⇒ 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 = \(\frac{1}{3}\).

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

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

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

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.

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

 \(\sum _{ n=1 }^{ 102 }{ { i }^{ n } } \) = (i¹+i²+i³+i⁴)+(i⁵+i⁶+i⁷+i⁸)+....+(i⁹⁷+i⁹⁸+i⁹⁹+i¹⁰⁰)+i¹⁰¹+i¹⁰²
= (i¹+i²⁺i³+i⁴)+(i¹+i²+i³+i⁴)+...+(i¹+i²+i³+i⁴)+i^-1+i^-2
= {i+(-1)+(-i)+1}+{i+(-1)+(-i)}+......+{i+(-1)+(-i)+1}+i+(-1)
= 0+0+...0+i-1
= -1+i

\(\theta =\frac { 2\pi }{ 3 } \)
When θ = \(\frac { \pi }{ 3 } \)
Roration of z is \(ze^{ i\theta }=ze^{ i2\frac { \pi }{ 3 } }\) [using (1)]
= \(2\sqrt { 2 } e^{ -i\frac { \pi }{ 4 } }.{ e }^{ i2\frac { \pi }{ 3 } }\)   [using (1)]
= \(2\sqrt { 2 } e^{ i\left( i\frac { \pi }{ 3 } -\frac { \pi }{ 4 } \right) }=2\sqrt { 2 } e^{ i5\frac { \pi }{ 12 } }\)

|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 \(\frac{8}{3}\).

\(\left| 2z+2-4i \right| =2\)
2|z+1-2i| = 2
⇒ |z-(-1+2i)| = 1
It is of the form |z - z₀| = r and so it represents a circle.
Its centre is (-1+2i) and radius is 1.

\(\frac { 3 }{ 2 } \left( cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 6 } \right) .6\left( cos\frac { 5\pi }{ 6 } +\frac { 5\pi }{ 6 } \right) \)
= \(\left( \frac { 3 }{ 2 } \right) \left( 6 \right) \left( cos\left( \frac { \pi }{ 3 } +\frac { 5\pi }{ 6 } \right) +isin\left( \frac { \pi }{ 3 } +\frac { 5\pi }{ 6 } \right) \right) \)
= \(9\left( cos\left( \frac { 7\pi }{ 6 } \right) +isin\left( \frac { 7\pi }{ 6 } \right) \right) \)
= \(9\left( cos\left( \pi +\frac { \pi }{ 6 } \right) +isin\left( \pi +\frac { \pi }{ 6 } \right) \right) \)
= \(9\left( -cos\left( \frac { \pi }{ 6 } \right) -isin\left( \frac { \pi }{ 6 } \right) \right) \)
= \(9\left( -\frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) =\frac { 9\sqrt { 3 } }{ 2 } -\frac { 9i }{ 2 } \)

Using the given values for z₁, z₂ and z₃ we get |z₁| = |3+4i| =\(\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } =5\)
|z₂| = |5-12i| = \(\sqrt { { 5 }^{ 2 }+(-12)^{ 2 } } =13\)
|z₃| = |6+8i| = \(\sqrt { { 6 }^{ 2 }+{ 8 }^{ 2 } } =10\)
|z₁+z₂| = |(3+4i)+(5-12i)| = |8-8i| = \(\sqrt { 128 } =8\sqrt { 2 } \)
|z₂-z₃| = |(5-12i)-(6+8i)| = |1-20i| = \(\sqrt { 401 } \)
|z₁+z₃| = |(3+4i)+(6+8i)| = |9+12i| = \(\sqrt { 225 } =15\)
Note that the triangle inequality is satisfied in all the cases
|z₁+z₃| = |z₁|+|z₃| = 15 

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
\({ z }^{ -1 }=\frac { \left( 5-i \right) }{ \left( 5+i \right) \left( 5-i \right) } =\frac { 5-i }{ { 5 }^{ 2 }+{ I }^{ 2 } } =\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)
\(\Rightarrow\)\({ z }^{ -1 }=\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)

Let z = x+iy ⇒ \(\bar { z } \) = x+iy
∴ \(\\ \frac { \bar { z } +1 }{ z-1 } =\frac { z-iy+1 }{ x-iy-i } =\frac { (x+1)iy }{ x-i(y+1) }\)
= \(\frac { (x+1)-iy }{ x-i(y+1) } \times \frac { x+i(y+1) }{ x+i(y+1) } \)
Choosing the real part alone we get,
\(\frac { x(x+1)+y(y+1) }{ { x }^{ 2 }+(y+1)^{ 2 } } \) = 0
⇒ x(x+1) + y(y+1) = 0
⇒ x²+x+y²+y = 0 which is the locus of z.

AB = |(3+2i) - (0+5i)| = |3-3i|
=\(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
BC = |(0+5i) - (-3+2i)| = |3+3i|
= \(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
CD = |(-3+2i) - (0-i) = |-3+i|
= \(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
DA = |(0-i) - (3+2i)| = |-3-3i|
=\(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
∴ AB = BC = CD = DA
Also AC = |(3+2i) - (-3+2i)|
= |6| = \(\sqrt { 36 } \) = 6
∴ AC = BD
Hence ABCD is a square

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\(\frac { 5\pi }{ 3 } \).

Given 2cos α = x+\(\frac { 1 }{ x } \)
⇒ 2cos α = \(\frac { { x }^{ 2 }+1 }{ x } \)
⇒ x²+1 = 2xcos α
⇒ x²-2x cos α+1 = 0
⇒ \(\frac { 2cos\alpha \pm \sqrt { (-2cos\alpha )^{ 2 }-4(1)(1) } }{ 2 } \)
= \(\frac { 2cos\alpha \pm \sqrt { 4cos^{ 2 }\alpha -4 } }{ 2 } \)  \(\left[ \because \frac { b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
= \(\frac { 2cos\alpha \pm \sqrt { -sin^{ 2 }\alpha } }{ 2 } \)
= \(\frac { 2cos\alpha \pm isin\alpha }{ 2 } \) [∵ sin²α+cos²α = 1]
⇒ x² = cos α ± sin α
Also, 2cos β = y+\(\frac { 1 }{ y } \)
⇒ 2cos β = \(\frac { { y }^{ 2 }+1 }{ y } \)
⇒ y²-2y cos β+1 = 0
⇒ \(\frac { 2cos\beta \pm \sqrt { (-2cos^{ 2 }\beta ^{ 2 }-4(1)(1) } }{ 2 } \)
=\(\frac { 2cos\beta \pm \sqrt { 4cos^{ 2 }\beta } -4 }{ 2 } =\frac { 2cos\beta \pm 2isin\beta }{ 2 } \)
⇒ y = cosβ ± i sinβ
\({ x }^{ m }{ y }^{ n }+\cfrac { 1 }{ { x }^{ m }{ y }^{ n } } =2cos(m\alpha +n\beta )\)
x^myⁿ = (cos α + i sin mα) (cos nβ + i sin nβ)
cos(mα+nβ)+i sin(mα+nβ)
\(\frac { 1 }{ { x }^{ m }{ y }^{ n } } \) = cos(mα+nβ)-i sin(mα+nβ)

= 2cos(mα+nβ)

|z-4| = 16
Given z = x + iy
|z - 4| = 16
⇒ |x + iy - 4| = 16
⇒ |(x- 4) + iy| = 16
⇒ \(\\ \sqrt { (x-4)^{ 2 }+{ y }^{ 2 } } \) = 16
⇒ (x - 4)² + y² = 16²
[Squaring both sides]
⇒ x²-8x + 16 + y² = 256
⇒ x²-8x + y²+ 16-256 = 0
⇒ x²-8x + y²-240 = 0 Which is the required Cartesian equation.
The locus of the point is a circle.

Given z = x + iy
Consider \(\left| \frac { z-4i }{ z+4i } \right| =1\Rightarrow \left| \frac { x+iy-4i }{ x+iy+4i } \right| \)=1
⇒ \(\left| \frac { x+i(y-4) }{ x+i(y+4) } \right| \)
⇒ \(\frac { \sqrt { { x }^{ 2 }+(y-4)^{ 2 } } }{ \sqrt { { x }^{ 2 }+((y+4)^{ 2 } } } \) = 1
⇒ \(\sqrt { { x }^{ 2 }+(y-4)^{ 2 } } =\sqrt { { x }^{ 2 }+(y+4)^{ 2 } } \)
Squaring both sides we get,
x²+(y-4)² = x²+(y+4)²
\(\Rightarrow \not x^{2}+\not y^{2}-8 y+\not 16=\not x^{2}+\not y^{2}+8 y+\not 16\)
⇒ 8y+8y = 0
⇒ 16y = 0
⇒ y = 0    [∵ 16 ≠ 0]
y = 0 is the equation of real axis Locus of z is the real axis.

Given z³+2\(\overline { z } \) = 0
⇒ z³ = -2\(\overline { z } \)
Taking modulus, |z³| = |-2\(\overline { z } \)|
⇒ |z|³ = 2|z|
⇒ |z| [|z|2-2] = 0
|z| = 0 or |z|²-2 = 0
⇒ |z| = 0
⇒ z = 0 is a solution ....(1)
|z²| = 2
⇒ |z²| = 2 ⇒ (z\(\overline { z } \))² = 2
⇒ z\(\overline { z } \) = \(\sqrt { 2 } \Rightarrow \overline { z } \frac { \sqrt { 2 } }{ z } \)
Given z + 2\(\overline { z } \) = 0
⇒ z³+2\(\frac { \sqrt { 2 } }{ z } \) = 0
⇒ z⁴+2\(\sqrt { 2 } \) = 0
If has 4 non-zero solutions
Hence from (1) and (2), z³+2\(\overline { z } \) has 5 solutions.

Let the points be A (10 - 8i), B (11 + 6i) and C(1-i)
Distance between A and C is |(10-8i)-(1+i)|
= |10-8i-1-i| = |9-9i|
= \(\sqrt { { 9 }^{ 2 }+{ (-9) }^{ 2 } } =\sqrt { 81+81 } =\sqrt { 2\times 81 } =\sqrt { 162 } \)
= 9\(\sqrt{2}\)
Distance between Band C is s |(11+6i)-(1+i)|
= |11 + 6i-1-i| = |10 + 5i|
=\(\sqrt { { 10 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 100+25 } =\sqrt { 125 } \)
=\(\sqrt { 25\times 5 } =5\sqrt { 5 } \)
Since \(5\sqrt { 5 } <9\sqrt {2 } \) , B is closest to C.
∴ 11 + 6i is closet to 1 + i.

Given v = 3-4i, w = 4+3i and \(\frac { 1 }{ u } =\frac { 1 }{ v } +\frac { 1 }{ w } \)
∴ \(\frac { 1 }{ u } =\frac { 1 }{ 3-4i } +\frac { 1 }{ 4+3i } \)
= \(\frac { 3+4i }{ (3-4i)(3+4i) } +\frac { 4-3i }{ (4+3i)(4-3i) } \)
= \(\\ \frac { 3+4i }{ 9-(4i)^{ 2 } } +\frac { 4-3i }{ 16-(3i)^{ 2 } } =\frac { 3+4i }{ 9+16 } +\frac { 4-3i }{ 16+9 } \)
= \(\frac { 3+4i }{ 25 } +\frac { 4-3i }{ 25 } =\frac { 3+4i+4-3i }{ 25 } \)
\(\frac { 1 }{ u } =\frac { 7+i }{ 25 } \)
∴ u = \(\frac { 25 }{ 7+i } \times \frac { 7-i }{ 7-i } =\frac { 25(7-i }{ 7^{ 2 }-({ i }^{ 2 }) } \) 
= \(\frac { 25(7-i) }{ 49+1 } =\frac { 25(7-i) }{ 50 } =\frac { 1 }{ 2 } \)(7-i)
∴ u = \(\frac { 1 }{ 2 } \)(7-i) or \(\frac { 7 }{ 2 } \) - \(\frac { i }{ 2 } \)

Let z =\(\left( \frac { 19+9i }{ 5-3i } \right) ^{ 15 }-\left( \frac { 8+i }{ I+2i } \right) ^{ 15 }\)
Here, \(\frac { 19+9i }{ 5-3i } =\frac { (19+9i)(5+3i) }{ (5-3i)(5+3i) } \)
= \(\frac { (95-27)+i(45+57) }{ { 5 }^{ 2 }+{ 3 }^{ 2 } } =\frac { 68+102i }{ 34 } \)
= 2 + 3i ................(1)
and \(\frac { 8+i }{ 1+2i } =\frac { (8+i)(1-2i) }{ (1+2i)(1-2i) } \)
= \(\frac { (8+2)+i(1-16) }{ { 1 }^{ 2 }+{ 2 }^{ 2 } } =\frac { 10-15i }{ 5 } \)
= 2 - 3i .............. (2)
Now z =\(\left( \frac { 19+9i }{ 5-3i } \right) ^{ 15 }-\left( \frac { 8+i }{ 1+2i } \right) ^{ 15 }\)
⇒ z = (2 + 3i)¹⁵ - (2 - 3i)¹⁵   (by (1) and (2))
Then by definition, \(\bar { z } =\left( \overline { (2+3i)^{ 15 }-(2-3i)^{ 15 } } \right) \)
= \(\left( \overline { 2+3i } \right) ^{ 15 }-\left( \overline { 2-3i } \right) ^{ 15 }\) (using properties of conjugates)
= (2 - 3i)¹⁵ - (2 + 3i)¹⁵ = -((2 + 3i)¹⁵ - (2-3i)¹⁵)
⇒ \(\\ \overline { z } \) = -z
Therefore, \(\left( \frac { 19+9i }{ 5-3i } \right) ^{ 15 }-\left( \frac { 8+i }{ 1+2i } \right) ^{ 15 }\) is purely imaginary.

arg(1+i) + arg(1-i) = arg[(1+i) (1-i)]
LHS = arg (1+i) + arg(1-i)
1+i = \(\sqrt { 2 } \left( \frac { 1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } \right) \)
= \(\sqrt { 2 } \left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \)
∴ arg (1+i) = π/4
-1+i =\(\sqrt { 2 } \left( \frac { -1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } \right) \)
= \(\sqrt { 2 } \left( cos3\frac { \pi }{ 4 } +isin3\frac { \pi }{ 4 } \right) \)
∴ (-1+i) = 3\(\frac { \pi }{ 4 } \)
∴ LHS = \(\frac { \pi }{ 4 } +\frac { 3\pi }{ 4 } =\frac { 4\pi }{ 4 } =\pi \)
RHS = arg[(1+i) (-1+i)]
= arg[-1-i + i + i²]
= (-1-i + i-1) = arg(-2)
= arg(2) - (1) = 2 arg(-1)
= 2 (cos π + isin π) = π
∴ LHS = RHS

LHS = 2 arg (-1)
= 2 arg (cos π + i sin π) = 2π
RHS = arg (-1)² = arg (1)
= arg (cos θ + isin θ) = 0
∴ LHS ≠ RHS

LHS = \(\left( \frac { i+\sqrt { 3 } }{ -i+\sqrt { 3 } } \right) ^{ 2\omega }+\left( \frac { i-\sqrt { 3 } }{ i+\sqrt { 3 } } \right) ^{ 2\omega }\)
= \(\left( \frac { \sqrt { 3 } +i }{ \sqrt { 3 } -i } \times \frac { \sqrt { 3 } +i }{ \sqrt { 3 } +i } \right) ^{ 2\omega }+\left( \frac { -\sqrt { 3 } +i }{ \sqrt { 3 } +i } \times \frac { \sqrt { 3 } -i }{ \sqrt { 3 } -1 } \right) ^{ 2\omega }\)
= \(\left( \frac { 3-1+2\sqrt { 3 } i }{ 3+1 } \right) ^{ 2\omega }+\left( \frac { -3+1+2\sqrt { 3 } i }{ 3+1 } \right) ^{ 2\omega }\)
= \(\left( \frac { 2+2\sqrt { 3 } i }{ 4 } \right) ^{ 2\omega }+\left( \frac { -2+2\sqrt { 3 } i }{ 4 } \right) ^{ 2\omega }\)
= \(\left( \frac { 1+\sqrt { 3 } i }{ 2 } \right) ^{ 2\omega }+\left( \frac { -2+2\sqrt { 3 } i }{ 2 } \right) ^{ 2\omega }\)
=\(\left[ -\left( \frac { -1-\sqrt { 3 } i }{ 2 } \right) \right] ^{ 2\omega }+\left[ \frac { -1+\sqrt { 3 } i }{ 2 } \right] ^{ 2\omega }\)
= (-ω²)^2ω+(ω)^2ω
[∴ ω = \(\frac { -1+i\sqrt { 3 } }{ 2 } \), ω² = \(\frac { -1-i\sqrt { 3 } }{ 2 } \)]
= ω^4ω+ω^2ω
= (ω³)¹³³. ω¹ + (ω³)⁶⁶.ω²
= 1.ω+1.ω²  [∴ 1+ω+ω² = 0 & ω³ = 1]
= ω + ω²
= -1 = RHS

Now, \(\frac { z-1 }{ z+1 } =\frac { x+iy-1 }{ x+iy+1 } =\frac { \left( x-1 \right) +iy }{ \left( x+1 \right) +iy } =\frac { \left[ \left( x-1 \right) +iy \right] \left[ \left( x+1 \right) -iy \right] }{ \left[ \left( x+1 \right) +iy \right] \left[ \left( x+1 \right) -iy \right] } \)
\(\Rightarrow \frac { z-1 }{ z+1 } =\frac { \left( { x }^{ 2 }+{ y }^{ 2 }-1 \right) +i\left( 2y \right) }{ \left( x+1 \right) ^{ 2 }+{ y }^{ 2 } } \)
Since, arg \(\left( \frac { z-1 }{ z+2 } \right) =\frac { \pi }{ 2 } \Rightarrow { tan }^{ -1 }\left( \frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 }-1 } \right) \)= \(\frac { \pi }{ 2 } \)
\(\Rightarrow \frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 }-1 } =tan\frac { \pi }{ 2 } \) ⇒ x²+ y² − 1 = 0
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }=1\)

Given z = x + iy
Im \(\left( \frac { 2z+1 }{ iz+1 } \right) \)= 0
⇒ Im\(\left( \frac { 2(x+iy)+1 }{ i(x+iy)+1 } \right) \)= 0
⇒ Im\(\left( \frac { (2x+1)+2iy }{ ix+i^{ 2 }y+1 } \right) \)
⇒ Im\(\left( \frac { (2x+1)+2iy }{ ix-y+1 } \right) \)
\(\left( \frac { (2x+1)+iy }{ (1-y)+ix } \right) \)
Multiply and divide by the conjugate of the denominator
We get Im\(\left( \frac { (2x+1)+2iy }{ (1-y)+ix } \times \frac { (1-y)-ix }{ (1-y)-ix } \right) \)=0
⇒ Im\(\left( \frac { (2x+1)+2iy\times (1-y)-ix }{ (1-y)^{ 2 }+{ x }^{ 2 } } \right) \)
Choosing the imaginably part we get,
\(\frac { (2x+1)(-x)+2y(1-y) }{ (1-y)^{ 2 }+{ x }^{ 2 } } \)
⇒ (2x+1)-x+2y(1-y) = 0
⇒ -2x²-x+2y-2y² = 0
⇒ 2x²+2y²+x-2y = 0
Hence, locus of z is 2x²+2y²+x-2y = 0