\(1+i\sqrt { 3 } \)
\(r=||z|=\sqrt { { 1 }^{ 2 }+\left( \sqrt { 3 } \right) ^{ 2 } } \)
|\(\theta ={ tan }^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right) =\frac { \pi }{ 3 } \)
Hence \(ang(z)=\frac { \pi }{ 3 } \)
Therefore, the polar form of \(1+i\sqrt { 3 } \) can be written as
\(1+i\sqrt { 3 } =2\left( cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } \right) \)
\(=2\left( cos\left( \frac { \pi }{ 3 } +2k\pi \right) +isin\left( \frac { \pi }{ 3 } +2k\pi \right) \right) ,k\varepsilon z\).

We consider \(\frac { 1+i }{ 1-i } =\frac { \left( 1+i \right) \left( 1+i \right) }{ \left( 1-i \right) \left( 1+i \right) } =\frac { 1+2i }{ 1+1 } =\frac { 2i }{ 2 } =i\)
and \(\frac { 1-i }{ 1+i } =\left( \frac { 1+{ i } }{ 1-i } \right) ^{ -1 }=\frac { 1 }{ i } =-i\)
Therefore,\(\left( \frac { 1+i }{ 1-i } \right) ^{ 3 }-\left( \frac { 1-i }{ 1-i } \right) ^{ 2 }\)= i³-(-i)³ = - i - i = -2i

⇒ 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 = \(\frac{1}{3}\).

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

(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

 \(\sum _{ n=1 }^{ 102 }{ { i }^{ n } } \) = (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

|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 \(\frac{8}{3}\).

(1-i)¹⁰
Let z = (1- i)¹⁰
|z| = |1- i|¹⁰ = \(\left[ \sqrt { { 1 }^{ 2 }+(-1)^{ 2 } } \right] ^{ 10 }\)
= \(\left[ \sqrt { 2 } \right] ^{ 10 }\) = 2^{1/2 x 10} = 2⁵ = 32

 \(\overline { 3i } +\frac { 1 }{ 2-i } \)
= - 3i + \(\frac { 1 }{ 2-i } \times \frac { 2+i }{ 2+i } \)
[∴ Conjugate of 3i is -3i]
= - 3i + \(\frac { 2+i }{ 2^{ 2 }-{ i }^{ 2 } } =-3i+\frac { 2+i }{ 4+1 } \)
= - 3i + \(\frac { 2+i }{ 5 } \)
= \(\frac { -15i+2+i }{ 5 } =\frac { -14i+2 }{ 5 } \)
\(=\frac { 2 }{ 5 } -\frac { 14}{ 5 }i \).

z-iw
= (5-2i) - i(-1+3i)
= (5-2i) + (+1-3i²)
5 - 2i + i - 3(-1) = 5 - i + 3 = 8 - i

ang \(z=\frac { -2 }{ 1+i\sqrt { 3 } } \)
= arg(-2)-arg \(\left( 1+i\sqrt { 3 } \right) \) \(\left( \because arg\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =arg{ z }_{ 1 }=g_{ 2 } \right) \)
= \(\left( \pi -{ tan }^{ -1 }\left( \frac { 0 }{ 2 } \right) \right) -tan^{ -1 }\left( \frac { \sqrt { 3 } }{ 1 } \right) \)
= \(\pi -\frac { \pi }{ 3 } =\frac { 2\pi }{ 3 } \)
This implies that one of the values of arg z is \(\frac { 2\pi }{ 3 } \)
Since \(\frac { 2\pi }{ 3 } \) lies between \(-\pi \), the principal argument Argz is \(\frac { 2\pi }{ 3 } \)

\(\left| z-2-i \right| =3\)
⇒ |z-(2+i)| = 3
It is of the form |z - z₀| = r and so it represents a circle.
Centre is (2, 1) and radius = 3 units.
Aliter:
Let z = x +iy
|z-2-i| = 3
|x + iy-2-i| = 3
\(|(x-2)+i(y-1)|=3\)
\(\sqrt{(x-2)^{2}+(y-1)^{2}}=3\)
Squaring on both sides
\( (x-2)^{2}+(y-1)^{2}=9 \)
\(x^{2}-4 x+4+y^{2}-2 y+1-9=0 \)
\(x^{2}+y^{2}-4 x-2 y-4=0 \)
Comparing with General form of equation of circle
\(a x^{2}+b y^{2}+2 g x+2 f y+c=0\)
we get a = 1, b = 1, g = -2, f = -1, c = -4
Centre (- g, - f) = (2, 1)
radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{4+1+4}=\sqrt{9}\)
= 3 units

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, \(|4+3 i|=\sqrt{4^{2}+3^{2}}=\sqrt{16+9}=\sqrt{25}=5\)
\(\therefore \sqrt{4+3 i}=\pm\left[\sqrt{\frac{5+4}{2}}+i \sqrt{\frac{5-4}{2}}\right]\)
\(=\pm\left[\frac{3}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right]\)

\(\frac { 2i }{ 3+4i } \)
Let z = \(\frac { 2i }{ 3+4i } \)
|z| = \(\left| \frac { 2i }{ 3+4i } \right| =\frac { |2i| }{ |3+4i| } =\frac { \sqrt { 2^{ 2 } } }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } =\frac { 2 }{ \sqrt { 9+16 } } \)
= \(\frac { 2 }{ \sqrt { 25 } } =\frac { 2 }{ 5 } \).

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