Question

\[z_1=\frac{-1+3i}{2}\]and \[z_2=\frac{-1-3i}{2}\]Find the value of \[z_1^3+z_2^3-3z_1z_2.\]

Its answer is given: -

\[\huge{\color{red}{-1.}}\]

Answer 1

Please help:)

Answer 2

I think there is typo in question
it must be like this: -
\[z_1=\frac{-1+\sqrt{3}i}{2}=\omega\]&\[z_2=\frac{-1-\sqrt{3}i}{2}=\omega^2.\]

Answer 3

give ur respective comments plz:)

Answer 4

z1*z2=(1-9)/4=-2

Answer 5

these are roots of unity. can you use exponential form?

Answer 6

because \(z_1^3=z_2^3=1\)

Answer 7

these aren't roots of unity

Answer 8

i think there is typo

Answer 9

no?

Answer 10

becoz there is only 3
nt sqrt 3

Answer 11

\[-\frac{1}{2}+\frac{\sqrt{3}}{2}i=e^{\frac{2\pi}{3}i}\]

Answer 12

oooh i see sorry!

Answer 13

yeah i was saying:)
np @satellite73 sir:)

Answer 14

ok if you grind it out you get it

Answer 15

oh sorry
z1^3+z2^3=(z1+z2)*(z1^2-z1z2+z2^2)
=(-1)[(z1+z2)^2-3z1z2]
=-1+3z1z2

Answer 16

thus
z1^3+z2^3-3z1z2=-1

Answer 17

u mean there is typo in question & i must correct it @satellite73 sir????????????

Answer 18

wait a minute @chanh_chung plz
i m seeing ur solution.

Answer 19

no no it is right and @chanh_chung solution is correct

Answer 20

can't get:(

Answer 21

@chanh_chung you are proving that this is true in general? \(z^3+\overline{z}^3-3z\overline{z}=-1\)?

Answer 22

I use
\[a ^{3}+b ^{3}=(a+b)(a ^{2}-ab+b ^{2})\]

Answer 23

yeah i got that

Answer 24

oh and here \(a+b=-1\)

Answer 25

and
\[(a+b)^{2}=a ^{2}+2ab+b ^{2}\]

Answer 26

but i didn't gt the last part

Answer 27

yes a+b=-1

Answer 28

one minute @chanh_chung wt is ur answer
please tell me

Answer 29

i wanna check

Answer 30

ok gt u @chanh_chung
thanx:)

Answer 31

ok