Question

Solve..

Answer 1

\[\sum(x^n + 1/x^n) \]

n = 5 \[x^2 - x + 1 =0\]

Answer 2

i think sattelite did it for u

Answer 3

But...I have Some...Doubts...

Answer 4

\[x=\frac{1\pm i\sqrt{3}}{2}\]choose one of them for example\[x=\frac{1+ i\sqrt{3}}{2}=e^{i\frac{\pi}{3}}\]

Answer 5

\[x = (1 \pm i\sqrt{3})/2\]

Answer 6

and it is a cube root of -1

Answer 7

-1 , -W , -W^2

Answer 8

leave the cube root...u just need to know its magnitude is 1

for example\[x+x^{-1}=e^{i\frac{\pi}{3}}+e^{-i\frac{\pi}{3}}=1\]

Answer 9

-1 * -W * -W^2 = -1

Answer 10

This is wat u mentioned..

Answer 11

see yahoo\[x=\frac{1\pm i\sqrt{3}}{2}\]ok?

Answer 12

Yes..

Answer 13

Continue..@mukushla

Answer 14

and |x|=1

Answer 15

ok?

Answer 16

Yup

Answer 17

now suppose that conjugate of x is \(\overline{x}\) then\[x \overline{x}=|x|^2=1\]

Answer 18

Yes....))

Answer 19

so \[\frac{1}{x}=\overline{x}\]and exaxtly the same result for other exponents\[\frac{1}{x^2}=\overline{x^2}\]

Answer 20

then....

Answer 21

so .. for example when u calculate \[x^3+\frac{1}{x^3}\]u have\[x^3+\frac{1}{x^3}=x^3+\overline{x^3}=2Re(x^3)\]\[x^3=(e^{i\frac{\pi}{3}})^3=e^{i\pi}=1\]so \[x^3+\frac{1}{x^3}=2\]do the same thing for other exponents

Answer 22

thxxx