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Mathematics 20 Online
OpenStudy (anonymous):

solve z^3=27i with explanation please

OpenStudy (anonymous):

z^3 = - 27i^3 z^3 + (3i)^3 = 0 => (z +3i)( z^2 +9i^2 - 3zi) = 0 Now solve for z

OpenStudy (anonymous):

how did u get the third step??

OpenStudy (anonymous):

Couldn't you also just take cube-root of both sides since both sides are perfect cubes?

OpenStudy (anonymous):

a3 + b^3 = (a+b)(a^2 + b^2 -ab)

OpenStudy (anonymous):

ok ty :)

OpenStudy (anonymous):

we need three roots here not one as it is z^3 @cliffedge

OpenStudy (anonymous):

Ah, true.. z^3 = (-3)^3*(i)^3 -> z = -3i Only finds one. I was actually first considering DeMoivre's Formula. That should work here, right?

OpenStudy (anonymous):

Not sure but you can try it if you want

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