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?
Thank you, I get it now. Sorry to keep bothering you, but could you perhaps explain the rest of the answer too? In particular, how you got to -2^30? And then if I then go back to the start and do the same for } \right)^{12} " alt="sqrt{3}} \right)^{12} " /> then multiply them together?