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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: -
\[\huge{\color{red}{-1.}}\]
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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?
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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
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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
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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:(
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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
\[(a+b)^{2}=a ^{2}+2ab+b ^{2}\]
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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:)
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OpenStudy (anonymous):
ok