Trigonometry
posted by Shin .
There are four complex fourth roots to the number 4−4√3i. These can be expressed in polar form as
z1=r1(cosθ1+isinθ1)
z2=r2(cosθ2+isinθ2)
z3=r3(cosθ3+isinθ3)
z4=r4(cosθ4+isinθ4),
where ri is a real number and 0∘≤θi<360∘. What is the value of θ1+θ2+θ3+θ4 (in degrees)?

let z = 44√3 i
then by De Moivre's theorem
tanØ = 4√3/4 = √3
so that Ø = 120° or Ø = 300°
z = 8(cos 120° + i sin120°) or z = 8(cos 300° + i sin300°)
case1:
then z^(1/4) = 8^(1/4) (cos 30° + i sin30°)
but for tanØ = 4√3/4 , recall that the period of tanØ is 180°
so adding 180° to our angle yields another solution
making z(1/4) = 8^(1/4) (cos 210° + i sin 210°)
so far we have Ø1=30° , Ø2 = 210°
cose2:
z^(1/4) = 8(1/2)( cos 300/4 + i sin 300/4) = 8(1/4) (cos 75 + i sin 75)
and with a period of 180° again,
z^(1/4) could also be 8^(1/4)(cos 255° + i sin 255°)
giving us Ø3 = 75 and Ø4 = 255
Ø1+Ø2+Ø3+Ø4 = 30+210+75+255 = 570° 
wrong man
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