Given 4x + i (3x - y) = 3 - 6i
Equating real and imaginary parts
\(4 x=3 \Rightarrow x=\frac{3}{4}\)
3x-y = -6
\(3\left(\frac{3}{4}\right)-y=-6\)
\(y=\frac{9}{4}+6=\frac{33}{4}\)

\(\mathrm{i}^{2}+\mathrm{i}^{4}+\mathrm{i}^{6}+\ldots(2 \mathrm{n}+1)\)
= (- 1) +1 - 1+ ... (2n+ 1) terms = -1

\(\frac{1}{i}-\frac{1}{i^{2}}+\frac{1}{i^{3}}-\frac{1}{i^{4}}=\frac{1}{i}+1-\frac{1}{i}-1\)
= 0

(i) \( i^{4 n+1} =(i)^{4 n+i} \)
\( =i^{4 n} i=\left(i^{4}\right)^{n} i\)
= (1)ⁿ i = i
(ii) \( \mathrm{i}^{3}+\mathrm{i}^{-3} =\mathrm{i}^{3}+\left(\mathrm{i}^{3}\right)^{-1} \)
\( =-\mathrm{i}+(-\mathrm{i})^{-1} \)
\( =-\mathrm{i}-\frac{1}{i}\)
= -i + i = 0

\( i^{57} =i^{56} i^{1}=\left(i^{4}\right)^{14} i \)
\( =(1)^{14} i=i \)
\(i^{125} =i^{124} i \)
\( =\left(i^{4}\right)^{31} i=(1)^{31} \cdot i=i\)
\( \therefore i^{57}+\frac{1}{i^{125}} =i+\left(\frac{1}{i} \times \frac{i}{i}\right) \)
\({\left[\because i^{2}=-1, i^{4}\right.} =1] \)
\( =i+\frac{i}{i^{2}} \)
= i - i = 0

⇒ 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 = \(\frac{1}{3}\).

|(1+3i)³| = |1+3i|³ = \(\left[ \sqrt { { 1 }^{ 2 }+{ 3 }^{ 2 } } \right] ^{ 3 }\) =\(\left( \sqrt { 10 } \right) ^{ 3 }\)
=\((\sqrt { 10 } )^{ 3 }=\sqrt { 10 } \times \sqrt { 10 } \times \sqrt { 10 } \times \sqrt { 10 } =10\sqrt { 10 } \).

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

i²⁵ = (i⁴)⁶ \(\times\) i¹ = i⁶ \(\times\) i = i
∴ |i²⁵| = |i| = 1

Z = \(\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 107 }+\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 107 }\)
\(\bar { z } =\left( \overline { \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } } \right) ^{ 107 }+\left( \overline { \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } } \right) ^{ 107 }\)
= \(\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 107 }+\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 107 }\) = z
Since z = \(\bar { z } \), Im(z) = 0

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

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

Let z₁ = a+ib and z₂ = c+id
Given |z₁| = Iz₂|
⇒ \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } } =\sqrt { { c }^{ 2 }+{ d }^{ 2 } } \)
Squaring both sides we get, a² + b² = c² + d²
This cannot imply that a = c and b = d
∴ z₁ and z₂ need not be equal