Given p+iq = \(\frac { (a+i)^{ 2 } }{ 2a-i } \) ...........(1)
Taking conjugate both sides we get,
p-iq =\(\frac { (a+i)^{ 2 } }{ 2a+i } \) ..........(2)
Multiplying (1) and (2) we get,
(p+iq)(p-iq) = \(\frac { (a+i)^{ 2 } }{ 2a-i } \times \frac { (a-i)^{ 2 } }{ 2a+i } \)
p²+q² = \(\frac { [(a+i)(a-i)^{ 2 }] }{ 4a^{ 2 }+1 } =\frac { ({ a }^{ 2 }+1)^{ 2 } }{ 4a^{ 2 }+1 } \).
Hence proved

Given  |3z - 5| = 3 |z + 1
⇒ |3(x+iy)-5| = 3|x+iy+1|
⇒ |(3x-5)+3y| = 3|(x+1)+iy|
⇒ \(\sqrt { (3x-5)^{ 2 }+3^{ 2 } } =3\left[ \sqrt { (x+1)^{ 2 }+{ y }^{ 2 } } \right] \)
Squaring both sides we get,
(3x - 5)² + 9 = 9 [(x + 1)² + y²]
⇒ 9x² - 30x + 25 + 9 = 9 [x² + 2x + 1 + y²]
⇒ 48x - 16 = 0
⇒ 3x-1 = 0

Let z = -2i = 2(-i)
= 2\(\left[ cos\left( -\frac { \pi }{ 2 } \right) +isin\left( -\frac { \pi }{ 2 } \right) \right] \)
[∵ cos(-θ) = cos θ and sin(-θ) = -sinθ]
Principal value of -2i c is\(\frac { \pi }{ 2 } \).

Let \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \) = r(cos θ + i sin θ)
r = \(\sqrt { \left( \frac { \sqrt { 3 } }{ 2 } \right) ^{ 2 }+\left( \frac { 1 }{ 2 } \right) ^{ 2 } } =\sqrt { \frac { 3 }{ 4 } +\frac { 1 }{ 4 } } =\sqrt { \frac { 4 }{ 4 } } \)=1
α = \(tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { \frac { 1 }{ 2 } }{ \frac { \sqrt { 3 } }{ 2 } } \right| =tan^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right) =\frac { \pi }{ 6 } \)
Since \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \) lies is the I quadrant, θ = α
∴ \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } =1\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \)
∴ \(\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 5 }=1^{ 5 }\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) ^{ 5 }\)
= \(cos\frac { 5\pi }{ 6 } +isin\frac { 5\pi }{ 6 } \)  ....(1) [De moivres theorem]
Similarly \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } =1\left( cos\frac { \pi }{ 6 } -isin\frac { \pi }{ 6 } \right) \)
∴ \(\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 5 }=1^{ 5 }\left[ cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right] ^{ 5 }\)
= \(cos\frac { 5\pi }{ 6 } -isin\frac { 5\pi }{ 6 } \) ....(2)
Adding (1) and (2) we get,
\(\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 5 }+\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 5 }\)

= \(2cos\frac { 5\pi }{ 6 } =2cos\left( \pi -\frac { \pi }{ 6 } \right) \)
= \(-2cos\ \frac { \pi }{ 6 } \) [∵ \(\frac { 5\pi }{ 6 } \) lies in the II quard]
= \(-2\left( \frac { \sqrt { 3 } }{ 2 } \right) =-\sqrt { 3 } \).

\(\left[ Re\left( iz \right) \right] ^{ 2 }=3\)
iz = i(x + iy) = ix + i²y = ix - y = -y + ix
⇒ Re(iz) = -y
[Re(iz)]² = -y
⇒ (-y)² = 3
⇒ y² = 3
Hence, the Cartesian equation is y² = 3

Given z₁ = 2 + 5i, z₂ = -3 - 4i and z₃ = 1 + i
Additive inverse of z₁ is
-z₁ = -(2 + 5i)
= -2 - 5i
Multiplicative inverse of z₁ is
\(\frac { 1 }{ { z }_{ 1 } } =\frac { 1 }{ 2+5i } \times \frac { 2-5i }{ 2-5i } \)
[Multiply and divide by the conjugate of denominator]
= \(\frac { 2-5i }{ { 2 }^{ 2 }-(5i)^{ 2 } } =\frac { 2-5i }{ 4-25^{ 2 } } =\frac { 2-5i }{ 4+25 } \)
(z₁)^-1 = \(\frac { 1 }{ 29 } \)(2- 5i)   [∴ i² = -1]
Additive inverse of z₂ is
-z₂ = -(3 - 4i)
= 3 + 4i
Multiplicative inverse of z₂ is
\(\frac { 1 }{ z_{ 2 } } =\frac { 1 }{ -3-4i } \times \frac { -3+4i }{ -3+4i } \)
= \(\frac { -3+4i }{ (-3)^{ 2 }-(4i)^{ 2 } } \)
= \(\frac { -3+4i }{ 9-16i^{ 2 } } =\frac { -3+4i }{ 9+16 } \)
(z₂)^-1 = \(\frac { 1 }{ 25 } \)(-3 + 4i)
Additive inverse of z₃ is
-z₃ = -(1 + i)
= -1- i
Multiplicative inverse of z₃ is
\(\frac { 1 }{ { z }_{ 3 } } =\frac { 1 }{ 1+i } \times \frac { 1-i }{ 1-i } =\frac { 1-i }{ { 1 }^{ 2 }-({ i }^{ 2 }) } \)
= \(\frac { 1-i }{ 1+i } \)
(z₃)^-1 \(=\frac { 1 }{ 2 } \)(1 - i)

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

⇒ 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

(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

let z = |4+3i|
= \(\\ \sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 16+9 } =\sqrt { 25 } \)
\(\sqrt { a+ib } =\pm \sqrt { \frac { |z|+a }{ 2 } } +i\frac { b }{ |b| } \sqrt { \frac { |z|-a }{ 2 } } \)
[Here |z| = 5, a = 4, b = 3]
\(\sqrt { 4+3i } =\pm \sqrt { \frac { 5+4 }{ 2 } } +i\frac { 3 }{ |3| } \sqrt { \frac { 5-4 }{ 2 } } \)
= \(\pm \sqrt { \frac { 9 }{ 2 } } +i\frac { 3 }{ 3 } \sqrt { \frac { 1 }{ 2 } } \)
= \(\pm \frac { 3 }{ \sqrt { 2 } } + \frac { i }{ \sqrt { 2 } } \)
Aliter :
Square root of 4 + 3i
Formula method
\(\sqrt{a+i b}=\pm\left[\sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}+i \frac{b}{|b|} \sqrt{\frac{\sqrt{a^{2}+b^{2}}-a}{2}}\right]\)
Now, \(|4+3 i|=\sqrt{4^{2}+3^{2}}=\sqrt{16+9}=\sqrt{25}=5\)
\(\therefore \sqrt{4+3 i}=\pm\left[\sqrt{\frac{5+4}{2}}+i \sqrt{\frac{5-4}{2}}\right]\)
\(=\pm\left[\frac{3}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right]\)

Let 2-2i = r(cosθ + isinθ)
r =\(\sqrt { { 2 }^{ 2 }+(-2)^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 } \)
The principal value α =tan^-1\(\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { -z }{ z } \right| |\)
= tan^-1(1) = \(\frac { \pi }{ 4 } \)
Since the complex number 2 - 2i lies in the quadrant
θ = -α = -\(\frac { \pi }{ 4 } \)
∴ 2-2i = \(2\sqrt { 2 } \left[ cos\left( -\frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \right] ^{ \frac { 1 }{ 3 } }\)
∴ \((2\sqrt { 2 } )^{ \frac { 1 }{ 3 } }\left[ cos\left( -\frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \right] ^{ \frac { 1 }{ 3 } }\)
= \(8^{ \frac { 1 }{ 6 } }\left[ cos\frac { 1 }{ 3 } \left( 2k\pi -\frac { \pi }{ 4 } \right) +isin\frac { 1 }{ 3 } \left( 2k\pi -\frac { \pi }{ 4 } \right) \right] \)
k = 0, 1, 2
The roots are
∴ When k = 0, \(8^{ \frac { 1 }{ 6 } }cis\left( -\frac { \pi }{ 12 } \right) \)
when k = 1, \(8^{ \frac { 1 }{ 6 } }cis\left( \frac { 7\pi }{ 12 } \right) \)
when k = 2, \(8^{ \frac { 1 }{ 6 } }cis\left( \frac { 15\pi }{ 12 } \right) \)
∴ The product of the root
= \(8^{ \frac { 1 }{ 6 } }cis\left( -\frac { \pi }{ 12 } +\frac { 7\pi }{ 12 } +\frac { 15\pi }{ 12 } \right) \)
= \(8^{ \frac { 1 }{ 6 } }cis\left( \frac { 21\pi }{ 12 } \right) =8^{ \frac { 1 }{ 6 } }cis\left( \frac { 7\pi }{ 12 } \right) \)
= \(8^{ \frac { 1 }{ 6 } }cis\left( 2\pi -\frac { \pi }{ 4 } \right) =8^{ \frac { 1 }{ 6 } }cis\left( -\frac { \pi }{ 4 } \right) \)
= \(8^{ \frac { 1 }{ 6 } }\left[ cos\left( -\frac { \pi }{ 4 } \right) +isin\left( -\frac { \pi }{ 4 } \right) \right] \)
= \(8^{ \frac { 1 }{ 6 } }\left[ cos\left( \frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \right] \)
= \(8^{ \frac { 1 }{ 6 } }\left[ \frac { 1 }{ \sqrt { 2 } } -\frac { i }{ \sqrt { 2 } } \right] =2^{ 3\times \frac { 1 }{ 6 } }\left( \frac { 1-i }{ \sqrt { 2 } } \right) =2^{ 1/2 }\left( \frac { 1-i }{ \sqrt { 2 } } \right) \)
= 1-i

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

Given z = x + iy and arg\(\left( \frac { z-i }{ z+2 } \right) =\frac { \pi }{ 4 } \)
⇒ arg(z-i) - arg(z+2) = \(\frac { \pi }{ 4 } \)
⇒ arg(x + iy-i) - arg(x+iy+2) = \(\frac { \pi }{ 4 } \)
⇒ arg(x+i(y-1)-arg((x+2)+iy) = \(\frac { \pi }{ 4 } \)
⇒ \(tan^{ -1 }\left( \frac { y-1 }{ x } \right) -tan^{ -1 }\left( \frac { y }{ x+2 } \right) \) = \(\frac { \pi }{ 4 } \)
⇒ \(tan^{ -1 }\left( \frac { \frac { y-1 }{ x } -\frac { y }{ x+2 } }{ 1+\frac { y-1 }{ x } .\frac { y }{ x+2 } } \right) \)
= \(\frac { \pi }{ 4 } \)\(\left[ \because tan^{ -1 }x-tan^{ -1 }y=tan^{ -1 }\left( \frac { x-y }{ 1+xy } \right) \right] \)
\(\Rightarrow \frac{\left(\frac{(x+2)(y-1)- x y}{\not {x (\not x+\not2)}}\right)}{\left(\frac{x(x+2)+y(y-1)}{\not x(\not x+\not 2)}\right)}=\tan \frac{\pi}{4}=1\)
⇒ \(\frac { (x+2)(y-1)-xy }{ x(x+2)+y(y-1) } \) = 1
⇒ -x + 2y-2 = x²+ 2x + y²-y
⇒ x² + 2x + y²-y + x-2y + 2 = 0
⇒ x² + y²+3x-3y + 2 = 0
Hence proved.