**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?

Convert them to polars:

\(\displaystyle |-2 + 2i| = \sqrt{(-2)^2 + 2^2} = 2\sqrt{2}\)

\(\displaystyle \arg{(-2 + 2i)} = \pi - \arctan{\frac{2}{2}} = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\).

\(\displaystyle |1 + \sqrt{3}i| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2\).

\(\displaystyle \arg{(1 + \sqrt{3}i)} = \arctan{\frac{\sqrt{3}}{1}} = \frac{\pi}{3}\).

Therefore:

\(\displaystyle (-2 + 2i)^{20}(1 + \sqrt{3}i) = \left(2\sqrt{2}e^{\frac{3\pi}{4}i}\right)^{20}\left(2e^{\frac{\pi}{3}i}\right)\)

\(\displaystyle = \left[\left(2\sqrt{2}\right)^{20}e^{15i}\right]\left(2e^{\frac{\pi}{3}i}\right)\)

\(\displaystyle = 2^{31}e^{15\pi i + \frac{\pi}{3}i}\)

\(\displaystyle = 2^{31}e^{\frac{46 \pi}{3}i}\)

\(\displaystyle = 2^{31}\left[\cos{\left(\frac{46\pi}{3}\right)} + i\sin{\left(\frac{46\pi}{3}\right)}\right]\)

\(\displaystyle = 2^{31}\left[\cos{\left(\frac{4\pi}{3}\right)} + i\sin{\left(\frac{4\pi}{3}\right)}\right]\)

\(\displaystyle = 2^{31}\left[-\cos{\left(\frac{\pi}{3}\right)} - i\sin{\left(\frac{\pi}{3}\right)}\right]\)

\(\displaystyle = 2^{31}\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)\)

\(\displaystyle = -2^{30} - 2^{30}\sqrt{3}i\).