Assume that 0.333 represents ⅓.
(-0.4168-0.9090i)⅓=(-1)(0.4168+0.9090i)⅓.
Let's assume that (0.4168+0.9090i)⅓=a(cosθ+isinθ)=aeiθ, where a is real.
a3e3iθ=0.4168+0.9090i=a3(cos(3θ)+isin(3θ)).
Equating real and imaginary components:
a3cos(3θ)=0.4168; a3sin(3θ)=0.9090.
tan(3θ)=0.9090/0.4168=2.1809 approx.
3θ=1.1409, θ=0.3803 approx.
sin(3θ)=0.9090 approx.
sinθ=0.3712, cosθ=0.9286 approx.
a3=0.9090/sin(3θ)=1, a=1.
So cosθ+isinθ=0.9286+0.3712i=(0.4168+0.9090i)⅓.
SOLUTION is -0.9286-0.3712i=(-0.4168-0.9090i)0.333, where 0.333 is taken to mean ⅓.