Use de moivre's theorem to find (-2 sqrt(3) -2i)^7. Express in standard form.?

Question

Just that. I have looked everywhere for how to do it and I can't figure it out. Please help.

Answer 1

-2 sqrt(3) -2i .. . . convert to polar form

x + yi = r CIS A ... where r^2 = x^2 + y^2 , cosA = x/r , sinA = y/r

thus

r = 4

cosA = -2sqrt(3)/4 = -sqrt(3)/2

sinA = -1/2

thus A = 210°

(-2 sqrt(3) -2i)^7 = (4 cis 210°)^7 = 4^7 cis(7*210°)

= 4^7 cis(1470°)  .. . .. . simplify

= 4^7 cis(30°)

= 4^7 (cos30° + i sin30°)

= 4^7 (sqrt(3)/2 + 1/2 i)

= 8192sqrt(3) + 8192 i

De Moivre's Theorem:

(r cisA)^n = r^n cis(nA)