⇒ 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 = \(\frac{1}{3}\).

z₁z₂ = (1 - i)(-2 + 4i) = -2 + 4i + 2i - 4i²
= -2 + 6i + 4 = 2 + 6i
\(\bar { { z }_{ 1 } } \) = 1+i
∴ \(\frac { { z }_{ 1 }{ z }_{ 2 } }{ \bar { { z }_{ 1 } } } =\frac { 2+6i }{ 1+6i } \times \frac { 1-i }{ 1-i } =\frac { 2(1-i+3i-i^{ 2 }) }{ 1+1 } \)
= 1 + 2i + 3
= 4 + 2i
∴ Im\(\left( \frac { { z }_{ 1 }{ z }_{ 2 } }{ \bar { { z }_{ 1 } } } \right) \) = 2

\(\theta =\frac { 2\pi }{ 3 } \)
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)]
= \(2\sqrt { 2 } e^{ i\left( i\frac { \pi }{ 3 } -\frac { \pi }{ 4 } \right) }=2\sqrt { 2 } e^{ i5\frac { \pi }{ 12 } }\)

\(\left| 2z+2-4i \right| =2\)
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.

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

 \(\overline { 3i } +\frac { 1 }{ 2-i } \)
= - 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 w
= (5-2i)(-1+3i)
= -5+15i+2i-6i²
= -5+17i-6(-1)
= -5+17i+6
= 1+17i

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

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

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

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
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }=1\)

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)

Given z = x + iy and arg\(\left( \frac { z-i }{ z+2 } \right) =\frac { \pi }{ 4 } \)
⇒ arg(z-i) - arg(z+2) = \(\frac { \pi }{ 4 } \)
⇒ arg(x + iy-i) - arg(x+iy+2) = \(\frac { \pi }{ 4 } \)
⇒ arg(x+i(y-1)-arg((x+2)+iy) = \(\frac { \pi }{ 4 } \)
⇒ \(tan^{ -1 }\left( \frac { y-1 }{ x } \right) -tan^{ -1 }\left( \frac { y }{ x+2 } \right) \) = \(\frac { \pi }{ 4 } \)
⇒ \(tan^{ -1 }\left( \frac { \frac { y-1 }{ x } -\frac { y }{ x+2 } }{ 1+\frac { y-1 }{ x } .\frac { y }{ x+2 } } \right) \)
= \(\frac { \pi }{ 4 } \)\(\left[ \because tan^{ -1 }x-tan^{ -1 }y=tan^{ -1 }\left( \frac { x-y }{ 1+xy } \right) \right] \)
\(\Rightarrow \frac{\left(\frac{(x+2)(y-1)- x y}{\not {x (\not x+\not2)}}\right)}{\left(\frac{x(x+2)+y(y-1)}{\not x(\not x+\not 2)}\right)}=\tan \frac{\pi}{4}=1\)
⇒ \(\frac { (x+2)(y-1)-xy }{ x(x+2)+y(y-1) } \) = 1
⇒ -x + 2y-2 = x²+ 2x + y²-y
⇒ x² + 2x + y²-y + x-2y + 2 = 0
⇒ x² + y²+3x-3y + 2 = 0
Hence proved.

Given |z₁| = 1, |z₂|= 2, |z₃| = 3, |z₁ + z₂ +  z₃| = 1
|z₁|² = 1² ⇒ z₁ \(\overline { { z }_{ 1 } } \) = 1 ⇒ z₁ = \(\frac { 1 }{ { z }_{ 1 } } \)
|z₂|² = 4 ⇒ z₂ \(\overline { { z }_{ 2 } } \) = 1 ⇒ z₂ = \(\frac { 4 }{ { z }_{ 2 } } \)
|z₃|² = 9 ⇒ z₃ \(\overline { { z }_{ 3 } } \) = 1 ⇒ z₃ = \(\frac { 9 }{ { z }_{ 3 } } \)
∴ \(\left| 9,\frac { 1 }{ \overline { { z }_{ 1 } } } .\frac { 4 }{ \overline { { z }_{ 2 } } } +4.\frac { 1 }{ \overline { { z }_{ 1 } } } .\frac { 9 }{ \overline { { z }_{ 3 } } } +\frac { 4 }{ \overline { { z }_{ 2 } } } .\frac { 9 }{ \overline { z_{ 3 } } } \right| \)
\(\left| \frac { 36 }{ \overline { { z }_{ 1 } } \overline { { z }_{ 2 } } } +\frac { 36 }{ \overline { { z }_{ 1 } } \overline { { z }_{ 3 } } } +\frac { 36 }{ \overline { { z }_{ 2 } } \overline { { z }_{ 3 } } } \right| =\left| 36\left( \frac { \overline { { z }_{ 3 } } +\overline { { z }_{ 2 } } +\overline { { z }_{ 1 } } }{ \overline { { z }_{ 1 } } \overline { { z }_{ 2 } } \overline { { z }_{ 3 } } } \right) \right| \)
\(\\ \left[ \because |\overline { z_{ 1 } } +\overline { { z }_{ 2 } } +\overline { { z }_{ 3 } } |=|\overline { { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } } | \right] \)
=\(\frac { 36|\overline { { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } } | }{ |\overline { { z }_{ 1 } } ||\overline { { z }_{ 2 } } ||\overline { { z }_{ 3 } } | } =36\frac { |\overline { \overline { { z }_{ 1 } } +\overline { z_{ 2 } } +\overline { z_{ 3 } } | } }{ |\overline { { z }_{ 1 } } ||\overline { { z }_{ 2 } } ||\overline { { z }_{ 3 } } | } \)
\(\left[ \because |\overline { { z }_{ 1 } } |=|{ z }_{ 1 }|,|\overline { { z }_{ 2 } } =|{ z }_{ 21 }|,|\overline { { z }_{ 3 } } |=|\overline { { z }_{ 3 } } | \right] \)
\(=\frac{36(1)}{1(2)(3)}=\frac{\not 36}{\not 6}=6\)
∴ |9z₁ + z₂ + 4z₁z₃ + z₂z₃| = 6

Given p+iq = \(\frac { (a+i)^{ 2 } }{ 2a-i } \) ...........(1)
Taking conjugate both sides we get,
p-iq =\(\frac { (a+i)^{ 2 } }{ 2a+i } \) ..........(2)
Multiplying (1) and (2) we get,
(p+iq)(p-iq) = \(\frac { (a+i)^{ 2 } }{ 2a-i } \times \frac { (a-i)^{ 2 } }{ 2a+i } \)
p²+q² = \(\frac { [(a+i)(a-i)^{ 2 }] }{ 4a^{ 2 }+1 } =\frac { ({ a }^{ 2 }+1)^{ 2 } }{ 4a^{ 2 }+1 } \).
Hence proved

Given  |3z - 5| = 3 |z + 1
⇒ |3(x+iy)-5| = 3|x+iy+1|
⇒ |(3x-5)+3y| = 3|(x+1)+iy|
⇒ \(\sqrt { (3x-5)^{ 2 }+3^{ 2 } } =3\left[ \sqrt { (x+1)^{ 2 }+{ y }^{ 2 } } \right] \)
Squaring both sides we get,
(3x - 5)² + 9 = 9 [(x + 1)² + y²]
⇒ 9x² - 30x + 25 + 9 = 9 [x² + 2x + 1 + y²]
⇒ 48x - 16 = 0
⇒ 3x-1 = 0

AB = |(3+2i) - (0+5i)| = |3-3i|
=\(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
BC = |(0+5i) - (-3+2i)| = |3+3i|
= \(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
CD = |(-3+2i) - (0-i) = |-3+i|
= \(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
DA = |(0-i) - (3+2i)| = |-3-3i|
=\(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
∴ AB = BC = CD = DA
Also AC = |(3+2i) - (-3+2i)|
= |6| = \(\sqrt { 36 } \) = 6
∴ AC = BD
Hence ABCD is a square

-1 = i² = i \(\times\) i =\(\sqrt { -1 } \times \sqrt { -1 } =\sqrt { (-1) } \times \sqrt { (-1) } \)
= \(\sqrt { 1 } \)
⇒ -1 = i
In the above proof we have used \(\sqrt { -1 } \times \sqrt { -1 } \)
= \(\sqrt { (-1)(-1) } \) which is wrong
Since \(\sqrt { ab } =\sqrt { a } .\sqrt { b } \) is true only at least one of a and b is non-negative.

Given that z = 3 + 2i.
Therefore, iz = i(3 + 2i) = −2 + 3i
z + iz = (3 + 2i) + i(3 + 2i) = 1 + 5i
Let A,B, and C be z, z + iz, and iz respectively
\({ AB }^{ 2 }={ \left| (z+iz)-z \right| }^{ 2 }={ \left| -2+3i \right| }^{ 2 }=13\)
\({ BC }^{ 2 }{ =\left| iz-(z+iz) \right| }^{ 2 }={ \left| -3-2i \right| }^{ 2 }=13\)
\({ CA }^{ 2 }={ \left| z-iz \right| }^{ 2 }={ \left| 5-i \right| }^{ 2 }=26\)
Since AB² + BC² = CA² and AB = BC ΔABC is an isosceles right triangle.

(1+ω)(1+ω²)(1+ω⁴)(1+ω⁸).........(1+ω²)¹¹ = 1
LHS = (1+ω)(1+ω²)(1+ω⁴)(1+ω⁸)...........2(1+ω²)¹¹
\((1+{ \omega }^{ 2^{ 0 } })^{ 6 }((1+{ \omega }^{ 2^{ 1 } })(1+{ \omega }^{ { 2 }^{ 2 } })((1+{ \omega }^{ { 2 }^{ 11 } })\)
[∵ ω⁴ = ω³.ωω⁸ = ω⁶.ω²]
There are 12 terms
= (1+ω)(1+ω²)(1+ω)(1+\({ \omega }^{ { 2 }^{ n } }\)).......... upto 12 terms
= (1+ω)⁶(1+ω²)⁶(-ω)⁶(-ω²)⁶
= (ω³)6 = 16 = 1 = RHS.

|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

(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)

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