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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
${( - 1 + i\sqrt 3 )^{20}}$ is equal to
  • A
    ${2^{20}}{( - 1 + i\sqrt 3 )^{20}}$
  • B
    ${2^{20}}{(1 - i\sqrt 3 )^{20}}$
  • C
    ${2^{20}}{( - 1 - i\sqrt 3 )^{20}}$
  • ✓
    None of these
✓

Answer

Correct option: D.
None of these
d
(d) Let $z = - 1 + i\sqrt 3 $, $r = \sqrt {1 + 3} = 2$
$\theta = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 }}{{ - 1}}} \right) = \frac{{2\pi }}{3}$
$\therefore \,z = 2\,\left( {\cos \frac{{2\pi }}{3} + i\sin \frac{{2\pi }}{3}} \right)$
$\therefore \,\,{(z)^{20}} = {\left[ {2\left( {\cos \frac{{2\pi }}{3} + i\sin \frac{{2\pi }}{3}} \right)} \right]^{20}}$
$ = {2^{20}}{\left( {\cos \frac{{2\pi }}{3} + i\sin \frac{{2\pi }}{3}} \right)^{20}}$$ = {2^{20}}{\left( { - \frac{1}{2} + i\frac{{\sqrt 3 }}{2}} \right)^{20}}$.

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