrt { 3 } \) lies in the IV quadrant, [∵ x ⟶ +ve y ⟶ -ve]
Its principal value θ = -α
⇒ θ = \(\frac { \pi }{ 6 } \)
∴ Its polar form is
\(3-i\sqrt { 3 } \) = 2\(\sqrt { 3 } \)\(\left[ cos\left( 2k\pi -\frac { \pi }{ 6 } \right) +isin\left( 2k\pi -\frac { \pi }{ 6 } \right) \right] ,k\in Z\)

\(\overline { z } \) = z-1
⇒ \(\overline { z } \) =\(\frac{1}{z}\)
⇒ z\(\overline { z } \) = 1
⇒ |z|² = 1
⇒ x² + y² = 1 which is the required Cartesian equation.
Aliter : 
\( \bar{z} =z^{-1} \)
\(x-i y =\frac{1}{x+i y} \)
\(x-i y =\frac{1}{x+i y} \times \frac{x-i y}{x-i y} \)
\(x-i y =\frac{x-i y}{x^{2}+y^{2}} \)
\(x^{2}+y^{2} =1\)

Let −1−i = \(r(cos\ \theta +i\ sin\ \theta )\)
We have r = \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 1+1 } =\sqrt { 2 } \)
\(\alpha =tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }1=\frac { \pi }{ 4 } \)
Since the complex number −1−i lies in the third quadrant, it has the principal value,
\(\theta =\alpha -\pi =\frac { \pi }{ 4 } -\pi =-\frac { 3\pi }{ 4 } \)
Therefore,\(-1-i=\sqrt { 2 } \left( cos\left( \frac { 3\pi }{ 4 } \right) +isin\left( \frac { 3\pi }{ 4 } \right) \right) \)
= \(\sqrt { 2 } \left( cos\frac { 3\pi }{ 4 } -isin\frac { 3\pi }{ 4 } \right) \)
\(-1-i=\sqrt { 2 } \left( cos\left( \frac { 3\pi }{ 4 } +2k\pi \right) -isin\left( \frac { 3\pi }{ 4 } +2k\pi \right) \right) \)
Depending upon the various values of k , we get various alternative polar forms.