A) - 64
B) - 32
C) - 16
D) \[\frac{1}{16}\]
Correct Answer: A
Solution :
\[{{2}^{15}}\left[ \frac{{{\left( -\frac{1}{2}+\frac{i\sqrt{3}}{2} \right)}^{15}}}{{{(1-i)}^{20}}}+\frac{{{\left( \frac{-1}{2}-\frac{i\sqrt{3}}{2} \right)}^{15}}}{{{(1+i)}^{20}}} \right]\] = \[{{2}^{15}}\left[ \frac{{{\omega }^{15}}}{{{(1-i)}^{20}}}+\frac{{{\omega }^{30}}}{{{(1+i)}^{20}}} \right]\]=\[{{2}^{15}}\left[ \frac{1}{{{(1-i)}^{20}}}+\frac{1}{{{(1+i)}^{20}}} \right]\] = \[{{2}^{15}}\left[ \frac{{{(1+i)}^{20}}+{{(1-i)}^{20}}}{{{(1-{{i}^{2}})}^{20}}} \right]\]=\[\frac{{{2}^{15}}}{{{2}^{20}}}[{{(1+i)}^{20}}+{{(1-i)}^{20}}]\] = \[\frac{1}{{{2}^{5}}}[{{(i-{{i}^{2}})}^{20}}+{{(1-i)}^{20}}]\] = \[\frac{1}{{{2}^{5}}}({{i}^{20}}+1)\,{{(1-i)}^{20}}\] \[=\frac{2}{{{2}^{5}}}{{(1-i)}^{20}}\] = \[\frac{1}{{{2}^{4}}}{{(1-i)}^{20}}\]= \[\frac{1}{{{2}^{4}}}{{[{{(1-i)}^{2}}]}^{10}}\] \[=\frac{1}{{{2}^{4}}}{{[1+{{i}^{2}}-2i]}^{10}}\]=\[\frac{1}{{{2}^{4}}}{{(-2i)}^{10}}\] = \[\frac{{{(-2)}^{10}}{{i}^{10}}}{{{2}^{4}}}=-{{2}^{6}}=-64\].
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