Complex Numbers And Quadratic Equations question 277
Question: $ \frac{{{(-1+i\sqrt{3})}^{15}}}{{{(1-i)}^{20}}}+\frac{{{(-1-i\sqrt{3})}^{15}}}{{{(1+i)}^{20}}} $ is equal to [AMU 2000]
Options:
A) - 64
B) - 32
C) - 16
D) $ \frac{1}{16} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ 2^{15}[ \frac{{{( -\frac{1}{2}+\frac{i\sqrt{3}}{2} )}^{15}}}{{{(1-i)}^{20}}}+\frac{{{( \frac{-1}{2}-\frac{i\sqrt{3}}{2} )}^{15}}}{{{(1+i)}^{20}}} ] $ = $ 2^{15}[ \frac{{{\omega }^{15}}}{{{(1-i)}^{20}}}+\frac{{{\omega }^{30}}}{{{(1+i)}^{20}}} ] $ = $ 2^{15}[ \frac{1}{{{(1-i)}^{20}}}+\frac{1}{{{(1+i)}^{20}}} ] $ = $ 2^{15}[ \frac{{{(1+i)}^{20}}+{{(1-i)}^{20}}}{{{(1-i^{2})}^{20}}} ] $ = $ \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 $ .
BETA
