Let x = \((-27)^{ \frac { 1 }{ 3 } }=3^{ 3\times \frac { 1 }{ 3 } }(-1)^{ \frac { 1 }{ 3 } }\)
= 3\((cos\pi +isin\pi )^{ \frac { 1 }{ 3 } }\)
= 3\(\left[ cos\frac { 1 }{ 3 } (2k\pi +\pi )+isin\frac { 1 }{ 3 } (2k\pi +\pi ) \right] \), k = 0, 1, 2..
∴ The roots of -27 are
When k = 0, 3\(\left[ cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } \right] =3\ c\ is\ \frac { \pi }{ 3 } \)
When k = 1, -3
When k = 2, 3 c is\(\frac { 5\pi }{ 3 } \).

-1 = i² = i \(\times\) i =\(\sqrt { -1 } \times \sqrt { -1 } =\sqrt { (-1) } \times \sqrt { (-1) } \)
= \(\sqrt { 1 } \)
⇒ -1 = i
In the above proof we have used \(\sqrt { -1 } \times \sqrt { -1 } \)
= \(\sqrt { (-1)(-1) } \) which is wrong
Since \(\sqrt { ab } =\sqrt { a } .\sqrt { b } \) is true only at least one of a and b is non-negative.

z₁(z₂ + z₃) = z₁z₂ + z₁z₃
Given z₁= 3, z₂ = -7i, z₃ = 5+4i
LHS = z₁(z₂ + z₃)
= 3 [-7i + 5 + 4i]
= 3[5-3i]
= 15-9i
RHS = z₁z₂ + z₁z₃
= 3(-7i) + 3(5 + 4i)
= -21i +15 +12i
= -9i +15
= 15-9i
LHS = RHS
∴ z₁(z₁ + z₃) = z₁z₂ + z₁z₃
Hence proved

(z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)
Given z₁= 1-3i, z₂ - 4i and z₃ = 5
LHS = (z₁+ z₂) + z₃
= [1- 3i + (- 4i)] + 5
[1-7i] + 5
= 6 -7i
RHS = z₁+ (z₂ + z₃)
= 1- 3i + (-4i + 5)
= 6 - 7i
LHS = RHS
∴ (z₁+ z₂)+ z₃ = z₁+(z₂+ z₃)

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 z₁ = (2+i)x + (1−i)y + 2i−3 = (2x+y−3) + i(x−y+ 2)and
z₂ = x+(−1+2i)y+1+i = (x−y+1) + i(2y+1)
Given that z₁ = z₂ 
Therefore (2x+y−3) + i(x−y+2) = (x−y+1) + i(2y+1).
Equating real and imaginary parts separately, gives
2x+y−3 = x−y+1 \(\Rightarrow\) x+2y = 4
x−y+2 = 2y +1 \(\Rightarrow\) x−3y = −1
Solving the above equations, gives
x = 2 and y = 1.

z₁z₂ = (1 - i)(-2 + 4i) = -2 + 4i + 2i - 4i²
= -2 + 6i + 4 = 2 + 6i
\(\bar { { z }_{ 1 } } \) = 1+i
∴ \(\frac { { z }_{ 1 }{ z }_{ 2 } }{ \bar { { z }_{ 1 } } } =\frac { 2+6i }{ 1+6i } \times \frac { 1-i }{ 1-i } =\frac { 2(1-i+3i-i^{ 2 }) }{ 1+1 } \)
= 1 + 2i + 3
= 4 + 2i
∴ Im\(\left( \frac { { z }_{ 1 }{ z }_{ 2 } }{ \bar { { z }_{ 1 } } } \right) \) = 2

(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

\(\left| \left( \overline { 1+i } \right) \left( 2+3i \right) \left( 4i-3 \right) \right| =\left| \left( \overline { 1+i } \right) \right| \left| 2+3i \right| \left| 4i-3 \right| \) (\(\because \) |z₁z₂z₃|=|z₁|z₂||z₃|)
= |1+i| |2+3i| |-3+4i|  \(\left( \because |z|=\left| \overline { z } \right| \right) \)
= \(\left( \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } \right) \left( \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 } } \right) \left( \sqrt { \left( 3 \right) ^{ 2 }+{ 4 }^{ 2 } } \right) \).
\(=(\sqrt{2})(\sqrt{13})(\sqrt{25})=5 \sqrt{26}\)

We have z = (2+3i)(1−i) = (2+3)+(3−2)i = 5+i
\(\Rightarrow\) \({ z }^{ -1 }=\frac { 1 }{ z } =\frac { 1 }{ 5+i } \)
Multiplying the numerator and denominator by the conjugate of the denominator, we get
\({ z }^{ -1 }=\frac { \left( 5-i \right) }{ \left( 5+i \right) \left( 5-i \right) } =\frac { 5-i }{ { 5 }^{ 2 }+{ I }^{ 2 } } =\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)
\(\Rightarrow\)\({ z }^{ -1 }=\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)

Using the given value for z₁ and z₂ the value of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } =\frac { 3-2i }{ 6+4 } =\frac { 3-2i }{ 6+4i } \times \frac { 6-4i }{ 6-4i } \)
= \(\frac { \left( 18-8 \right) +i\left( 12-12 \right) }{ { 6 }^{ 2 }+{ 4 }^{ 2 } } =\frac { 10-24i }{ 52 } =\frac { 10 }{ 52 } =\frac { 24i }{ 52 } \)
= \(\frac { 5 }{ 26 } -\frac { 6 }{ 13 } i\)

We have = \(\frac { z+3 }{ z-5i } =\frac { 1+4i }{ 2 } \)
\(\Rightarrow\) 2(z + 3) = (1 + 4i) (z− 5i)
\(\Rightarrow\) 2z + 6 = (1 + 4i)z + 20−5i
\(\Rightarrow\) (2−1−4i)z = 20− 5i− 6 
\(\Rightarrow\) \(z=\frac { 14-5i }{ 1-4i } =\frac { \left( 14-5i \right) \left( 1+4i \right) }{ \left( 1-4i \right) \left( 1+4i \right) } =\frac { 34+51i }{ 17 } =2+3i\)

(1+5ω²+ω⁴)(1+5ω+ω²)(5+ω+ω²)
= (1+5ω²+ω)(1+5ω+ω²)(5+ω+ω²)
[∴ ω⁴ = ω³.ω¹= ω]
= (1+ω+5ω²)(1+ω²+5ω)(5+ω+ω²)
= (-ω²+5ω²)(-ω+5ω)(5-1)
= (4ω²)(4ω)(4) = 64 ω³
= 64(1) = 64   [∴ ω³ = 1]
RHS
Hence proved