Find the real and imaginary parts of (-2+2i)^20(1+isquareRt(3))I don't really know how to solve this question, but I've tried and I haven't actually got a complex number. Could anybody tell me where I've gone wrong if it I've just done it the wrong way altogether. Sorry it looks quite messy.

(-2+2i)^20

sqroot(8) (cos(pi/4) + isin(pi/4)) -- found modulus and principle argument

sqroot(8)^20 (cos(20pi/4) + isin(20pi/4)) -- multiply modulus and principle argument to power of 20

sqroot(8)^20 (cos(pi) + isin(pi)) -- subtract principle argument by 360 until 0 < pi < 360

sqr(8)^20 (-1 + 0) = -(sqr(8)^20) -- find cos + sin values)

(1+iSqr(3))^12

2(cos(pi/3) + isin(pi/3))

4096(cos(12pi/3) + isin(12pi/3))

4096(cos(2pi) + isin(2pi))

4096(1 + 0)

4096 . -(sqr(8)^20) = -4.39 x 10^12?