\( z^{4}=1-\sqrt{3 i} \)
\( z=(1-\sqrt{3 i}) 1 / 4 \)
\( z=[r(\cos \theta+i \sin \theta)]^{1 / 4} \)
\( r=\sqrt{1+3}=2 \)
\(\theta\) Lies on IV Quadrant
\( \theta =-\alpha \)
\(\alpha =\tan ^{-1} \frac{|y|}{|x|} \)
\( =\tan ^{-1} \frac{\sqrt{3}}{1} \)
\(Q =\pi / 3 \)
\(0 =-\pi / 3 \)
\(z =\left[2(\cos (-\pi / 3)+\mathrm{i} \sin (-\pi / 3)]^{1 / 4}\right. \)
\(=2^{1 / 4}\left[(\cos (2 \mathrm{k} \pi-\pi / 3)+\mathrm{i} \sin (2 \mathrm{k} \pi-\pi / 3)]^{1 / 4}\right.\)
\(=2^{1 / 4}\left[\cos \left(\frac{6 \mathrm{k} \pi-\pi}{12}\right)+\mathrm{i} \sin \left(\frac{6 \mathrm{k} \pi-\pi}{12}\right)\right.\)
put k = 0, 1, 2, 3.
\(Z=2^{1 / 4} \operatorname{cis}(-\pi / 12) ; 2^{1 / 4} \operatorname{cis}\left(\frac{5 \pi}{12}\right)\)
\(2^{1 / 4} \operatorname{cis}\left(\frac{11 \pi}{12}\right), 2^{1 / 4} \operatorname{cis}\left(\frac{\pi \pi}{12}\right)\)

\(2^{15}\left[\frac{\left(-\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)^{15}}{(1-i)^{20}}+\frac{\left(-\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)^{15}}{(1+i)^{20}}\right]\)
\( =2^{15}\left[\frac{\omega^{15}}{(1-i)^{20}}+\frac{\omega^{30}}{(1+i)^{20}}\right] \)
\( =2^{15}\left[\frac{1}{(1-i)^{20}}+\frac{1}{(1+i)^{20}}\right] \)
\( =2^{15}\left[\frac{(1+i)^{20}+(1-i)^{20}}{\left(1-i^{2}\right)^{20}}\right] \)
\( =\frac{2^{15}}{2^{20}}\left[(1+i)^{20}+(1-i)^{20}\right] \)
\( =\frac{1}{2^{5}}\left[\left(1-i^{2}\right)^{10}+(1-i)^{10}\right] \)
\( =\frac{1}{2^{5}}\left[(2 i)^{10}+(-2 i)^{10}\right] \)
\( \left.=\frac{1}{2^{5}} 2^{10}\left(i^{10}+i^{10}\right)\right] =2^{5}(-1-1) \)
= -64

Given \( \cos \alpha+\cos \beta+\cos \gamma=0 \)........... (1)
\( \sin \alpha+\sin \beta+\sin \gamma=0 \) ..............(2)
Let \( \mathrm{a}=\cos \alpha+i \sin \alpha \)
\( \mathrm{b}=\cos \beta+i \sin \beta \)
\( \mathrm{c}=\cos \gamma+i \sin \gamma\)
\(\Rightarrow \quad a+b+c=0\)
Given from (1), (2) and (3)
Now \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\( =(\cos \alpha+i \sin \alpha)^{-1} +(\cos \beta+i \sin \beta)^{-1} +(\cos \gamma+i \sin \gamma)^{-1} \)
\(=\cos \alpha-i \sin \alpha+\cos \beta-i \sin \beta+\cos \gamma-i \sin \gamma\) ............... (4)
Squaring both sides of equation (3)
We get \(a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a=0\) (or)
\(a^{2}+b^{2}+c^{2}=0\) [from (4)]
\( \therefore(\cos \alpha+i \sin \alpha)^{2}+(\cos \beta+i \sin \beta)^{2} +(\cos \gamma+i \sin \gamma)^{2}=0 \)
By de Moivre's theorem
\( \cos 2 \alpha+i \sin 2 \alpha+\cos 2 \beta +i \sin 2 \beta +\cos 2 \gamma+i \sin 2 \gamma=0 \)
Equating real and imaginary parts
\( \Rightarrow \cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma =0 \text { and } \)...............(5)
\(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma =0\) ................(6)
From (5)
\( \Rightarrow 1-2 \sin ^{2} \alpha+1-2 \sin ^{2} \beta+1-2 \sin ^{2} \gamma=0 \)
\(\therefore \sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=\frac{3}{2} \)
Hence proved.

\( \frac{b}{c} =\frac{\cos \beta+i \sin \beta}{\cos \gamma+i \sin \gamma} \)
\( =\cos (\beta-\gamma)+i \sin (\beta-\gamma) \) ..............(1)
Similarly \(\frac{c}{a}=\cos (\gamma-\alpha)+i \sin (\gamma-\alpha)\) ...........(2)
and \(\frac{a}{b}=\cos (\alpha-\beta)+i \sin (\alpha-\beta)\) ..............(3)
Now (1) + (2) + (3) 
\( {[\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)]+} \mathrm{i}[\sin (\beta-\gamma)+\sin (\gamma-\alpha)+\sin (\alpha-\beta)]=1=1+\mathrm{i} 0 \)
Equating real and imaginary parts
\(\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=1\)
Hence proved.

z = x +iy
Then \( \frac{z-1}{z+1} =\frac{(\mathrm{x}-1)+\mathrm{iy}}{(\mathrm{x}+1)+\mathrm{iy}} \times \frac{(\mathrm{x}+1)-\mathrm{iy}}{(\mathrm{x}+1)-i y} \)
\( =\frac{\left(\mathrm{x}^{2}+\mathrm{y}^{2}-1\right)+2 i y}{(\mathrm{x}+1)^{2}+y^{2}} \)
\(\Rightarrow \frac{z-1}{z+1}=\left[\frac{x^{2}+y^{2}-1}{(\mathrm{x}+1)^{2}+y^{2}}\right]+i\left[\frac{2 y}{(\mathrm{x}+1)^{2}+y^{2}}\right]\)
Let '\(\theta\)' be the argument, then
\( \tan \theta=\frac{2 y}{x^{2}+y^{2}-1} \)
Given that arg \( \left(\frac{z-1}{z+1}\right) \ is\ \frac{\pi}{4} , i.e., \theta=\frac{\pi}{4} \)
\( \therefore \tan \frac{\pi}{4} =\frac{2 y}{x^{2}+y^{2}-1} \)
\(1 =\frac{2 y}{x^{2}+y^{2}-1} \)
\( x^{2}+y^{2}-1 =2 y \)
\(x^{2}+y^{2}-2 y-1 =0 \)
Which represent a circle.

Given \( \overline{z_{1}}+i \overline{\bar z_{2}} =0 \)
\(\overline{z_{1}} =-i \overline{z_{2}} \)
Taking conjugate on both sides
\( \Rightarrow \left(\overline{\left.z_{1}\right)}\right. =\left(-i z_{2}\right) \)
\(\Rightarrow z_{1} =1 z_{2} \)
\(\Rightarrow z_{2} =\frac{1}{i} z i \)
\( \Rightarrow z_{2} =\frac{i}{i z_{1}} \)
\( \Rightarrow \arg \left(z_{2}\right) =\arg \left(-iz_{1}\right) \)
\(\Rightarrow \arg \left(z_{2}\right) =\arg (-\mathrm{i})+\arg \left(z_{1}\right) \)
\(\Rightarrow \ \arg \left(z_{2}\right) =-\frac{\pi}{2}+\arg \left(z_{1}\right)\)
Also given \(\arg \left(z_{1} z_{2}\right)=\pi\)
\( \Rightarrow \ \arg \left(z_{1}\right)+\arg \left(z_{2}\right)=\pi \)
\( \Rightarrow \ \arg \left(z_{1}\right)+\arg \left(z_{1}\right)-\frac{\pi}{2}=\pi \)
\( \Rightarrow \ 2 \arg \left(z_{1}\right)=\pi+\frac{\pi}{2}=3 \frac{\pi}{2} \)
\( \Rightarrow \ \arg \left(z_{1}\right)=3 \frac{\pi}{4}\)

Let z = x+iy
Now, \( \frac{2 z+1}{i z+1} =\frac{2(x+i y)+1}{i(x+i y)+1} \)
\(=\frac{(2 x+1)+2 i y}{(1-y)+i x} \times \frac{(1-y)-i x}{(1-y)-i x} \)
\( =\left[\frac{(2 x+1)(1-y)+2 x y}{(1-y)^{2}+x^{2}}\right]+i\left[\frac{2 y(1-y)-x(2 x+1)}{(1-y)^{2}+x^{2}}\right] \)
Given Im \(\left(\frac{2 z+1}{i z+1}\right)=-2 \)
\(\Rightarrow \frac{2 y(1-y)-x(2 x+1)}{(1-y)^{2}+x^{2}}=-2\)
Simplifying, we get x + 3y - 2 = 0
Which is straight line.

Let z = x+ iy, then \(\bar{z}=x-\mathrm{iy}\)
Now \( z \bar{z}-(2+3 i) z-(2-3 i) \bar{z}+9=0\)
\( (x+\mathrm{iy})(x-\mathrm{iy})-(2+3 \mathrm{i})(x+\mathrm{iy})-(2-3 \mathrm{i})(x-\mathrm{iy})+9=0 \)
\(x^{2}-(\mathrm{iy})^{2}-(2 x+\mathrm{i} 2 \mathrm{y}+\mathrm{i} 3 x-3 \mathrm{y})- \) \((2 x-\mathrm{i} 2 \mathrm{y}-\mathrm{i} 3 x-3 \mathrm{y})+9=0\)
\( x^{2}+y^{2}-4 x+6 y+9=0\) is a circle
Comparing this with general form of a circle
\(x^{2}+y^{2}+2 g x+2 f y+c=0\)
then g \(=-2, f=3, c=9\)
radius \( =\sqrt{g^{2}+f^{2}-c} \)
\(=\sqrt{(-2)^{2}+(3)^{2}-9} \)
\( =\sqrt{4+9-9}=\sqrt{4}=2 \text { units } \)
centre = (-g, -f) = (2, -3) .

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.

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

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

(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)