\(\theta =\frac { \pi }{ 3 } \)
Given z = 2-2i
θ = \(\frac { \pi }{ 3 } \)
Let z = 2-2i = r(cos θ + i sin θ)
r =\(\sqrt { { 2 }^{ 2 }+(-2)^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 } \)
α = \(\\ tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { -2 }{ 2 } \right| \)
= tan^-1(1) = \(\frac { \pi }{ 4 } \)
The complex number 2-2i lie in the IV quadrant
∴ θ = -α = - \(\frac { \pi }{ 4 } \) [∵ x is +ve, y is -ve]
∴ 2-2i = 2\(\sqrt{2}\)\(\left[ cos\left( -\frac { \pi }{ 4 } \right) +isin\left( -\frac { \pi }{ 4 } \right) \right] \)
Z = \(2\sqrt { 2 } { e }^{ -i\frac { \pi }{ 4 } }\)   ......... (1) [By uler'e formula]
Th rotation of z by θ radians in the counter clockwise direction about the origin in ze^iθ
∴ Rotaton of z is \(z{ e }^{ i\frac { \pi }{ 3 } }\)
= \(2\sqrt { 2 } { e }^{ -i\frac { \pi }{ 4 } }.{ e }^{ i\frac { \pi }{ 3 } }\)     [using (1)]
= \(2\sqrt { 2 } \left[ { e }^{ \left( i\frac { \pi }{ 3 } -\frac { \pi }{ 4 } \right) } \right] =2\sqrt { 2 } e^{ i\frac { \pi }{ 12 } }\)

Since α, β, and γ are the roots of the equation x³+ px²+ qx + r = 0, we have
Σ₁ α + β + γ = -p and Σ₃ αβγ = -r
\(\Sigma \frac { 1 }{ \beta \gamma } =\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } +\frac { 1 }{ \alpha \beta } =\frac { \alpha +\beta +\gamma }{ \alpha \beta \gamma } =\frac { -p }{ -r } =\frac { p }{ r } \).

Since α and β are the roots of 17x²+ 43x −73 = 0 , we have α + β =\(\frac { -43 }{ 17 } \) and  αβ =\(\frac { -73 }{ 17 } \).
We wish to construct a quadratic equation with roots α + 2 and  β + 2. Thus, to construct such a quadratic equation, calculate
the sum of the roots =  α + β + 4 = \(\frac { -4 }{ 17 } +4=\frac { 25 }{ 17 } \) and
the product of the roots = αβ + 2(α+β)+4 = \(\frac { -73 }{ 17 } +2\left( \frac { -43 }{ 17 } \right) +4=\frac { -91 }{ 17 } \)
Hence a quadratic equation with required roots is x²-\(\frac { 25 }{ 17 } x-\frac { 91 }{ 17 } \) = 0
Multiplying this equation by 17, gives 17x²−25x−91 = 0
which is also a quadratic equation having roots α + 2 and β + 2

LHS = \(\left( \cfrac { 1+sin\frac { \pi }{ 10 } +icos\frac { \pi }{ 10 } }{ 1+sin\frac { \pi }{ 10 } -icos\frac { \pi }{ 10 } } \right) ^{ 10 }\)
Let z = \(sin\frac { \pi }{ 10 } +icos\frac { \pi }{ 10 } \)
∴ \(\frac { 1 }{ z } =sin\frac { \pi }{ 10 } -icos\frac { \pi }{ 10 } \)
∴ LHS =\(\left[ \frac { 1+z }{ 1+\frac { 1 }{ z } } \right] ^{ 10 }=\left[ \frac { 1+z }{ \frac { z+1 }{ z } } \right] ^{ 10 }\)= z¹⁰
= \(\left[ sin\frac { \pi }{ 10 } +icos\frac { \pi }{ 10 } \right] ^{ 10 }\)
= \({ i }^{ 10 }\left[ cos\frac { \pi }{ 10 } -isin\frac { \pi }{ 10 } \right] ^{ 10 }\)
= \({ i }^{ 10 }\left[ cos10\frac { \pi }{ 10 } -isin10\frac { \pi }{ 10 } \right] \) [By De Moivers theorum]
= i¹⁰[cos π- i sin π] = -1(-1-i(0)) = 1
Aliter: 
Let \(z=\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}\)
\(\frac{1}{z}=\sin \frac{\pi}{10}-i \cos \frac{\pi}{10}\)
\(\left[\frac{1+\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}}{1+\sin \frac{\pi}{10}-i \cos \frac{\pi}{10}}\right]^{10}=\left[\frac{1+z}{1+1 / z}\right]\)
\( =\left[\frac{(1+z)}{(z+1)} \cdot z\right]^{10} =z^{10} \)
\( =\left(\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}\right)^{10} \)
\( =\left[\cos \left(\frac{\pi}{2}-\frac{\pi}{10}\right)+i \sin \left(\frac{\pi}{2}-\frac{\pi}{10}\right)\right]^{10} \)
\( =\left[\cos \frac{4 \pi}{10}+i \sin \frac{4 \pi}{10}\right]^{10} \)
\( =\cos \frac{4 \pi}{10}(10)+i \sin \frac{4 \pi}{10}(10) \)
\( =\cos 4 \pi+i \sin 4 \pi \)   [\( \because\) By de Moiwe's theorem]  
= 1+ i(0) = 1

Given equation are \({ x }^{ 2 }px+q=0\)  ............(1)
and \({ x }^{ 2 }+p'x+q'=0\)    ..........(2)
Let ∝ be the common root for (1) and (2)
∴ ∝² + p∝ + q = 0   .........(3)
and ∝²+ p'∝ + q' = 0    .............(4)
Solving (3) and (4) by cross multiplication method we get
p  q 1 p
p' q' 1 p'
\(\Rightarrow \frac { { \alpha }^{ 2 } }{ pq'-p'q } =\frac { \alpha }{ q-q' } =\frac { 1 }{ p'-p } \)
consider \(\frac { { \alpha }^{ 2 } }{ pq'-p'q } =\frac { \alpha }{ q-q' } \)
\(\Rightarrow \frac { { \alpha }^{ 2 } }{ \alpha } =\frac { pq'-p'q }{ q-q' } \)
\(\alpha =\frac { pq'-p'q }{ q-q } \)
Consider \(\frac { \alpha }{ q-q' } =\frac { 1 }{ p'-p } \Rightarrow \alpha =\frac { q-q' }{ p'-p } \)
Hence its roots are \(\frac { pq'-p'q }{ q-q' } or\quad \frac { q-q' }{ p'-p } \)

Let ∝, β, ૪ be the roots of the equation 
Given \(\frac { \alpha }{ \beta } =\frac { 3 }{ 2 } \Rightarrow 2\alpha =3\beta \Rightarrow \alpha =\frac { 3 }{ 2 } \beta \)
\(\therefore \frac { 3 }{ 2 } \beta ,\beta ,\gamma \) are the roots of the given equation
Then by Vieta's formula,
\(\frac { 3 }{ 2 } \beta +\beta +\gamma =\frac { -b }{ a } =\frac { -(-9) }{ 1 } =9\)
\(\frac { 5 }{ 2 } \beta +\gamma =9\Rightarrow \gamma =9-\frac { 5 }{ 2 } \beta \)
\(\Rightarrow \gamma =\frac { 18-5\beta }{ 2 } ...(2)\)
Also \(\frac { 3 }{ 2 } \beta (\beta )+\beta \gamma +\left( \frac { 3 }{ 2 } \beta \right) \gamma =\frac { c }{ a } =\frac { 14 }{ 1 } =14\)
\(\Rightarrow \frac { 3 }{ 2 } { \beta }^{ 2 }+\frac { 5 }{ 2 } \beta \left( \frac { 18-5\beta }{ 2 } \right) =14\ [using\ (2)]\)
\(\Rightarrow \frac { 3 }{ 2 } { \beta }^{ 2 }+\frac { 90\beta }{ 4 } -\frac { 25{ \beta }^{ 2 } }{ 4 } =14\)
Multiplying by \(4,6{ \beta }^{ 2 }+90{ \beta }-25{ \beta }^{ 2 }=56\)
\(19{ \beta }^{ 2 }-90{ \beta }+56=0\)
\(\Rightarrow ({ \beta }-4)(19{ \beta }-14)=0\)
\(\Rightarrow { \beta }=4\)
\({ \beta }=\frac { 14 }{ 19 } \)
When \(\beta\) = 4, the other roots are \(\frac { 3 }{ 2 } (4),4,\frac { 18-5 }{ 2 } (4)\)
\(\Rightarrow 6,4,-1\)

When \(\\ \beta =\frac { 14 }{ 19 } ,\) the other roots are \(\frac { 3 }{ 2 } \beta ,\beta \frac { 18-5\beta }{ 2 } [by(2)]\)
\(\Rightarrow \frac { 3 }{ 2 } \left( \frac { 14 }{ 19 } \right) ,\frac { 14 }{ 19 } ,\frac { 18-5\left( \frac { 14 }{ 19 } \right) }{ 2 } \Rightarrow \frac { 21 }{ 19 } ,\frac { 14 }{ 19 } ,\frac { 136 }{ 19 } \)

Given 2cos α = x + \(\frac { 1 }{ x } \)
⇒ 2cos α = \(\frac { { x }^{ 2 }+1 }{ x } \)
⇒ x²+1 = 2x cos α
⇒ x²-2x cos α+1 = 0
⇒ \(\frac { 2cos\alpha \pm \sqrt { (-2cos\alpha )^{ 2 }-4(1)(1) } }{ 2 } \)
= \(\frac { 2cos\alpha \pm \sqrt { 4cos^{ 2 }\alpha -4 } }{ 2 } \)  \(\left[ \because \frac { b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
= \(\frac { 2cos\alpha \pm \sqrt { -sin^{ 2 }\alpha } }{ 2 } \)
= \(\frac { 2cos\alpha \pm isin\alpha }{ 2 } \) [∵ sin²α + cos²α = 1]
⇒ x² = cos α ± sin α
Also, 2cos β = y + \(\frac { 1 }{ y } \)
⇒ 2cos β = \(\frac { { y }^{ 2 }+1 }{ y } \)
⇒ y²-2y cos β + 1 = 0
⇒ \(\frac { 2cos\beta \pm \sqrt { (-2cos^{ 2 }\beta ^{ 2 }-4(1)(1) } }{ 2 } \)
= \(\frac { 2cos\beta \pm \sqrt { 4cos^{ 2 }\beta } -4 }{ 2 } =\frac { 2cos\beta \pm 2isin\beta }{ 2 } \)
⇒ y = cos β ±i sin β
\(\frac { { x }^{ m } }{ { y }^{ n } } -\frac { { y }^{ n } }{ { x }^{ m } } =2isin\left( m\alpha -n\beta \right) \)
x^m = (cos α+sin α)^m = cos mα + i sin mα  [By De moivre's theorem]
yⁿ = (cos β + i sin β)ⁿ = cos nβ+i sin nβ
∴ \(\frac { { x }^{ m } }{ { y }^{ n } } =\frac { cosm\alpha +isinm\alpha }{ cosn\beta +isinn\beta } \)
= cos(mα-nβ) +i sin(mα+nβ)
and \(\frac { { y }^{ n } }{ { x }^{ m } } =\frac { \frac { 1 }{ { x }^{ m } } }{ { y }^{ n } } \)
= cos(mα-nβ)-i sin(mα-nβ)

Given 2i + 3 is a root
∴ Its conjugate 3 - 2i is also a root of the polynomial equation.
∴ Sum of the roots 3 + 2i + 3 - 2i = 6
Product of the roots = (3 + 2i) (3 - 2i)
= 3² + 2² = 9+ 4 = 13
∴ The polynomial equation of minimum degree with rational co-efficients is
x² - x (sum of the roots) + product of the roots = 0
⇒x² -x (6) + 13 = 0
⇒ x²- 6x + 13 = 0

Given polynomial equation is 
2x⁴+ 5x³−7x² + 8 = 0
Here a = 2, b = 5, c = -7, d = 0, e = 8
By Vieta's formula, 
\(\alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -5 }{ 2 } \)
\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { -7 }{ 2 } \)
\(\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =0\)
\(\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { 8 }{ 2 } =4\)
Given roots of quadratic equation are 
∝ + β + ૪ + \(\delta \) and ∝β૪\(\delta \)
∴ sum of the roots = (∝+β+૪+\(\delta \)) (∝β૪\(\delta \))
\(=\left( \frac { -5 }{ 2 } +4 \right) =\frac { -5+8 }{ 2 } =\frac { 3 }{ 2 } \)
\(=\left( \alpha +\beta +\gamma +\delta \right) (\alpha \beta \gamma \delta )\)
\(=\left( \frac { -5 }{ 2 } \right) (4)=\frac { -20 }{ 2 } =-10\)
∴ The required quadratic equation is x²-x 
(sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x\left( \frac { 3 }{ 2 } \right) -10=0\)
\(\Rightarrow { 2x }^{ 2 }-3x-20=0\)