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

i¹⁹⁴⁷+ i¹⁹⁵⁰
i¹⁹⁴⁷+i¹⁹⁵⁰ = i¹⁹⁴⁴.i³+i¹⁹⁴⁸.i²
[∴ 1944 is a multiple of 4, or 1948 is also a multiple of 4]
= (i⁴)486.i².i¹+(i⁴)487.i² [i⁴ = 1]
= (1⁴⁸⁶)(-1) + (1)⁴⁸⁷(-1) [i²= -1]
= -i-1
= -(1- i)

(i)⁷= (i)⁴⁺³ = (i)³ = -i

Let z = x+iy be the given complex number
∴ \(\bar { z } \) = x-iy
z\(\bar { z } \) = (x+iy) (x-iy) = x²+y²
∴ z\(\bar { z } \) -(2+3i)z -(2-3i)\(\bar { z } \)+9
⇒ x²+y²-(2+3i)(x+iy)-(2-3i)(x-iy)+9 = 0
⇒ x²+y²-[2x+2iy+3ix+i²y] - [2x-2iy-3ix+3i²y]+9 = 0
\(\Rightarrow x^{2}+y^{2}-2 x -\not 2 i y-\not 3i x +3 y-2 x+\not 2 i y+\not 3 i x+3 y+9=0 \)
⇒ x²+y²-4x+6y+9 = 0
Here 2u = -4 ⇒ u = -2
2v = 6 ⇒ v = 3 and d = 9
∴ Centre of the circle is (-u, -v) = (2, -3)
Radius =\(\sqrt { { u }^{ 2 }+{ v }^{ 2 }-d } =\sqrt { 4+9-9 } \)
=\(\sqrt { 4 } \) = 2 units
Hence, the centre of the circle is (2, -3) and radius is 2 units.

(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

Let z = \((-1)^{ \frac { 1 }{ 4 } }\)
⇒ z = (cosπ+i sinπ)^1/4
[∵ cos π = -1 and sin π = 0]
⇒ z = \(cos\frac { 1 }{ 4 } (2k\pi +\pi )+isin\frac { 1 }{ 4 } (2k\pi +\pi )\)
k = 0, 1, 2, 3
When k = 0
z = \(cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } (1+i)\)
When k = 1
z = \(cos\frac { 3\pi }{ 4 } +isin\frac { 3\pi }{ 4 } =cos\left( \pi -\frac { \pi }{ 4 } \right) +isin\left( \pi -\frac { \pi }{ 4 } \right) \)
= \(-cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 }\)
[∵ \(\left( \pi -\frac { \pi }{ 4 } \right) \) is in the II quad and cosθ is -ve and sinθ is +ve]
= \(-\frac { 1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } (-1+i)\)
When k = 2,
z = \(cos5\frac { \pi }{ 4 } +isin\frac { \pi }{ 4} \)
= \(cos\left( \pi +\frac { \pi }{ 4 } \right) +isin\left( \pi +\frac { \pi }{ 4 } \right) \)
= \(-cos\frac { \pi }{ 4\\ } -isin\frac { \pi }{ 4\\ } \)
[∵ \(\left( \pi +\frac { \pi }{ 4 } \right) \) is in the III quadrant Where cosθ and sinθ are -ve]
= \(-\frac { 1 }{ \sqrt { 2 } } -\frac { i }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } (-1-i)\)
When k = 3
z = \(cos7\frac { \pi }{ 4\\ } +isin7\frac { \pi }{ 4\\ } \)
= \(cos\left( 2\pi \frac { \pi }{ 4 } \right) -isin\left( 2\pi -\frac { \pi }{ 4 } \right) \)
= \(cos\frac { \pi }{ 4 } -isin\frac { \pi }{ 4 } \)
[∵ \(\left( 2\pi -\frac { \pi }{ 4 } \right) \) is in the IV quadrant Where cosθ is +ve and sinθ is -ve]
\(-\frac { 1 }{ \sqrt { 2 } } -\frac { i }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } (1-i)\)
Hence the four roots are
\(\frac { 1 }{ \sqrt { 2 } } (1+i),\frac { 1 }{ \sqrt { 2 } } (-1+i),\frac { 1 }{ \sqrt { 2 } } (-1-i),\frac { 1 }{ \sqrt { 2 } } (1-i)\)
\(\pm \frac { 1 }{ \sqrt { 2 } } \)(1±i)

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

Let z = x + iy be a complex number
∴ \(\left| \frac { x+iy-3 }{ x+iy+3 } \right| \)  = 2
⇒ \(\left| \frac { x+iy-3 }{ x+iy+3 } \right| \) = 2
⇒ |(x-3) + iy| = 2|(x+3)+iy|
⇒ \(\sqrt { (x-3)^{ 2 }+{ y }^{ 2 } } =2\sqrt { (x+3)^{ 2 }+{ y }^{ 2 } } \)
Squaring both sides we get,
(x - 3)² + y² = 4[(x + 3)² + y²]
⇒ x² + 9 - 6x + y² = 4 [x²+ 9 + 6x + y²]
x² + 9 - 6x + 1 = 4x² + 36 + 24x + 4y²
⇒ 3x² + 3y² + 30x + 27 = which represent a circle.

Let z = x + iy
Then \(\frac { 1+z }{ 1-z } \) = cos2θ + i sin 2θ
⇒ \(\frac { 1+x+iy }{ 1-x-iy } \) = cos 2θ + i sin 2θ ....(1)
Taking modulus,
\(\left| \frac { 1+x+iy }{ 1-x-iy } \right| \) = |cos 2θ+i sin 2θ| ⇒ \(\frac { |1+x+iy| }{ |1-x-iy| } \)
=\(\sqrt { cos^{ 2 }2\theta +sin^{ 2 }2\theta } \) = 1
⇒ |1 + x + iy| = |1 - x - iy|
⇒ \(\sqrt { (1+x)^{ 2 }+{ y }^{ 2 } } =\sqrt { (1+x)^{ 2 }+{ y }^{ 2 } } \)
⇒ (1 + x)²+ y² = (1-x)²+ y²

⇒ 4x = 0 ⇒ x = 0
From (1) \(\frac { (1+x)+iy }{ (1-x)-iy } \times \frac { (1-x)+iy }{ (1-x)+iy } \)
= cos2θ + isin 2θ
Choosing the imaginary part alone we get,
\(\frac { y(1+x)+y(1-x) }{ (1-x)^{ 2 }+{ y }^{ 2 } } \)= sin 2θ

\(\frac { 2y }{ 1+y^{ 2 } } \) = sin 2θ
⇒ \(\frac { 2tan\theta }{ 1+tan^{ 2 }\theta } \) = sin 2θ
∴ y must be equal to tan θ
⇒ y = tan θ
z = x+ iy
∴ z = 0 + i tan θ
⇒ z = tan θ

x-iy

i