How do I use DeMoivre's theorem to find ${(1 - i)}^{10}$ $(1-i)^10$ ?

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How do I use DeMoivre's theorem to find ${(1 - i)}^{10}$ $(1-i)^10$ ?

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Answer 1 · 2015-11-04T20:05:18

$- 32 i$ $-32i$

Explanation:

First write this complex number in polar form and then apply De Moivre :

${(1 - i)}^{10} = {(\sqrt{2} ∠ - \frac{π}{4})}^{10} = {[\sqrt{2} (cos (- \frac{π}{4}) + i sin (- \frac{π}{4}))]}^{10}$ $(1-i)^10=(sqrt2/_-pi/4)^10=[sqrt2(cos(-pi/4)+isin(-pi/4))]^10$

$= {(\sqrt{2})}^{10} [cos (- 10 \frac{π}{4}) + i sin (- 10 \frac{π}{4})]$ $=(sqrt2)^10[cos(-10pi/4)+isin(-10pi/4)]$

$= - 32 i$ $=-32i$

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How do I use DeMoivre's theorem to find ${(1 - i)}^{10}$ $(1-i)^10$ ?

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