Find all complex cube roots of 3−i. Give your answers in a+bi form, separated by commas.

Answer 1

Put into polar form after normalizing: sqrt(10) (3/sqrt(10) - 1/sqrt(10) * i) is the r(cosθ + i sinθ) form where θ is the inv. cos of 3/sqrt(10) and inv. sine of -1/sqrt(10) = -18.43495

Demoivre's Theorem for cube root is r^1/3cis(θ/3+ 2πk/3) for k = 0,1,2 This can all be calculated, but it will not be a pretty answer. cis is short for cos + i sin.

Answer 2

(3-i)^1/3

same as √10*(cos((tan^-1-1/3)+2πk)+isin((tan^-1-1/3)+2πk)^1/3,,,,,where k=0,1,2

tan^-1-1/3=-18.43 degrees and 2π=360 degrees

(3-i)^1/3= √10*(cos(-6.14)+isin(-6.14)),

& (3-i)^1/3= √10*(cos(113.86)+isin(113.86)),

& (3-i)^1/3= √10*(cos(233.86)+isin(233.86))

cos and sin are given with degree values as operands.

correction, replace √10 with 10^1/6,, thanks jacques