Question

solve this ........​

Answer 1

Solution:- This question can be solved by using two methods

1) Using cube root of unity

2)Using Simple mathematical calculation

First method

Solution:-given equation=z^2+z+1

$z = \frac{ - 1 + - \sqrt{ {1 - 4} } }{2} \\ z = \frac{ - 1 + - \sqrt{ - 3} }{2} = > \frac{ - 1 + - \sqrt{3}i }{2} \\ by \: using \: cube \: root \: of \: unity \: we \: can \: say \: \: that \\ z = w \: {w}^{2} \\ we \: know \: that \\ 1 + w + {w}^{2} = 0 \\ putting \: z = w \: we \: get \\ {(w + \frac{1}{w} })^{2} +( { {w}^{2} + \frac{1}{ {w}^{2} } })^{2} + ( { {w}^{3} + \frac{1}{ {w}^{3} } })^{2} + \\ { {(w}^{4} + \frac{1}{ {w}^{4} } })^{2} + ( { {w}^{5} + \frac{1}{ {w}^{5} } })^{2} + {( {w}^{6} + \frac{1}{ {w}^{6} }) }^{2} \\ = > ( { \frac{ {w}^{2} + 1 }{w} })^{2} + ( { \frac{ {w}^{4} + 1}{ {w}^{2} }) }^{2} + ( { \frac{ {w}^{6 } + 1}{ {w}^{3} }) }^{2} + \\ ( { \frac{ {w}^{8} + 1}{ {w}^{4} } )}^{2} + ( { \frac{ {w}^{10} + 1 }{ {w}^{5} }) }^{2} + ( { \frac{ {w}^{12} + 1}{ {w}^{6} } )}^{2}$

Using the relations

w^3=1

1+w=-w^2

w^2=-(1+w)

we can write

$( { - 1)}^{2} + ( { - 1})^{2} + {( - 1)}^{2} + { (- 1)}^{2} + 4 + 4 \\ = 12$

Second method

Given eqaution:-z^2+z+1=0

=>z^2+z=-1

=>(z+1)=-1/z

=>z+1/z=-1.....i)

square both side

=>z^2+1/z^2 =-1...ii)

multiply equation I) and ii)

z^3+1/z+z+1/z^3=1

z^3+1/z^3-1=1

z^3+1/z^3=2....iii)

multiplying equation i) and iii) we get

z^4+1/z^2 +z^2+1/z^4=-2

z^4+1/z^4 -1=-2

z^4+1/z^4=-1.....iv)

multiply equation I) and iv) we get

z^5+1/z^3+z^3+1/z^5=1

z^5+1/z^5+2=1

z^5+1/z^5=-1 ......v)

simlilarly multiplying v) and i) equation we get

z^6+1/z^6=-2......vi)

now,

putting the value of these eqautions we get

(-1)^2+(-1)^2+(2)^2+(-1)^2+(-1)^2+(-2)^2

1+1+4+1+1+4=12

Answer 2

Answer is 12.

Formula used : 1 + ω + ω² = 0

Where, ω & ω² are the complex cube root of unity.

Hope it helps...

^___°