Using DeMoivre's Theorem?

Question

Use DeMoivre's Theorem to write each expression in the standard form a + bi.

(1) [3(cos80degrees + isin80degrees)]^3

(2) (sqrt{3} - i)^6

Answer 1

deMoiver's Theorem states that

[cos(x) + i sin(x)]^n = cos(nx) + i sin(nx)
_____________

1) [3(cos80° + i sin80°)]³ = 27((cos80° + i sin80°)³
= 27[cos(3*80°) + i sin(3*80°]
= 27[cos(240°) + i sin(240°]
= 27[-1/2 - √3/2] = (-27/2)(1 + √3)

2) (√3 - i)^6 = (2^6)(√3/2 - i/2)^6
= 64[cos(-30°) + i sin(-30°)]^6
= 64[cos(6*(-30°)) + i sin(6*(-30°))]
= 64[cos(-180°) + i sin(-180°)]
= 64(-1 + 0) = -64

Answer 2

Ans: [3(Cos80 + iSin80)]^3

[a(Cos x + i Sin x)]^n = a^n [Cos nx + i Sin nx]

So, [3(Cos80 + iSin80)]^3 = 3^3 (Cos80 + iSin80)^3
= 27 (Cos 80*3 + i Sin 80*3)
= 27 (Cos240 + i Sin240)
= 27 (Cos (180+60) + i Sin (180+60))
= 27 (-Cos 60 - i Sin 60)
= -27(Cos60 + i Sin60)
= -27 (0.5 + i sqrt(3) /2)


{sqrt(3) - i}^6 = {2 [ sqrt(3) / 2 - i 1/2 ] }^6
= {2 [Cos30 - i Sin 30]}^6
= 2^6 [ Cos 180 - i Sin 180]
= 64 [-1 - i (0)] = -64

Answer 3

a million+i SQRT3 rcosx=a million rsinx=SQRT3 squaring and including r^2=4 r=2 cosx=a million/2 sinx=(SQRT3)/2 x=pi/6 so (a million+isquareroot3)^12= 2^12[cos(pi/6)+i sin(pi/6)]^12 4096[cos(2pi)+i sin(2pi)] utilising DeMoivres Theorem 4096[a million+0] 4096