Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
You could just use the binomial theorem and expand it out then simplify it
BUT ... the chances of messing up are pretty good ... so I think it's way easier to use De Moivre's Theorem to do it
Drawing it on an argand diagram -√3 + i is in the 2nd quadrant
mod (-√3 + i) = √(3 + 1) = 2
arg (-√3 + i) = π - arctan (1 / √3) = π - (π/6) = 5π/6
so (-√3 + i)^6 = {2 [cos (5π/6) + i sin (5π/6)]}^6
= 2^6 [cos (5π) + i sin (5π)] ... [using De Moivre's theorem [r (cos θ + i sin θ]^n = r^n [cos (nθ) + i sin (nθ)]
= 64 [-1 + 0 i]
= -64
BUT ... the chances of messing up are pretty good ... so I think it's way easier to use De Moivre's Theorem to do it
Drawing it on an argand diagram -√3 + i is in the 2nd quadrant
mod (-√3 + i) = √(3 + 1) = 2
arg (-√3 + i) = π - arctan (1 / √3) = π - (π/6) = 5π/6
so (-√3 + i)^6 = {2 [cos (5π/6) + i sin (5π/6)]}^6
= 2^6 [cos (5π) + i sin (5π)] ... [using De Moivre's theorem [r (cos θ + i sin θ]^n = r^n [cos (nθ) + i sin (nθ)]
= 64 [-1 + 0 i]
= -64
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.