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solve z^3=27i with explanation please
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z^3 = - 27i^3 z^3 + (3i)^3 = 0 => (z +3i)( z^2 +9i^2 - 3zi) = 0 Now solve for z
how did u get the third step??
Couldn't you also just take cube-root of both sides since both sides are perfect cubes?
a3 + b^3 = (a+b)(a^2 + b^2 -ab)
ok ty :)
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we need three roots here not one as it is z^3 @cliffedge
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?
Not sure but you can try it if you want
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