Rectangular form of (-2+ i)^8

My answer is (-559, 280i)
The book has an answer of (-506, 367i)
Who is right?

I get (−527, 336)


−2 + i

r = √((−2)²+1²) = √5
tanθ = 1/−2 = −1/2 (θ in Q2)
θ = 153.434948823°

−2 + i = √5 (cos(153.434948823°) + i sin(153.434948823°))

Use De moivre's formula:

(−2 + i)⁸ = (√5)⁸ (cos(8*153.434948823°) + i sin(8*153.434948823°))
. . . . . . . = 625 (−0.8432 + 0.5376 i)
. . . . . . . = −527 + 336 i

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Or we could just expand it algebraically

(−2 + i)² = 4 − 4i + i² = 3 − 4i

(−2 + i)⁴ = ((−2 + i)²)² = (3 − 4i)² = 9 − 24i + 16i² = −7 − 24i

(−2 + i)⁸ = ((−2 + i)⁴)² = (−7 − 24i)² = 49 + 336i + 576i² = −527 + 336i


Or let WolframAlpha calculate it:

http://www.wolframalpha.com/input/?i=%28…


Whichever method I use, I still get (−527, 336)

its (625+336i)