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 2cos α = x + \(\frac { 1 }{ x } \)
⇒ 2cosα = \(\frac { { x }^{ 2 }+1 }{ x } \)
⇒ 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
\(xy-\frac { 1 }{ xy } =2isin\left( \alpha +\beta \right) \)
xy = (cos α + i sin α) (cos β + i sin αβ)
= cos(α+β) + i isn(α+β)
\(\frac { 1 }{ xy } \) = cos(α+β)-i isn(α+β)
\(\therefore x y-\frac{1}{x y}=\cos { (\not \alpha +\not \beta)}+i \sin (\alpha+\beta)-\cos (\not \alpha+\not \beta)+i \sin (\alpha+\beta)\)
= 2i sin(α + β)

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 +\(\frac { 1 }{ y } \)
⇒ 2cosβ = \(\frac { { y }^{ 2 }+1 }{ y } \)
⇒ 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
\(\frac { x }{ y } +\frac { y }{ x } =2cos\left( \alpha -\beta \right) \)
\(\frac { x }{ y } =\frac { cos\alpha +isin\alpha }{ cos\beta +isin\beta } \)
= cos(α-β) + i sin(α-β) .........(1)
and \(\frac { y }{ x } =\frac { 1 }{ x } \) = cos(α-β) + i sin(α-β) .......(2)
(1) + (2) ⟶ = \(\frac { x }{ y } +\frac { y }{ x } \)
= cos(α-β) + i sin(α-β) + cos(α-β) - i sin(α-β)
= 2cos(α-β)

\( |z|=2 \)
\(|z+6+8 i|=|z|+|6+8 i| \)
\(=2+\sqrt{6^{2}+8^{2}} \)
\(=2+\sqrt{100} \)
= 2 + 10
= 12
\( \therefore|z+6+8 i| \leq 12 \) ............. (1)
\(|z+6+8 i| \geq|| z|-|-6-8 i|| \)
\(=|2-10| \)
\(=|-8|\)
= 8
\(|z+6+8 i| \geq 8\) .............(2)
From 1 and 2 we get
\(8 \leq|z+6+8 i| \leq 12\)
Hence proved

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

\(\frac { cos\frac { \pi }{ 6 } -isin\frac { \pi }{ 6 } }{ 2\left( cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } \right) } \)
=\(\frac { 1 }{ 2 } \left[ \frac { cos\left( \frac { -\pi }{ 6 } \right) +isin\left( \frac { -\pi }{ 6 } \right) }{ cos\left( \frac { \pi }{ 3 } \right) +isin\left( \frac { \pi }{ 3 } \right) } \right] \)
[∵ cos(-θ)=cosθ and sin(-θ)=-sinθ
\(=\frac { 1 }{ 2 } \left[ cos\left( \frac { -\pi }{ 6 } -\frac { \pi }{ 3 } \right) +isin\left( \frac { -\pi }{ 6 } -\frac { \pi }{ 3 } \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ cos\left( \frac { -\pi -2\pi }{ 6 } \right) +isin\left( \frac { -\pi -2\pi }{ 6 } \right) \right] \)
\(\frac { 1 }{ 2 } \left[ cos\left( \frac { -3\pi }{ 6 } \right) +isin\left( -\frac { 3\pi }{ 6 } \right) \right] \)
\(\\ \frac { 1 }{ 2 } \left[ cos\left( -\frac { \pi }{ 2 } \right) +isin\left( -\frac { \pi }{ 2 } \right) \right] \\ \)
\(\frac { 1 }{ 2 } \left[ cos\left( \frac { \pi }{ 2 } \right) -isin\left( \frac { \pi }{ 2 } \right) \right] \)
\(\frac { 1 }{ 2 } [0-i(1)]=\frac { -i }{ 2 } \).

|z-4|²-|z-1|² = 16
|x+iy-4|² - |x+iy-1|² = 16
⇒ |(x-4)+iy|² - |(x-1)+iy2|² = 16
⇒ [(x-4)²+y²] - [(x-1)²+y²] = 16
⇒ x²-8x+16+y²-[x²-2x+1+y²] = 16
\(\Rightarrow \not x^{2}-8 x+16+\not y^{2}-\not x^{2}+2 x-1-\not y^{2}=16\)
⇒ -6x+15-16 = 0
⇒ -6x-1 = 0
⇒ 6x+1 = 0 Which is the required Cartesian equation.
The locus of the point is a straight line.

(z₁ + z₂)z₃ = z₁z₃ + z₂ + z₃
LHS = (z₁ + z₂)z₃
= (3-7i)(5 + 4i)
15+12i-35i-28i²
= 15-23i + 28
= 43 - 23i
RHS = z₁z₃ + z₂z₃
3(5+4i)+(-7i)(5+4i)
= 15+12i -35i -28i²
= 15-23i + 28
= 43 - 23i
LHS = RHS
∴ (z₁+z₂)z₃ = z₁z₃ + z₂z₃

(z₁+ z₂₎z₃ = z₁(z₂z₃)
LHS = (z₁ z₂)z₃
= [(1-3i)(-4i)]5
= [-4i + 12i²]5
= (-4i-12)5
= -20i - 60
RHS = z₁(z₂z₃)
= (1-3i) [-4i)5
= (1-i) (-20i)
= -20i + 60i²
= -20i - 60
LHS = RHS
∴ (z₁z₂)z₃ = z₁(z₂z₃)

z = 2+3t
-iz = -i(2+3i)
= 2i -3i² = -2i+3
= 3 - 2i
z - iz = 2+ 3i-3+2i
= -1+5i

We have, \(sin\frac { \pi }{ 6 } +icos\frac { \pi }{ 6 } =i\left( cos\frac { \pi }{ 6 } -isin\frac { \pi }{ 6 } \right) \)
Raising the power 18 on both sides,
\(\left( sin\frac { \pi }{ 6 } +icos\frac { \pi }{ 6 } \right) ^{ 18 }=\left( i \right) ^{ 18 }\left( cos\frac { \pi }{ 6 } -isin\frac { \pi }{ 6 } \right) \)
= \(\left( -1 \right) \left( cos{ \frac { 18\pi }{ 6 } -isin\frac { 18\pi }{ 6 } } \right) ^{ 18 }\)
= \(-\left( cos3\pi -isin3\pi -isin3\pi \right) =1+0i\)
Therefore, \(\left( sin\frac { \pi }{ 6 } +icos\frac { \pi }{ 6 } \right) ^{ 18 }=1\)

Let \(z=(cos\ \theta +isin\ \theta )\)
By de Moivre’s theorem ,
\(zn=\left( cos\ \theta +isin\ \theta \right) ^{ n }=cos\ n\theta +isin\ n\theta \)
\(\frac { 1 }{ { z }^{ n } } ={ z }^{ -n }=cos\ n \theta -isin\ n \theta \)
Therefore,\(z^{ n }+\frac { 1 }{ { z }^{ n } } =\left( cos\ n\theta +isin\ n\theta \right) +(cos\ n\theta -isin\ n\theta )\)
\({ z }^{ n }+\frac { 1 }{ { z }^{ n } } =2cosn\theta \)
Similarly,
\({ z }^{ n }-\frac { 1 }{ { z }^{ n } } =\left( cos\ n\theta +isin\ n\theta \right) -\left( cos\ n\theta -isin\ n\theta \right) \)
\({ z }^{ n }-\frac { 1 }{ { z }^{ n } } =2isin\ n\theta \)

Let −1−i = \(r(cos\ \theta +i\ sin\ \theta )\)
We have r = \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 1+1 } =\sqrt { 2 } \)
\(\alpha =tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }1=\frac { \pi }{ 4 } \)
Since the complex number −1−i lies in the third quadrant, it has the principal value,
\(\theta =\alpha -\pi =\frac { \pi }{ 4 } -\pi =-\frac { 3\pi }{ 4 } \)
Therefore,\(-1-i=\sqrt { 2 } \left( cos\left( \frac { 3\pi }{ 4 } \right) +isin\left( \frac { 3\pi }{ 4 } \right) \right) \)
= \(\sqrt { 2 } \left( cos\frac { 3\pi }{ 4 } -isin\frac { 3\pi }{ 4 } \right) \)
\(-1-i=\sqrt { 2 } \left( cos\left( \frac { 3\pi }{ 4 } +2k\pi \right) -isin\left( \frac { 3\pi }{ 4 } +2k\pi \right) \right) \)
Depending upon the various values of k , we get various alternative polar forms.

Consider the equation |z+2−i| = 2.
This can be written as |z−(−2+i) | = 2.
The above equation represents the circle with centre z₀ = -2+i and radius r = 2. Therefore z + 2−i<2 represents all points inside the circle with centre at −2+i and radius 2 as shown in figure.

The given equation |3z −5+i- 4| can be written as

3\(\left| z-\frac { 5-i }{ 3 } \right| =4\Rightarrow \left| z-\left( \frac { 5 }{ 3 } -\frac { i }{ 3 } \right) \right| =\frac { 4 }{ 3 } \)
It is of the form |z−z₀| = r and so it represents a circle, whose centre and radius are \(\left( \frac { 5 }{ 3 } ,\frac { i }{ 3 } \right) \) and \(\frac { 4 }{ 3 } \) respectively.

\(c\ is\frac { 2\pi }{ 9 } +c\ is\frac { 4\pi }{ 9 } +c\ is\frac { 6\pi }{ 9 } +c\ is\frac { 8\pi }{ 9 } +c\ is\left( \frac { 10\pi }{ 9 } \right) +c\ is\frac { 12\pi }{ 9 } +c\ is\frac { 14\pi }{ 9 } +c\ is\frac { 16\pi }{ 9 } \)
=\(\ c\ is\left( \frac { 2\pi }{ 9 } +\frac { 4\pi }{ 9 } +\frac { 6\pi }{ 9 } +\frac { 8\pi }{ 9 } +\frac { 10\pi }{ 9 } +\frac { 12\pi }{ 9 } +\frac { 14\pi }{ 9 } +\frac { 16\pi }{ 9 } \right) \)

= \(\left[ \because 1+2+3+....+n=\frac { n(n+1) }{ 2 } \right] \)
= c is 8π = [cos(8π)+i sin 8π]
= -1 + i(0) [∴ cos8π = -1 and sin 8π = 0 = -1

Given 2cos α = x+\(\frac { 1 }{ x } \)
⇒ 2cos α = \(\frac { { x }^{ 2 }+1 }{ x } \)
⇒ x²+1 = 2xcos 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 = cosβ ± i sinβ
\({ x }^{ m }{ y }^{ n }+\cfrac { 1 }{ { x }^{ m }{ y }^{ n } } =2cos(m\alpha +n\beta )\)
x^myⁿ = (cos α + i sin mα) (cos nβ + i sin nβ)
cos(mα+nβ)+i sin(mα+nβ)
\(\frac { 1 }{ { x }^{ m }{ y }^{ n } } \) = cos(mα+nβ)-i sin(mα+nβ)

= 2cos(mα+nβ)

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]
= \(-2\left( \frac { \sqrt { 3 } }{ 2 } \right) =-\sqrt { 3 } \).

LHS =  \(\cfrac { a+b\omega +c{ \omega }^{ 2 } }{ b+c\omega +{ a\omega }^{ 2 } } +\cfrac { a+b\omega +{ c\omega }^{ 2 } }{ c+a\omega +b{ \omega }^{ 2 } } \)
\(\cfrac { a\omega ^{ 3 }+b\omega +c{ \omega }^{ 2 } }{ b+c\omega +{ a\omega }^{ 2 } } +\cfrac { a\omega ^{ 3 }+b\omega ({ \omega }^{ 3 })+{ c\omega }^{ 2 } }{ c+a\omega +b{ \omega }^{ 2 } } \) [∵ ω³ = 1]

= ω + ω² = -1              [ ∵ 1 + ω + ω² = 0]

\(\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \left( cos\frac { \pi }{ 12 } +isin\frac { \pi }{ 12 } \right) \)
[∵ (cosθ₁+isinθ₁)(cosθ₂+isonθ₂)
= cos(θ₁+θ₂)+isin(θ₁+θ₂)
= \(cos\left( \frac { 2\pi +\pi }{ 12 } \right) +isin\left( \frac { 2\pi +\pi }{ 12 } \right) \)
\(cos\left( \frac { 3\pi }{ 12 } \right) +isin\left( \frac { 3\pi }{ 12 } \right) \)
\(cos\left( \frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \)
\(\frac { 1 }{ \sqrt { 2 } } +i\frac { 1 }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \)(1+i)
Aliter:
\( \left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right) \)
\( =\cos \left(\frac{\pi}{6}+\frac{\pi}{12}\right)+i \sin \left(\frac{\pi}{6}+\frac{\pi}{12}\right) \)
\( =\left(\cos \frac{3 \pi}{12}+i \sin \frac{3 \pi}{12}\right) \)
\( =\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \)
\( =\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}=\frac{1+i}{\sqrt{2}} \)

|z-4| = 16
Given z = x + iy
|z - 4| = 16
⇒ |x + iy - 4| = 16
⇒ |(x- 4) + iy| = 16
⇒ \(\\ \sqrt { (x-4)^{ 2 }+{ y }^{ 2 } } \) = 16
⇒ (x - 4)² + y² = 16²
[Squaring both sides]
⇒ x²-8x + 16 + y² = 256
⇒ x²-8x + y²+ 16-256 = 0
⇒ x²-8x + y²-240 = 0 Which is the required Cartesian equation.
The locus of the point is a circle.

Given z³+2\(\overline { z } \) = 0
⇒ z³ = -2\(\overline { z } \)
Taking modulus, |z³| = |-2\(\overline { z } \)|
⇒ |z|³ = 2|z|
⇒ |z| [|z|2-2] = 0
|z| = 0 or |z|²-2 = 0
⇒ |z| = 0
⇒ z = 0 is a solution ....(1)
|z²| = 2
⇒ |z²| = 2 ⇒ (z\(\overline { z } \))² = 2
⇒ z\(\overline { z } \) = \(\sqrt { 2 } \Rightarrow \overline { z } \frac { \sqrt { 2 } }{ z } \)
Given z + 2\(\overline { z } \) = 0
⇒ z³+2\(\frac { \sqrt { 2 } }{ z } \) = 0
⇒ z⁴+2\(\sqrt { 2 } \) = 0
If has 4 non-zero solutions
Hence from (1) and (2), z³+2\(\overline { z } \) has 5 solutions.

Given \(\left| z-\frac { 2 }{ z } \right| \) = 2
Consider |z| = \(\left| z-\frac { 2 }{ z } +\frac { 2 }{ z } \right| \)
≤ \(\left| z-\frac { 2 }{ z } \right| +\left| \frac { 2 }{ z } \right| \)   [Triangle law of in equality]
|z| ≤ \(\frac { 2|z|+2 }{ |z| } \)  [∵ \(\left| z-\frac { 2 }{ z } \right| \) = 2]
≤ \(\frac { 2|z|+2 }{ |z| } \) ⇒ |z|² ≤ 2|z|+2 ⇒ |z|²-2|z|≤ 2
Adding 1 both sides.
|z|²-2|z|+1 ≤ 2+1
⇒ [|z|-1]² ≤ 3 ⇒ |z|-1 ≤ ±\(\sqrt { 3 } \)
⇒ |z| ≤ ±\(\sqrt { 3 } \)+1
∴ The greatest value of |z| is \(\sqrt { 3 } \)+1 and the least value pf |z| is 1-\(\sqrt { 3 } \) respectively.

|z²-3| ≤ |z²|+|-3| [Triangle law of inequality]
≤ |z|²+3≤1+3  [∴ |z| = 1]
|z²-3| ≤ 4 ..............(1)
Also, |z²- 3| ≥ ||z²|-|-3||
≥ ||z|²-3|   [∵ |-3| = 3]
≥ |1²-3|  [∵ |z| = 1]
≥ |-2| .
|z²-3| ≥ 2.............(2)
From (1) and (2) we get 2 ≤ |z²-3| ≤ 4
Hence proved.

Let the points be A (10 - 8i), B (11 + 6i) and C(1-i)
Distance between A and C is |(10-8i)-(1+i)|
= |10-8i-1-i| = |9-9i|
= \(\sqrt { { 9 }^{ 2 }+{ (-9) }^{ 2 } } =\sqrt { 81+81 } =\sqrt { 2\times 81 } =\sqrt { 162 } \)
= 9\(\sqrt{2}\)
Distance between Band C is s |(11+6i)-(1+i)|
= |11 + 6i-1-i| = |10 + 5i|
=\(\sqrt { { 10 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 100+25 } =\sqrt { 125 } \)
=\(\sqrt { 25\times 5 } =5\sqrt { 5 } \)
Since \(5\sqrt { 5 } <9\sqrt {2 } \) , B is closest to C.
∴ 11 + 6i is closet to 1 + i.

Given |z₁| = |z₂| = 1
⇒ z₁ \(\bar { z } \) = 1
⇒ z₁ = \(\frac { 1 }{ \bar { { z }_{ 1 } } } \)
and z₁ z₂ ≠ -1
Also z₂\(\bar { { z }_{ 2 } } \) = 1
⇒ z₂ = \(\frac { 1 }{ \bar { { z }_{ 2 } } } \)
Consider \(\frac { z_{ 1 }+{ z }_{ 2 } }{ 1+{ z }_{ 1 }{ z }_{ 2 } } \)
∴ \(\frac { z_{ 1 }+{ z }_{ 2 } }{ 1+{ z }_{ 1 }{ z }_{ 2 } } =\frac { \frac { 1 }{ \bar { { z }_{ 1 } } } +\frac { 1 }{ \bar { { z }_{ 2 } } } }{ 1+\frac { 1 }{ \bar { { z }1 } } .\frac { 1 }{ \bar { { z }_{ 2 } } } } =\frac { \frac { \bar { { z }_{ 2 } } +\bar { { z }_{ 1 } } }{ \bar { { z }_{ 1 } } \bar { { z }_{ 2 } } } }{ \frac { \bar { { z }_{ 1 } } \bar { { z }_{ 2 } } +1 }{ \bar { { z }_{ 1 } } \bar { { z }_{ 2 } } } } \)
= \(\frac { \bar { { z }_{ 2 } } +\bar { { z }_{ 1 } } }{ \bar { { z }_{ 1 } } \bar { { z }_{ 2 } } +1 } \)
= \(\left( \frac { \overline { { z }_{ 1 }{ +z }_{ 2 } } }{ 1+{ z }_{ 1 }{ z }_{ 2 } } \right) \)
∴ \(\frac { z_{ 1 }+{ z }_{ 2 } }{ 1+{ z }_{ 1 }{ z }_{ 2 } } =\frac { \frac { 1 }{ \bar { { z }_{ 1 } } } +\frac { 1 }{ \bar { { z }_{ 2 } } } }{ 1+\frac { 1 }{ \bar { { z }1 } } .\frac { 1 }{ \bar { { z }_{ 2 } } } } =\frac { \frac { \bar { { z }_{ 2 } } +\bar { { z }_{ 1 } } }{ \bar { { z }_{ 1 } } \bar { { z }_{ 2 } } } }{ \frac { \bar { { z }_{ 1 } } \bar { { z }_{ 2 } } +1 }{ \bar { { z }_{ 1 } } \bar { { z }_{ 2 } } } } \) is real [∵ z = \(\bar { z } \) ⇒ is real]

It is enough to prove that the sides of the triangle are equal.
Let z₁ = 1, \({ z }_{ 2 }=\frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \) and \({ z }_{ 3 }=\frac { -1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \)
The length of the sides of the triangles are
\(\left| { z }_{ 1 }-{ z }_{ 2 } \right| =\left| 1-\left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) \right| =\left| \cfrac { 3 }{ 2 } -\cfrac { \sqrt { 3 } }{ 2 } i \right| =\sqrt { \frac { 9 }{ 4 } +\frac { 3 }{ 4 } } =\frac { 2\sqrt { 3 } }{ 2 } =\sqrt { 3 } \)
\(\left\lfloor { z }_{ 2 }-{ z }_{ 3 } \right\rfloor =\left| \left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) -\left( \frac { -1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \right) \right| =\sqrt { \left( \sqrt { 3 } \right) ^{ 2 } } =\sqrt { 3 } \)
\(\left| { z }_{ 3 }-{ z }_{ 1 } \right| =\left| \left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) -1 \right| =\left| \frac { -3 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } i \right| =\sqrt { \frac { 9 }{ 4 } +\frac { 3 }{ 4 } } =\sqrt { 3 } \)
Since the sides are equal, the given points form an equilateral triangle

\(\left| z+3+4i \right| \le \left| z \right| +\left| 3+4i \right| =2+5=7\)
\(\left| z+3+4i \right| \le 7\) .............. (1)
\(\left| z+3+4i \right| \ge \left| \left| z \right| -\left| 3+4i \right| \right| =\left| 2-5 \right| =3\)
\(\left| z+3+4i \right| \ge 3\) ............ (2)
From (1) and (2) we get, \(3\le \left| z+3+4i \right| \le 7\) 

Given v = 3-4i, w = 4+3i and \(\frac { 1 }{ u } =\frac { 1 }{ v } +\frac { 1 }{ w } \)
∴ \(\frac { 1 }{ u } =\frac { 1 }{ 3-4i } +\frac { 1 }{ 4+3i } \)
= \(\frac { 3+4i }{ (3-4i)(3+4i) } +\frac { 4-3i }{ (4+3i)(4-3i) } \)
= \(\\ \frac { 3+4i }{ 9-(4i)^{ 2 } } +\frac { 4-3i }{ 16-(3i)^{ 2 } } =\frac { 3+4i }{ 9+16 } +\frac { 4-3i }{ 16+9 } \)
= \(\frac { 3+4i }{ 25 } +\frac { 4-3i }{ 25 } =\frac { 3+4i+4-3i }{ 25 } \)
\(\frac { 1 }{ u } =\frac { 7+i }{ 25 } \)
∴ u = \(\frac { 25 }{ 7+i } \times \frac { 7-i }{ 7-i } =\frac { 25(7-i }{ 7^{ 2 }-({ i }^{ 2 }) } \) 
= \(\frac { 25(7-i) }{ 49+1 } =\frac { 25(7-i) }{ 50 } =\frac { 1 }{ 2 } \)(7-i)
∴ u = \(\frac { 1 }{ 2 } \)(7-i) or \(\frac { 7 }{ 2 } \) - \(\frac { i }{ 2 } \)

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

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

(z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)
Given z₁= 1-3i, z₂ - 4i and z₃ = 5
LHS = (z₁+ z₂) + z₃
= [1- i + (- 4i)] + 5
[1-7i] + 5
= 6 -7i
RHS = z₁+ (z₂ + z₃)
= 1- 3i + (-4i 5)
= 6 - 7i
LHS = RHS
∴ (z₁+ z₂)+ z₃ = z₁+(z₂+ z₃)

Let z₁ = (2+i)x + (1−i)y + 2i−3 = (2x+y−3) + i(x−y+ 2)and
z₂ = x+(−1+2i)y+1+i = (x−y+1) + i(2y+1)
Given that z₁ = z₂ 
Therefore (2x+y−3) + i(x−y+2) = (x−y+1) + i(2y+1).
Equating real and imaginary parts separately, gives
2x+y−3 = x−y+1 \(\Rightarrow\) x+2y = 4
x−y+2 = 2y +1 \(\Rightarrow\) x−3y = −1
Solving the above equations, gives
x = 2 and y = 1.

\(\theta =\frac { 3\pi }{ 2 } \)
When θ = \(\frac { 3\pi }{ 2 } \)
Rotation of z is \(ze^{ i\theta }=ze^{ i\frac { 3\pi }{ 2 } }\)
= \(2\sqrt { 2 } e^{ -\frac { \pi }{ 4 } }.e^{ i\frac { 3\pi }{ 2 } }=2\sqrt { 2 } e^{ i\left( i\frac { \pi }{ 2 } -\frac { \pi }{ 4 } \right) }\)
= \(2\sqrt { 2 } e^{ i5^{ \frac { \pi }{ 4 } } }\)

\(\theta =\frac { 2\pi }{ 3 } \)
When θ = \(\frac { \pi }{ 3 } \)
Roration of z is \(ze^{ i\theta }=ze^{ i2\frac { \pi }{ 3 } }\) [using (1)]
= \(2\sqrt { 2 } e^{ -i\frac { \pi }{ 4 } }.{ e }^{ i2\frac { \pi }{ 3 } }\)   [using (1)]
= \(2\sqrt { 2 } e^{ i\left( i\frac { \pi }{ 3 } -\frac { \pi }{ 4 } \right) }=2\sqrt { 2 } e^{ i5\frac { \pi }{ 12 } }\)

|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 \(\frac{8}{3}\).

\(\left| 2z+2-4i \right| =2\)
2|z+1-2i| = 2
⇒ |z-(-1+2i)| = 1
It is of the form |z - z₀| = r and so it represents a circle.
Its centre is (-1+2i) and radius is 1.

\(\frac { 3 }{ 2 } \left( cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 6 } \right) .6\left( cos\frac { 5\pi }{ 6 } +\frac { 5\pi }{ 6 } \right) \)
= \(\left( \frac { 3 }{ 2 } \right) \left( 6 \right) \left( cos\left( \frac { \pi }{ 3 } +\frac { 5\pi }{ 6 } \right) +isin\left( \frac { \pi }{ 3 } +\frac { 5\pi }{ 6 } \right) \right) \)
= \(9\left( cos\left( \frac { 7\pi }{ 6 } \right) +isin\left( \frac { 7\pi }{ 6 } \right) \right) \)
= \(9\left( cos\left( \pi +\frac { \pi }{ 6 } \right) +isin\left( \pi +\frac { \pi }{ 6 } \right) \right) \)
= \(9\left( -cos\left( \frac { \pi }{ 6 } \right) -isin\left( \frac { \pi }{ 6 } \right) \right) \)
= \(9\left( -\frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) =\frac { 9\sqrt { 3 } }{ 2 } -\frac { 9i }{ 2 } \)

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 \(\frac { 2\pi }{ 3 } \)

\(\theta =\frac { \pi }{ 3 } \)
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)]
= \(2\sqrt { 2 } \left[ { e }^{ \left( i\frac { \pi }{ 3 } -\frac { \pi }{ 4 } \right) } \right] =2\sqrt { 2 } e^{ i\frac { \pi }{ 12 } }\)

\(\left| z-2-i \right| =3\)
⇒ |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

Using the given values forr z₁, z₂ and z₃ we get |z₁| = |3+4i| =\(\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } =5\)
|z₂| = |5-12i| = \(\sqrt { { 5 }^{ 2 }+(-12)^{ 2 } } =13\)
|z₃| = |6+8i| = \(\sqrt { { 6 }^{ 2 }+{ 8 }^{ 2 } } =10\)
|z₁+z₂| = |(3+4i)+(5-12i)| = |8-8i| = \(\sqrt { 128 } =8\sqrt { 2 } \)
|z₂-z₃| = |(5-12i)-(6+8i)| = |1-20i| = \(\sqrt { 401 } \)
|z₁+z₃| = |(3+4i)+(6+8i)| = |9+12i| = \(\sqrt { 225 } =15\)
Note that the triangle inequality is satisfied in all the cases
|z₁+z₃| = |z₁|+|z₃| = 15 

We have z = (2+3i)(1−i) = (2+3)+(3−2)i = 5+i
\(\Rightarrow\) \({ z }^{ -1 }=\frac { 1 }{ z } =\frac { 1 }{ 5+i } \)
Multiplying the numerator and denominator by the conjugate of the denominator, we get
\({ z }^{ -1 }=\frac { \left( 5-i \right) }{ \left( 5+i \right) \left( 5-i \right) } =\frac { 5-i }{ { 5 }^{ 2 }+{ I }^{ 2 } } =\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)
\(\Rightarrow\)\({ z }^{ -1 }=\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)

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

We have = \(\frac { z+3 }{ z-5i } =\frac { 1+4i }{ 2 } \)
\(\Rightarrow\) 2(z + 3) = (1 + 4i) (z− 5i)
\(\Rightarrow\) 2z + 6 = (1 + 4i)z + 20−5i
\(\Rightarrow\) (2−1−4i)z = 20− 5i− 6 
\(\Rightarrow\) \(z=\frac { 14-5i }{ 1-4i } =\frac { \left( 14-5i \right) \left( 1+4i \right) }{ \left( 1-4i \right) \left( 1+4i \right) } =\frac { 34+51i }{ 17 } =2+3i\)

To find the real and imaginary parts of \(\frac { 3+4i }{ 5-12i } \) first it should be expressed in the rectangular form x + iy .
To simplify the quotient of two complex numbers, multiply the numerator and denominator by the conjugate of the denominator to eliminate i in the denominator
\(\frac { 3+4i }{ 5-12i } =\frac { (3-4i)(5+12i) }{ (5-12i)(5+12i) } \)
= \(\frac { (15-48)+(20+36)i }{ { 5 }^{ 2 }+{ 12 }^{ 2 } } \)
= \(\frac { -33+56i }{ 169 } =-\frac { 33 }{ 169 } +i\frac { 56 }{ 169 } \)
Therefore,\(\frac { 3+4i }{ 5-12i } =-\frac { 33 }{ 169 } +i\frac { 56 }{ 169 } \) This is in the x + iy form.
Hence real part is \(-\frac { 33 }{ 169 } \) and imaginary part is \(\frac { 56 }{ 169 } \)

\(\sum _{ r=1 }^{ n }{ tan^{ -1 } } \left( \frac { { y }_{ r } }{ { x }_{ r } } + \right) ={ tan }^{ -1 }\left( \frac { b }{ a } \right) +2k\pi ,k\epsilon Z\)
(x₁ + iy₁)(x₂ + iy₂) ....(x_n + iy_n) = a + ib
Taking argument
arg(x₁ + iy₁)(x₂ + iy₂) ....(x_n + iy_n)) = arg(a+ib)
⇒ arg(x₁+ iy₁) + arg(x₂ + iy₂)+...+ arg(x_n + iy_n) = arg(a + ib)
(∵ arg(z₁z₂....z_n) = argz₁+arg z₂+...+ argz_n)
⇒ \(tan^{ -1 }\left( \frac { { y }_{ 1 } }{ { x }_{ 1 } } \right) +tan^{ -1 }\left( \frac { { y }_{ 2 } }{ { x }_{ 2 } } \right) +...+tan^{ -1 }\left( \frac { { y }_{ n } }{ { x }_{ n }\\ } \right) \)
= \(tan^{ -1 }\left( \frac { b }{ a } \right) +2k\pi\) \( k\in Z\)
⇒ \(\overset { n }{ \underset { r=1 }{ \Sigma } } tan^{ -1 }\left( \frac { { y }_{ r } }{ { x }_{ r } } \right) =tan^{ -1 }\left( \frac { b }{ a } \right) +2k\pi \ k\in Z\).

Let \(z=cos2\theta +isin2\theta \)
As |z| = |z|² = z\(\bar { z } \) = 1, we get \(\bar { z } =\frac { 1 }{ z } =cos2\theta -isin2\theta \)
Therefore, \(\frac { 1+cos2\theta +isin2\theta }{ 1+cos2\theta -isin2\theta } =\frac { 1+z }{ 1+\frac { 1 }{ z } } =\frac { \left( 1+z \right) z }{ z+1 } =z\)
Therefore, \(\left( \frac { 1+cos2\theta +isin2\theta }{ 1+cos2\theta -isin2\theta } \right) ^{ 30 }={ z }^{ 30 }=\left( cos2\theta +isin2\theta \right) ^{ 30 }\)
= \(cos60\ \theta +isin60\ \theta \)

We have to find \((\sqrt{3}+1)^{\frac{1}{3}}\). Let \(z=(\sqrt{3}+i)^{\frac{1}{3}}\). Then \({ z }^{ 3 }=\sqrt { 3 } +i=r\left( cos\theta +isin\theta \right) \)
Then, \(r=\sqrt { 3+1 } =2\) and \(\alpha =\theta =\frac { \pi }{ 6 } \) (\(\because \sqrt{3}+i\) lies in the first quadrant)
Therefore, \({ z }^{ 3 }=\sqrt { 3 } +i=2\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \)
\(\Rightarrow z=\sqrt [ 3 ]{ 2 } \left( cos\left( \frac { \pi +12k\pi }{ 18 } \right) +isin\left( \frac { \pi +12k\pi }{ 18 } \right) \right) \), k = 0, 1, 2.
Taking k = 0, 1, 2, we get
k = 0, z \(={ 2 }^{ \frac { 1 }{ 3 } }\left( cos\frac { \pi }{ 18 } +sin\frac { \pi }{ 18 } \right) \)
k = 1, \(z={ z }^{ \frac { 1 }{ 3 } }\left( cos\frac { \pi }{ 18 } +sin\frac { \pi }{ 18 } \right) \)
k = 2, \(z={ 2 }^{ \frac { 1 }{ 3 } }\left( cos\frac { 25\pi }{ 18 } +sin\frac { 25\pi }{ 18 } \right) ={ 2 }^{ \frac { 1 }{ 3 } }\left( -cos\frac { 7\pi }{ 18 } -sin\frac { 7\pi }{ 18 } \right) \)

Now, \(\frac { z-1 }{ z+1 } =\frac { x+iy-1 }{ x+iy+1 } =\frac { \left( x-1 \right) +iy }{ \left( x+1 \right) +iy } =\frac { \left[ \left( x-1 \right) +iy \right] \left[ \left( x+1 \right) -iy \right] }{ \left[ \left( x+1 \right) +iy \right] \left[ \left( x+1 \right) -iy \right] } \)
\(\Rightarrow \frac { z-1 }{ z+1 } =\frac { \left( { x }^{ 2 }+{ y }^{ 2 }-1 \right) +i\left( 2y \right) }{ \left( x+1 \right) ^{ 2 }+{ y }^{ 2 } } \)
Since, arg \(\left( \frac { z-1 }{ z+2 } \right) =\frac { \pi }{ 2 } \Rightarrow { tan }^{ -1 }\left( \frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 }-1 } \right) \)= \(\frac { \pi }{ 2 } \)
\(\Rightarrow \frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 }-1 } =tan\frac { \pi }{ 2 } \) ⇒ x²+ y² − 1 = 0
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }=1\)

z³ = -27 = (-1 \(\times\) 3)³ = -1 \(\times\) 33
z = \((-1)^{ \frac { 1 }{ 3 } }\times 3^{ 3\times \frac { 1 }{ 3 } }=(-1)^{ \frac { 1 }{ 3 } }\)\(\times\) 3
∴ z = 3\(\left[ cos\pi +isin\pi \right] ^{ \frac { 1 }{ 3 } }\)
[∵ cos π = -1 and sin π = 0]
= 3\(\left[ cos\frac { 1 }{ 3 } (2k\pi +\pi )isin\frac { 1 }{ 3 } (2k\pi +\pi ) \right] \)
k = 0, 1, 2
When k = 0,
z = 3\(\left[ cos\frac { 1 }{ 3 } (\pi )isin\frac { 1 }{ 3 } (\pi ) \right] =3cos\frac { \pi }{ 3 } \)
When k = 1
z = 3\(\left[ cos\frac { 1 }{ 3 } (3\pi )isin\frac { 1 }{ 3 } (3\pi ) \right] \)
= 3[cos π + i sin π] = 3(-1+0) 
When k = 2
z = 3\(\left[ cos\frac { 1 }{ 3 } (5\pi )isin\frac { 1 }{ 3 } (5\pi ) \right] =3\left[ cos5\frac { \pi }{ 3 } \right] \)
Hence, the roots are 3 cis\(\frac { \pi }{ 3 } \), -3, 3 c is 5\(\frac { \pi }{ 3 } \)

Given (x₁ + iy₁)(x₂ + iy₂) ....(x_n + iy_n) = a + ib
Taking modulus
|(x₁+ iy₁)(x₂ + iy₂) ....(x_n + iy_n)| = |a + ib|
|x₁ + iy₁| + |x₂ + iy₂|+...+ |x_n + iy_n| = |a + ib|
\( \sqrt{x_{1}^{2}+y_{1}^{2}} \sqrt{x_{2}^{2}+y_{2}^{2}} \sqrt{x_{3}^{2}+y_{3}^{2}} \cdots \sqrt{x_{n}{ }^{2}+y_{n}^{2}} =\sqrt{a^{2}+b^{2}} \)
Squaring on both sides
\( \left(x_{1}^{2}+\mathrm{y}_{1}^{2}\right)\left(x_{2}^{2}+\mathrm{y}_{2}^{2}\right)\left(x_{3}^{2}+\mathrm{y}_{3}^{2}\right) \ldots\left(x_{\mathrm{n}}^{2}+\mathrm{y}_{\mathrm{n}}^{2}\right) =\mathrm{a}^{2}+\mathrm{b}^{2} \)
Hence proved.