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Mathematics 53 Online
OpenStudy (maheshmeghwal9):

\[z_1=\frac{-1+3i}{2}\]and \[z_2=\frac{-1-3i}{2}\]Find the value of \[z_1^3+z_2^3-3z_1z_2.\] Its answer is given: - \[\huge{\color{red}{-1.}}\]

OpenStudy (maheshmeghwal9):

Please help:)

OpenStudy (maheshmeghwal9):

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

OpenStudy (maheshmeghwal9):

give ur respective comments plz:)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (maheshmeghwal9):

these aren't roots of unity

OpenStudy (maheshmeghwal9):

i think there is typo

OpenStudy (anonymous):

no?

OpenStudy (maheshmeghwal9):

becoz there is only 3 nt sqrt 3

OpenStudy (anonymous):

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

OpenStudy (anonymous):

oooh i see sorry!

OpenStudy (maheshmeghwal9):

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

OpenStudy (anonymous):

ok if you grind it out you get it

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (maheshmeghwal9):

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

OpenStudy (maheshmeghwal9):

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

OpenStudy (anonymous):

no no it is right and @chanh_chung solution is correct

OpenStudy (maheshmeghwal9):

can't get:(

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yeah i got that

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (maheshmeghwal9):

but i didn't gt the last part

OpenStudy (anonymous):

yes a+b=-1

OpenStudy (maheshmeghwal9):

one minute @chanh_chung wt is ur answer please tell me

OpenStudy (maheshmeghwal9):

i wanna check

OpenStudy (maheshmeghwal9):

ok gt u @chanh_chung thanx:)

OpenStudy (anonymous):

ok

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