Asked by sue
are there solutions to x^3+27i=0 is so what are they?
Answers
Answered by
Reiny
x^3 = -27i
x = -3(i)^(1/3)
let's look at the i^(1/3)
i = 0 + 1i
changing that to polar form
r = 1
the hard part is to find the angle, since tanØ = y/x = 1/0 or undefined
but we know that tan π/2 is undefined, so
Ø = π/2 or 90°
so i = 1(cosπ/2 + i sinπ/2)
then i^(1.3)
= (1(cosπ/2 + i sinπ/2))^(1/3)
= cos (1/3)(π/2) + i sin (1/3)(π/2)) by De Moivre's Theorem
= cos π/6 + i sin π/6
so x = -3(cos π/6 + i sin π/6)
there are other answers, which can be obtained by using the fact that
tan 3π/2 or tan 5π/2 etc are also undefined
so you could write i in other polar forms , then use De Moivre's Theorem on those new ones.
x = -3(i)^(1/3)
let's look at the i^(1/3)
i = 0 + 1i
changing that to polar form
r = 1
the hard part is to find the angle, since tanØ = y/x = 1/0 or undefined
but we know that tan π/2 is undefined, so
Ø = π/2 or 90°
so i = 1(cosπ/2 + i sinπ/2)
then i^(1.3)
= (1(cosπ/2 + i sinπ/2))^(1/3)
= cos (1/3)(π/2) + i sin (1/3)(π/2)) by De Moivre's Theorem
= cos π/6 + i sin π/6
so x = -3(cos π/6 + i sin π/6)
there are other answers, which can be obtained by using the fact that
tan 3π/2 or tan 5π/2 etc are also undefined
so you could write i in other polar forms , then use De Moivre's Theorem on those new ones.
Answered by
Anonymous
X-27i
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