Question
$\frac{{{{( - 1 + i\sqrt 3 )}^{15}}}}{{{{(1 - i)}^{20}}}} + \frac{{{{( - 1 - i\sqrt 3 )}^{15}}}}{{{{(1 + i)}^{20}}}}$ =
= ${2^{15}}\left[ {\frac{{{\omega ^{15}}}}{{{{(1 - i)}^{20}}}} + \frac{{{\omega ^{30}}}}{{{{(1 + i)}^{20}}}}} \right]$=${2^{15}}\left[ {\frac{1}{{{{(1 - i)}^{20}}}} + \frac{1}{{{{(1 + i)}^{20}}}}} \right]$
= ${2^{15}}\left[ {\frac{{{{(1 + i)}^{20}} + {{(1 - i)}^{20}}}}{{{{(1 - {i^2})}^{20}}}}} \right]$=$\frac{{{2^{15}}}}{{{2^{20}}}}[{(1 + i)^{20}} + {(1 - i)^{20}}]$
= $\frac{1}{{{2^5}}}[{(i - {i^2})^{20}} + {(1 - i)^{20}}]$ = $\frac{1}{{{2^5}}}({i^{20}} + 1)\,{(1 - i)^{20}}$
$ = \frac{2}{{{2^5}}}{(1 - i)^{20}}$ = $\frac{1}{{{2^4}}}{(1 - i)^{20}}$= $\frac{1}{{{2^4}}}{[{(1 - i)^2}]^{10}}$
$ = \frac{1}{{{2^4}}}{[1 + {i^2} - 2i]^{10}}$=$\frac{1}{{{2^4}}}{( - 2i)^{10}}$
= $\frac{{{{( - 2)}^{10}}{i^{10}}}}{{{2^4}}} = - {2^6} = - 64$
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