Let z1=2(cos 2pi/3+isin 2pi/3)
what are z1^(1/3)?
4 answers
Are you familiar with DeMoivre's Theorem?
no not at all
Too bad,
ok, the long way then .....
I am just going to call z1 as z
let z^(1/3) = a + bi
then z = (a+bi)^3
= a^3 + 3a^2 bi + 3a b^2 i^2 + b^3 i^3
= a^3 + 3a^2 b i - 3ab^2 -b^3 i
= (a^3 - 3ab^2) + i (3a^2b - b^3)
but z = 2(cos 2π/3 + i sin 2π/3) = -1 + √3i
so by comparison
a^3 - 3ab^2 = -1 , and
3a^2 - b^3 = √3
nasty , nasty to solve, so I ran it through Wolfram
using x and y instead of a and b
http://www.wolframalpha.com/input/?i=+x%5E3+-+3xy%5E2+%3D+-1+%2C+3x%5E2+-+y%5E3+%3D+%E2%88%9A3
notice we have more than one solution, one such solution would be
.86363 + .796609 i
ok, the long way then .....
I am just going to call z1 as z
let z^(1/3) = a + bi
then z = (a+bi)^3
= a^3 + 3a^2 bi + 3a b^2 i^2 + b^3 i^3
= a^3 + 3a^2 b i - 3ab^2 -b^3 i
= (a^3 - 3ab^2) + i (3a^2b - b^3)
but z = 2(cos 2π/3 + i sin 2π/3) = -1 + √3i
so by comparison
a^3 - 3ab^2 = -1 , and
3a^2 - b^3 = √3
nasty , nasty to solve, so I ran it through Wolfram
using x and y instead of a and b
http://www.wolframalpha.com/input/?i=+x%5E3+-+3xy%5E2+%3D+-1+%2C+3x%5E2+-+y%5E3+%3D+%E2%88%9A3
notice we have more than one solution, one such solution would be
.86363 + .796609 i
what are those numbers in radians
1 answer