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STD 11 - 4.1 complex nubers — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 4.1 complex nubers1 Mark
MCQ
$\frac{{{{( - 1 + i\sqrt 3 )}^{15}}}}{{{{(1 - i)}^{20}}}} + \frac{{{{( - 1 - i\sqrt 3 )}^{15}}}}{{{{(1 + i)}^{20}}}}$ is equal to
  • ✓
    $-64$
  • B
    $-32$
  • C
    $-16$
  • D
    $\frac{1}{{16}}$
✓

Answer

Correct option: A.
$-64$
a
(a)${2^{15}}\left[ {\frac{{{{\left( { - \frac{1}{2} + \frac{{i\sqrt 3 }}{2}} \right)}^{15}}}}{{{{(1 - i)}^{20}}}} + \frac{{{{\left( {\frac{{ - 1}}{2} - \frac{{i\sqrt 3 }}{2}} \right)}^{15}}}}{{{{(1 + i)}^{20}}}}} \right]$
= ${2^{15}}\left[ {\frac{{{\omega ^{15}}}}{{{{(1 - i)}^{20}}}} + \frac{{{\omega ^{30}}}}{{{{(1 + i)}^{20}}}}} \right]$=${2^{15}}\left[ {\frac{1}{{{{(1 - i)}^{20}}}} + \frac{1}{{{{(1 + i)}^{20}}}}} \right]$
= ${2^{15}}\left[ {\frac{{{{(1 + i)}^{20}} + {{(1 - i)}^{20}}}}{{{{(1 - {i^2})}^{20}}}}} \right]$=$\frac{{{2^{15}}}}{{{2^{20}}}}[{(1 + i)^{20}} + {(1 - i)^{20}}]$
= $\frac{1}{{{2^5}}}[{(i - {i^2})^{20}} + {(1 - i)^{20}}]$ = $\frac{1}{{{2^5}}}({i^{20}} + 1)\,{(1 - i)^{20}}$
$ = \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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