Find the indicated power using De Moivre's Theorem. (Express your fully simplified answer in the form a + bi. ) (1 − i )2 I need a lot of help on this because its not working for me

DeMoivre's Theorem states: (a+bi)

^{n}= (r cisθ)^{n}= r^{n}cis(nΘ)r cisθ means r (cosΘ + isinΘ) where r = √(a

^{2}+ b^{2}) and θ = arctan(b/a) evaluated in the correct quadrant.So, in your example, a=1 and b=-1, so r = √2, and, since a and b are equal in magnitude, and in quadrant IV, we can find θ = 315

^{o}or -45^{o}. (if you are using radians, it would be 7π/4, or -π/4), and n=2.Therefore, (1-i)

^{2}= (√2)^{2}cis(2*315^{o}) = 2 cis(630^{o}) = 2 cis(270^{o}) = 2(cos 270^{o}+ i sin270^{o}) = 2(0 - i) = -2i.I hope this helps.

## Comments

^{2}using DeMoivre's theorem.