Calculate (√3 - i)6 using DeMoivre's theorem.

Question

Calculate (√3 - i)⁶ using DeMoivre's theorem. State the answer in the rectangular form of a complex number. To make it easier to grade. write your work where indicated:

1. Write √3 - i in trigonometric form.
2. Do the calculation.
3. Write the answer in rectangular form.

Answer 1

r = l √3 - i l = √ [(√3)² + (-1)²] = 2

θ = 11π / 6

√3 - i = r(cosθ + isinθ) = 2[cos(11π/6) + isin(11π/6)]

By DeMoivre's Theorem,

(√3 - i )⁶ = 2⁶[cos(6(11π/6) + isin(6(11π/6))] = 64[cos(11π) + isin(11π)] = -64 + 0i

Answer 2

(sqr3 -i)^6 = -64

(sqr3-i)^2 = 3-2isqr3-1 = 2-2isqr3= 2(1-isqr3)

(sqr3-i)^4 = 4(1-isqr3)^2 = 4(1-2isqr3-3)= -8(1+isqr3)

(sqr3-i)^6 = 2(1-isqr3)(-8)(1+isqr3)=-16(1+3)=-16(4)=-64

angle = 330 degrees = 11pi/6 radians, polar coordinates: (2,330)

rectangular coordinates= (sqr3,-1)

z =r(cosT+isinT)= 2(cos11pi/6 +isin11pi/6) = 2(sqr3/2 -i/2) = sqr3 -i

z^n = r^n(cosnT+isinnT), n=6, r=2, T=11pi/6

=2^6(cos11pi+isin11pi)

=64(-1+0) = -64