Complex question (1 Viewer)

[Jackson94]

Find all complex numbers z such that z^3 = 8i
Thanks in advance guys

 

[slyhunter]

z^3 = 8i

=>z^3 - 8i = 0

=>(z)^3 + (2i)^3 = (z + 2i)(z^2 - 2zi - 4) = 0

so z + 2i = 0 or

z^2 - 2zi - 4 = 0

when z + 2i = 0

z = -2i

when z^2 -2zi - 4 = 0

z = [2i +/- sqrt(-4 +16)]/2

z = [2i +/- sqrt(12)]/2

z = [2i +/- 2sqrt(3)]/2

z = i + sqrt(3) or i - sqrt(3)

So three roots are -2i, [i + sqrt(3)], [i - sqrt(3)]



Too lazy to latex.

 

[AAEldar]

z^3 = 8i

=>z^3 - 8i = 0

=>(z)^3 + (2i)^3 = (z + 2i)(z^2 - 2zi - 4) = 0

so z + 2i = 0 or

z^2 - 2zi - 4 = 0

when z + 2i = 0

z = -2i

when z^2 -2zi - 4 = 0

z = [2i +/- sqrt(-4 +16)]/2

z = [2i +/- sqrt(12)]/2

z = [2i +/- 2sqrt(3)]/2

z = i + sqrt(3) or i - sqrt(3)

So three roots are -2i, [i + sqrt(3)], [i - sqrt(3)]



Too lazy to latex.

You wrote down the wrong factor in the third line, should be and hence the first root is .

 

[slyhunter]

though because