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Mathematics 15 Online
OpenStudy (anonymous):

If\[ x + iy = (1 - i \sqrt{3})^{100}\] then (x,y) is?

OpenStudy (anonymous):

\[ x + iy = (1 - i \sqrt{3})^{100}\]

OpenStudy (anonymous):

FInd X and Y

OpenStudy (anonymous):

\[x + iy = \frac{2^{100}}{2^{100}}(1 - i \sqrt{3})^{100}=2^{100}(\frac{1}{2} - i \frac{\sqrt{3}}{2})^{100}\]

OpenStudy (anonymous):

ok?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

\[\frac{1}{2} - i \frac{\sqrt{3}}{2}=e^{-i\frac{\pi}{3}}\]

OpenStudy (anonymous):

yes..))

OpenStudy (anonymous):

well u continue

OpenStudy (anonymous):

Any Hint:

OpenStudy (anonymous):

\[(e^{i\theta})^n=e^{in\theta}\]

OpenStudy (anonymous):

lol.....that...doesnt help... 2^100 ( cos6000 - i sin6000)

OpenStudy (anonymous):

lol....))

OpenStudy (anonymous):

cos6000= -1/2

OpenStudy (anonymous):

i got...it thxxx...)))

OpenStudy (anonymous):

\[\boldsymbol{MUKHSALA THXXX} \]

OpenStudy (anonymous):

\[(e^{-i\frac{\pi}{3}})^{100}=e^{-i\frac{100\pi}{3}}= \cos \frac{100 \pi}{3}+i \sin \frac{100 \pi}{3}\]

OpenStudy (anonymous):

\[\huge \color\yellow{LoL}\]

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