\(\sqrt { 3 } -i\)

r = 2 and \(-\alpha =\frac { \pi }{ 6 } \)
Since the complex number lies in the fourth quadrant, has the principal value
\(\theta =-\alpha =-\frac { \pi }{ 6 } \)
Therefore, the modulus and principal argument of
\(-\sqrt { 3 } -i\) are 2 and \(\frac { \pi }{ 6 } \)

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

r = 2 and \(\alpha =\frac { \pi }{ 6 } \)
Since the complex number lies in the fourth quadrant, has the principal value
\(\theta =\alpha -\pi =\frac { \pi }{ 6 } -\pi =-\frac { 5\pi }{ 6 } \)
Therefore, the modulus and principal argument of \(-\sqrt { 3 } -i\)  are 2 and -\(\frac { 5\pi }{ 6 } \) respectively.

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

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.

 \(\left| \frac { i\left( 2+i \right) ^{ 2 } }{ \left( 1+i \right) ^{ 2 } } \right| =\frac { \left| i \right| \left| \left( 2+i \right) ^{ 3 } \right| }{ \left| \left( 1+i \right) ^{ 2 } \right| } =\frac { \left( \sqrt { 4+1 } \right) ^{ 3 } }{ \left( \sqrt { 2 } \right) ^{ 2 } } \) \(\left( \because \left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| =\left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| ,{ z }_{ 2 }\neq 0 \right) \)
= \(\frac { \left( \sqrt { 5 } \right) ^{ 3 } }{ 2 } =\frac { 5\sqrt { 5 } }{ 2 } \).

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

i²i³...i⁴⁰ = i^1+2+3...+40 = \(i^{\frac{40 \times 41}{2}}=i^{820}=i^{0}=1 .\)

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

(i)^-1924+ (i)²⁰¹⁸ = (i)^{-1924 + 0} + (i)^{2016 + 2} = (i)⁰ + (i)² = 1 - 1 = 0

 i¹⁷²⁹ = i¹⁷²⁸ i¹ = i

-5 -12i
Let z = -5 -12i
⇒ \(\sqrt { { (-5) }^{ 2 }+(-12)^{ 2 } } \)
= \(\sqrt { 25+144 } =\sqrt { 169 } \) = 13
\(\sqrt { a+ib } =\pm \left( \sqrt { \frac { |z|+a }{ 2 } } +i\frac { b }{ |b| } \sqrt { \frac { |z|-a }{ 2 } } \right) \)
\(\sqrt { -5-12i } \pm \left( \sqrt { \frac { 13-5 }{ 2 } } +i\frac { (-12) }{ |-12| } \sqrt { \frac { 13+5 }{ 2 } } \right) \)
\(\pm \left( \sqrt { \frac { 8 }{ 2 } } +i\sqrt { 9 } \right) \)
[∵ |-12| = 12]
\(\pm (\sqrt { 4 }- i3)\) = ±(2 - 3i)
Aliter :
Square root of -5 -12i
Let a + ib = -5 -12i
a = -5, b = -12
\(|z|=\sqrt{5^{2}+12^{2}}=\sqrt{169}=13\)
\(\left.\sqrt{a+i b}=\pm \sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}+i \frac{b}{|b|} \sqrt{\frac{\sqrt{a^{2}+b^{2}}-a}{2}}\right]\)
\(=\pm\left[\sqrt{\frac{13-5}{2}}-i \sqrt{\frac{13+5}{2}}\right]\) [\(\because\) b is negative]
\(=\pm(2-3 i)\)

Let z = -6+8i
|z| =\(\sqrt { (-6)^{ 2 }+8^{ 2 } } \)
= \(\sqrt { 36+64 } =\sqrt { 100 } \) = 10
\(\sqrt { a+ib } =\pm \left( \sqrt { \frac { |z|+a }{ 2 } } +i\frac { b }{ |b| } \sqrt { \frac { |z|-a }{ 2 } } \right) \)
[Here |z| = 10, a = -6, b = 8]
\(\sqrt { -6+8i } \pm \left( \sqrt { \frac { 10-6 }{ 2 } } +i\frac { 8 }{ |8| } \sqrt { \frac { 10+6 }{ 2 } } \right) \)
= \(\pm \left( \sqrt { \frac { 4 }{ 2 } } +i\sqrt { \frac { 16 }{ 2 } } \right) \)
= \(\pm (\sqrt { 2 } +i\sqrt { 8 } )\)
= \(\\ \pm (\sqrt { 2 } +i2\sqrt { 2 } )\)
Aliter :
Square root of -6 + 8i
Let a + ib = - 6 + 8i
a = -6, b = 8
\(|z|=\sqrt{6^{2}+8^{2}}=\sqrt{100}=10\)
\(\sqrt{a+i b}=\pm\left[\sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}+i \frac{b}{|b|} \sqrt{\frac{\sqrt{a^{2}+b^{2}}-a}{2}}\right]\)
\(=\pm\left[\sqrt{\frac{10-6}{2}}+i \sqrt{\frac{10+6}{2}}\right]\)
\(=\pm[\sqrt{2}+i \quad 2 \sqrt{2}]\)

2i(3−4i)(4−3i)
Let z = 2i(3−4i)(4−3i).
∴ |z| = |2i(3-4i)(4-3i)|
= |2i| |3-4i| |4-3i|
= \(\sqrt { { 2 }^{ 2 } } \sqrt { { 3 }^{ 2 }+(-4)^{ 2 } } \sqrt { { 4 }^{ 2 }+(-3)^{ 2 } } \)
= \(\\ 2.\sqrt { 9+16 } \sqrt { 16+9 } =2.\sqrt { 25 } .\sqrt { 25 } \)
= 2(5)(5) = 50

(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

\(\frac { 2-i }{ 1+i } +\frac { 1-2i }{ 1-i } \)
Let z = \(\frac { 2-i }{ 1+i } +\frac { 1-2i }{ 1-i } \)
= \(\frac { (2-i)(1-i)+(1-2i)(1+i) }{ (1+i)(1-i) } \)
= \(\frac { 2-2i-i+{ i }^{ 2 }+1+i-2i-2i^{ 2 } }{ { 1 }^{ 2 }-{ i }^{ 2 } } \)
= \(\\ \frac { 2-i-1+1-i+2 }{ 2 } =\frac { 4-4i }{ 2 } \)
= \(\frac { 2(2-2i) }{ 2 } \) = 2 - 2i
∴ |z| = \(\sqrt { { 2 }^{ 2 }+(-2)^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 } \)

\(Re\left( z \right) =\frac { z+\bar { z } }{ 2 } \) and Im\(\left( z \right) =\frac { z-\bar { z } }{ 2i } \)
Let z = x+iy where x is the Re(x) and y is the Im (z)
Then \(\bar { z } \) = x-iy
z+\(\bar { z } \) = x + iy + x - iy = 2x
∴ \(\frac { z+\bar { x } }{ 2 } \) = x
\(\frac { z+\bar { x } }{ 2 } \) = Re(z)
Also z-\(\bar { z } \) = x+iy-(x-iy)
= x+iy-x+iy = 2iy
\(\frac { z-\bar { z } }{ 2i } \) = y
∴ \(\frac { z-\bar { z } }{ 2i } \) = Im(z)
Hence proved

 Im(3z+4\(\bar { z } \)−4i)
= Im (3(x+iy)+4(x-iy)-4i)
= Im (3x+i3y+4x-i4y-4i)
= Im (3x+4x+i(3y-4y-4)
= Im (7x+i(-y-4))
∴ Imaginary part is -y - 4

Re(i\(\bar { z } \))
= Re (i(x - iy))
[∵ when z = x + iy, \(\bar { z } \) = x - iy]
= Re (ix - i²y)
= Re (ix + y)    [∵ i² = -1]
= Re (y + ix)
∴ Real part is y

 \(\overline { 3i } +\frac { 1 }{ 2-i } \)
= - 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 + \(\frac { 2+i }{ 5 } \)
= \(\frac { -15i+2+i }{ 5 } =\frac { -14i+2 }{ 5 } \)
\(=\frac { 2 }{ 5 } -\frac { 14}{ 5 }i \).

\(\cfrac { 10-5i }{ 6+2i } \)
\(\frac { 10-5i }{ 6+2i } \times \frac { 6-2i }{ 6-2i } \)
[Multiply and divided by the conjugate of the denominator]
= \(\frac { 60-20i-30i+10{ i }^{ 2 } }{ { 6 }^{ 2 }-(2i)^{ 2 } } =\frac { 60-50i-10 }{ 36+4 } \)
= \(\frac { 50-50i }{ 40 } =\frac { 50(1-i) }{ 40 } \)
\(=\frac { 5 }{ 4 } \)(1 - i)

(z+w)²
= [(5-2i)+(-1+ 3i)]² = (4+ i)²
= 16 + i²+ 8i =16 -1+8i
= 15+ 8i

z²+ 2zw +w²
(5-2i)²+2(5-2i)(-1+3i)+(-1+3i)²
= 25 + 4i²- 20i + 2 [-5 + 15i + 2i - 6i²] + 1 + 9i²-6i [∴ i² = -1]
= 25 - 4 - 20i + 2(-5 + 17i + 6) +1-9 -6i
= 21- 20i + 2(1+17i) -8 -6i
= 21- 20i + 2 + 34i -8 -6i
= 15 + 8i

z w
= (5-2i)(-1+3i)
= -5+15i+2i-6i²
= -5+17i-6(-1)
= -5+17i+6
= 1+17i

2z+3w
= 2(5-2i)+3(-1+3i)
= 10-4i-3+9i
= (10-3)+ i(-4+9)
= 7+5i

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

i¹⁺⁵⁰ + i²⁺⁵⁰ + ....+ i¹⁰⁺⁵⁰
= i⁵¹+ i⁵²+ ....+ i⁶⁰
Taking i⁵⁰ common we get,
i⁵⁰ [i + i²+ i³+ i⁴) + (i⁵+ i⁶+ i⁷+ i⁸) + i⁹+ i¹⁰]
= i⁵⁰[0 + (i⁴⁺¹+ i⁴⁺²+ i⁴⁺³+ i⁴⁺⁴) + (i⁸⁺¹ + i⁸⁺²)]
= i⁵⁰ [0 + 0 + i + i²] [∵ i + i²+ i³+ i⁴ = 0]
= i⁵⁰ [i-1] = i⁴⁸⁺²(i-1)
= i²(i-1)  [∴ i⁴⁸ = 1]
= -1(i-1) = -i+1 = 1-i

i i² i³ ....i²⁰⁰⁰
= i^{1+2+3+.....+2000}
= \({ i }^{ \frac { 2000\times 2001 }{ 2 } }\)
[∴ 1+2+3+....n = \(\frac { n(n+1) }{ 2 } \)]
= i^{1000 x 2001}
= i^2001000
= 1
[∴ 2001000 is divisible by 4 as its last two digits are divisible by 4]

\({ i }^{ 59 }+\frac { 1 }{ { i }^{ 59 } } \)
i^{4 \(\times\) 14 + 3} + i^{-(4 \(\times\) 14 + 3)}
= (i⁴)¹⁴.i³ + (i⁴)^-14.i^-3
= 1.i³+1.i^-3  [∵ i⁴ = 1]
= -i + i  [∴ i³ = -i and i^-3= i]
= 0

\(\overset { 12 }{ \underset { n-1 }{ \Sigma } } \)iⁿ
\(\overset { 12 }{ \underset { n-1 }{ \Sigma } } \)iⁿ = (i¹+i²+i³+i⁴)+(i⁵+i⁶+i⁷+i⁸)+(i⁹+i¹⁰+i¹¹+i¹²)
= (i-1-i+1)+(i⁴⁺¹+i⁴⁺²+i⁴⁺³+(i⁴)²+(ⁱ⁸⁺¹+i⁸⁺²+i⁸⁺³+(i⁴)³)
= 0+(i+i²+i³+i⁴)+(i¹+i²+i³+i⁴)
[∴ i² = -1, i³ = -i, i⁴ = 1]
= 0 +(i-1-i+1)+(i-1-i+1)
= 0+0+0 = 0

i¹⁹⁴⁸-i^-1869
(i⁴)⁴⁸⁷-[i^-1868.i^-1]
= 1⁴⁸⁷-\(\left[ (i^{ 4 })^{ -467 }.\frac { 1 }{ i } \right] \) [i⁴ = 1]
= 1 - [1.(-i)] [∴ i^-1 = \(\frac { 1 }{ i } \) = -i]
[One power any number is 1]
= 1 + i

\(\sqrt { 3 } +i\)

Modulus = \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { \left( \sqrt { 3 } \right) ^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 3+1 } =2\)
\(\alpha =tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }\frac { 1 }{ \sqrt { 3 } } =\frac { \pi }{ 6 } \)
Since the complex number \(\sqrt { 3+ } i\) lies in the first quadrant, has the principal value
\(\theta =\alpha =\frac { \pi }{ 6 } \)
Therefore, the modulus and principal argument of \(\sqrt { 3+ } i\) are 2 and \(\frac { \pi }{ 6 } \) respectively.

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

Given |z| = 3
Show that 7 ≤ |z+6-8i| ≤ 13
|z+6-8i| ≤ |z|+|6-8i|
[Triangle law of inequality]
\(\le 3\sqrt { { 6 }^{ 2 }+({ -8) }^{ 2 } } \le 3+\sqrt { 36+64 } \le +\sqrt { 1w } \)
|z+6-8i| ≤3+10 ≤ 13 ............... (1)
Also ||z+6-8i| ≥ |x|-|6-8i| ≥ \(|3-\sqrt { { 6 }^{ 2 }+(-8)^{ 2 } } |\)
≥ \(|3-\sqrt { 36+84 } |\) ≥ |3-10| ≥ |-7| ≥ 7 ............... (2)
From (1) and (2) we get,
7 ≤ |z+6-8i| ≤ 13.

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

We compute \(\left| 6-8i \right| =\sqrt { { 6 }^{ 2 }+\left( -8 \right) ^{ 2 } } =10\)
and applying the formula for square root, we get
\(\sqrt { 6-8i } =\pm \left( \sqrt { \frac { 10+6 }{ 2 } } -i\sqrt { \frac { 10-6 }{ 2 } } \right) \) (\(\therefore\) b is negative\( \frac{b}{|b|}=-1 \))
= \(\pm \left( \sqrt { 8 } +i\sqrt { 2 } \right) \)
= \(\pm \left( 2\sqrt { 2 } -i\sqrt { 2 } \right) \)

\(\left| \frac { 2+i }{ -1+2i } \right| =\frac { \left| 2+i \right| }{ \left| -1+2i \right| } =\frac { \sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 } } }{ \sqrt { \left( -1 \right) ^{ 2 }+{ 2 }^{ 2 } } } =1\) \(\left( \because \left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| =\left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| ,{ z }_{ 2 }\neq 0 \right) \)

Prove the following properties
Z is real if and only if z =\(\bar { z } \)
Let z = x+iy
Then \(\bar { z } \) = x+iy
z = \(\bar { z } \)
⇔ x + iy = x-iy
⇔ x + iy - x + iy = 0
⇔ 2iy = 0
⇔ y = 0
[∴ 2 and i are constants]
when y = 0, z = x which is real
∴ z is purely ⇔ z = \(\bar { z } \)

Given z₁= 2 - i and z₂= -4+3i
z₁z₂ = (2-i)(-4+3i)
= -8 + 6i + 4i - 3i²
= -8 +10i - 3(-1)
= -8 +10i + 3 = -5 +10i
Inverse of z₁z₁ is \(\frac { 1 }{ { z }_{ 1 }{ z }_{ 2 } } \)
= \(\frac { 1 }{ -5+10i } \times \frac { -5-10i }{ -5-10i } \)
= \(\frac { -5-10i }{ (-5)^{ 2 }-(10i)^{ 2 } } \)
= \(\frac { -5-10i }{ 25-100i^{ 2 } } \)
= \(\frac { -5-10i }{ 25+100 } \)  [∵ i² = -1]
\(=\frac{\not{5}(-1-2 i)}{\not{5}\langle(2 5)}=\frac{-1-2 i}{25}\)
∴ Inverse of z₁z₂ is \(\frac { 1 }{ 25 } \) (-1-2i)
Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } \) is \(\frac { 1 }{ \frac { { z }_{ 1 } }{ { z }_{ 2 } } } =\frac { { z }_{ 2 } }{ { z }_{ 2 } } \)
∴ Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } =\frac { { z }_{ 2 } }{ { z }_{ 1 } } =\frac { -4+3i }{ 2-i } \times \frac { 2+i }{ 2+i } \)
= \(\frac { -4+3i }{ 2-i } \times \frac { 2+i }{ 2+i } \)
= \(\frac { -8-4i+6i+3i^{ 2 } }{ 2^{ 2 }-({ i }^{ 2 }) } \)
= \(\frac { -8+2i-3 }{ 4+1 } =\frac { -11+2i }{ 5 }\)
\( =\frac { 1 }{ 5 } \)(-11 + 2i)
∴ Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } \) is \(\frac { -11+2i }{ 5 }\)or \(\frac { 1 }{ 5 } \)(-11 + 2i)

\(\overline { \left( 5+9i \right) +\left( 2-4i \right) } \)
= \(\overline { (5+2)+(9i-4i) } =\overline { 7+5i } \)
= 7 - 5i [∵ Conjugate of 7 + 5i is 7 - 5i]

Represent z, iz and z + iz in the Argand diagram.
z = 2 + 3i can be represented as (2,3)
iz = i(2 + 3i)
= 2i + 3i²
= 2i-3
= -3 + 2i can be represented as (-3, 2)
z + iz = 2 + 3i - 3 + 2i = -1 + 5i can be represented as (-1, 5) in the argand diagram.

(z+w)
= (5-2i) + (-1+3i)
= (5-1) + i(-2+3)
= 4+i(1)
= 4+i

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

We have |2z-3-i| = 3
|2(x + iy)-3 - i| = 3
Squaring on both sides, we get
|(2x - 3) + (2y - 1)i|² = 9
\(\Rightarrow\) (2x - 3)² + (2y - 1)² = 9
\(\Rightarrow\) 4x² + 4y² -12x - 4y + 1 = 0, the locus of z in Cartesian form

\(\frac { i-1 }{ cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } } \)
Let x+iy = \(\frac { i-1 }{ cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } } \)
Consider i-1 = -1+i = r
(cos θ+isin θ)
r = \(\sqrt { (-1)^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 2 } \)
θ = \(tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { 1 }{ -1 } \right| \)
= tan^-1(1) = \(\frac { \pi }{ 4 } \)
Since (-1+i) lies in the II quadrant
[x is -ve and y is +ve]
Its principal value θ = \(\\ \pi -\frac { \pi }{ 4 } =\frac { 3\pi }{ 4 } \)
∴ i-1 = \(\sqrt { 2 } \left[ cos\left( \frac { 3\pi }{ 4 } \right) +isin\left( \frac { 3\pi }{ 4 } \right) \right] \)
∴ x+iy = \(\frac { \sqrt { 2 } \left[ cos\left( \frac { 3\pi }{ 4 } \right) +isin\left( \frac { 3\pi }{ 4 } \right) \right] }{ cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 3 } } \)
[∵ \(\frac { cos{ \theta }_{ 1 }+isin{ \theta }_{ 1 } }{ cos\theta _{ 2 }+isin\theta _{ 2 } } \) = cos(θ₁ - θ₂) + i sin(θ₁ - θ₂)]
= \(\sqrt { 2 } \left[ cos \right] \left( \frac { 3\pi }{ 4 } -\frac { \pi }{ 3 } \right) +isin\left( \frac { 3\pi }{ 4 } -\frac { \pi }{ 3 } \right) \)
= \(\sqrt { 2 } \left[ cos\frac { 5\pi }{ 12 } +isin\frac { 5\pi }{ 12 } \right] \)
Hence the polar form is math where
\(\sqrt { 2 } \left[ cos\left( 2k\pi +\frac { 5\pi }{ 12 } \right) +isin\left( 2k\pi +\frac { 5\pi }{ 12 } \right) \right] \)where k \(\in \) Z.

2-i2
Let x + iy = -2-2i = r(cos θ + isin θ)
r =\(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { (-2)^{ 2 }+(-2)^{ 2 } } =\sqrt { 4+4 } \)
= \(\sqrt { 8 } =2\sqrt { 2 } \)
α = \(tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { -2 }{ -2 } \right| =tan^{ -1 }(1)=\frac { \pi }{ 4 } \)
Since the complex number -2-2i lies in the III quadrant [x is -ve, y is -ve]
Its principal value θ = α-π
⇒ θ =\(\frac { \pi }{ 4 } -\pi =\frac { \pi -4\pi }{ 4 } =-\frac { 3\pi }{ 4 } \)
Its polar form is
-2-2i = 2\(\sqrt { 2 } \)\(\left[ cos\left( 2k\pi -\frac { 3\pi }{ 4 } \right) +isin\left( 2k\pi -\frac { 3\pi }{ 4 } \right) \right] ,k\in Z\).

\(3-i\sqrt { 3 } \)
Let x + iy = \(3-i\sqrt { 3 } \)
= r(cos θ + i sin θ)
r = \(\\ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { 3^{ 2 }+(\sqrt { 3 } )^{ 2 } } =\sqrt { 9+3 } \)
= \(\sqrt { 12 } =2\sqrt { 3 } \)
α = \(tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { -\sqrt { 3 } }{ 3 } \right| =tan^{ -1 }\left| \frac { 1 }{ \sqrt { 3 } } \right| =\frac { \pi }{ 6 } \)
Since the complex number \(3-i\sqrt { 3 } \) lies in the IV quadrant, [∵ x ⟶ +ve y ⟶ -ve]
Its principal value θ = -α
⇒ θ = \(\frac { \pi }{ 6 } \)
∴ Its polar form is
\(3-i\sqrt { 3 } \) = 2\(\sqrt { 3 } \)\(\left[ cos\left( 2k\pi -\frac { \pi }{ 6 } \right) +isin\left( 2k\pi -\frac { \pi }{ 6 } \right) \right] ,k\in Z\)

\(\overline { z } \) = z-1
⇒ \(\overline { z } \) =\(\frac{1}{z}\)
⇒ z\(\overline { z } \) = 1
⇒ |z|² = 1
⇒ x² + y² = 1 which is the required Cartesian equation.
Aliter : 
\( \bar{z} =z^{-1} \)
\(x-i y =\frac{1}{x+i y} \)
\(x-i y =\frac{1}{x+i y} \times \frac{x-i y}{x-i y} \)
\(x-i y =\frac{x-i y}{x^{2}+y^{2}} \)
\(x^{2}+y^{2} =1\)

|z+i| = |z-1|
⇒ |x + iy +i| = |x + iy-1|
⇒ |x + i(y + 1)| = |(x - 1) + iy|
⇒ \(\sqrt { { x }^{ 2 }+(y+1)^{ 2 } } =\sqrt { (x-1)^{ 2 }+y^{ 2 } } \)
⇒ x² + (y + 1)² =(x- 1)² + y²
[ squaring both sides]
\(\Rightarrow \not x^{2}+ \not y^{2}+2 y+ \not1= \not x^{2}-2x+ \not 1+ \not y^2\)
⇒ 2y + 2x = 0
⇒ x + y = 0
Hence, the Cartesian  equation is x + y = 0

Im[(1−i)z + 1] = 0
(1-i)z + 1 = (1-i)( x + iy) +1
= x + iy-ix-i²y+1
= x+iy-ix+y+1
= (x + y + 1) + i(y - x)
∴ Im[(1-i)z + 1] = y-x = 0
⇒ x - y = 0
Hence, the Cartesian equation is x - y = 0

Since z is purely imaginary
z = -\(\overline { z } \)
∴ 2n\(\left[ cos\frac { n\pi }{ 6 } +isin\frac { n\pi }{ 6 } \right] \)
= -2n\(\left[ cos\frac { n\pi }{ 6 } +isin\frac { n\pi }{ 6 } \right] \) [From (1) & (2)]
\(\Rightarrow \cos \frac{n \pi}{6}+ i{\not \sin \frac{m \pi}{6}}=-\cos \frac{n \pi}{6}+i {\not\sin \frac{m \pi}{6}}\)
⇒ 2cos\(\frac { n\pi }{ 6 } \) = 0
⇒ \(cosn\frac { \pi }{ 6 } =0=cos\frac { \pi }{ 2 } \)
[∴ cos \(\frac { \pi }{ 2 } \) = 0]

⇒ n = \(\frac{6}{2}\)
⇒ n = 3
Hence z is purely imaginary

We have |z| - |z - i|
\(\Rightarrow\)|x + iy| = |x + iy - i|
\(\Rightarrow\) \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { { x }^{ 2 }+\left( y-1 \right) ^{ 2 } } \)
\(\Rightarrow x^{2}+y^{2}=x^{2}+y^{2}-2 y+1\)
\(\Rightarrow\) 2y -1 = 0

2 +i2\(\sqrt { 3 } \)
Let 2+i2\(\sqrt { 3 } \)  = x + iy = r (cosθ + i sinθ)
r = modulus =\(\\ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)
=\(\\ \sqrt { { 2 }^{ 2 }+(2\sqrt { 3 } )^{ 2 } } \)
= \(\sqrt { 4+12 } =\sqrt { 16 } \) = 4
α = tan-1\(\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { 2\sqrt { 3 } }{ 2 } \right| \)
= \(tan^{ -1 }(\sqrt { 3 } )=\frac { \pi }{ 3 } \)
Since the complex number 2+i2 \(\sqrt { 3 } \) lies in the I quadrant, [x, y both +ve] its principal value θ = α = \(\frac { \pi }{ 3 } \)
∴ Its polar form is 2+i2\(\sqrt { 3 } \)
= 4\(\left[ cos\left( 2k\pi +\frac { \pi }{ 3 } \right) +isin\left( 2k\pi +\frac { \pi }{ 3 } \right) \right] ,k\in Z\).

\(\left[ Re\left( iz \right) \right] ^{ 2 }=3\)
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 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.

We have, \({ z }^{ 2 }=\bar { z } \)
\(\Rightarrow|z|^{2}=|z|\)
\(\left| z \right| \left( \left| z \right| -1 \right) =0\)
\(\Rightarrow \left| z \right| =0,or\left| z \right| =1\) 
\(|z|=0 \Rightarrow z=0\) is a solution \(\left| z \right| =1\Rightarrow z\bar { z } =1\Rightarrow \bar { z } =\frac { 1 }{ z } \)
Given \({ z }^{ 2 }=\bar { z } \Rightarrow { z }^{ 2 }=\frac { 1 }{ z } \Rightarrow { z }^{ 3 }=1\)
It has 3 non-zero solutions. Hence including zero solution, there are four solutions.

The distance between origin to z = i, −2 + i, and 3 are
|z| = |i| = 1
|z| = |−2+i| = \(\sqrt { \left( -2 \right) ^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 5 } \)
|z| = |3| = 3
Since \(1<\sqrt { 5 } <3\), the farthest point from the origin is 3 .

Let x = \(\left( \sqrt { 3 } +i \right) ^{ n }\)
x = \((\sqrt { 3 } +i)^{ n }=\left[ 2\left( \frac { \sqrt { 3 } +i }{ 2 } \right) \right] ^{ n }\)
= 2ⁿ\(\left[ \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right] ^{ n }\)
[Multiply and divide by 2]
= 2ⁿ\(\left[ cos\frac { \pi }{ 6 } +isinn\frac { \pi }{ 6 } \right] ^{ n }\)
= 2ⁿ\(\left[ cosn\frac { \pi }{ 6 } +isinn\frac { \pi }{ 6 } \right] \)...(1)
\(\overline { z } \) = 2ⁿ\(\left[ cos\frac { \pi }{ 6 } -isinn\frac { \pi }{ 6 } \right] \) ....(2)
Since z is real, z = \(\overline { z } \)
⇒ 2ⁿ\(\left[ cos\ n\frac { \pi }{ 6 } +i\ sin\ n\frac { \pi }{ 6 } \right] \)
= 2ⁿ\(\left[ cos\ n\frac { \pi }{ 6 } -i\ sin \ n\frac { \pi }{ 6 } \right] \)
[From (1) and (2)]
⇒ 2i sin n\(\frac { \pi }{ 6 } \)= 0 ⇒ sin n \(\frac { \pi }{ 6 } \)= 0
⇒ sin n\(\frac { \pi }{ 6 } \) = sinπ    [∴ sin π = 0]
⇒ \(\frac { n\pi }{ 6 } =\pi \Rightarrow \frac { n }{ 6 } \)=1 ⇒ n = 6

Since,\(\left| { z }_{ 1 } \right| =\left| { z }_{ 2 } \right| =\left| { z }_{ 3 } \right| =1\)
\(\left| { z }_{ 1 } \right| ^{ 2 }=1\Rightarrow { z }\bar { { z }_{ 1 } } =1,\left| { z }_{ 2 } \right| ^{ 2 }=1\Rightarrow { z }_{ 2 }\bar { { z }_{ 2 } } =1\ \)
Therefore, \(\bar { { z }_{ 1 } } =\frac { 1 }{ { z }_{ 1 } } ,\bar { { z }_{ 2 } } =\frac { 1 }{ { z }_{ 3 } } \) and hence
\(\left| \frac { 1 }{ { z }_{ 1 } } +\frac { 1 }{ { z }_{ 2 } } +\frac { 1 }{ { z }_{ 3 } } \right| =\left| \bar { { z }_{ 1 } } +\bar { { z }_{ 2 } } +{ \bar { z } }_{ 3 } \right| \)
= \(\left| \overline { { z }_{ 1 }+\left| { z }_{ 2 }+{ z }_{ 3 } \right| } \right| ={ { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } }=1\)