Posted by **Shin** on Monday, May 27, 2013 at 7:06am.

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)?

- Trigonometry -
**Reiny**, Monday, May 27, 2013 at 9:14am
let z = 4-4√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°**

- Trigonometry -
**K**, Sunday, June 2, 2013 at 7:57pm
wrong man

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