How do you find the fifth roots of $128 (- 1 + i)$ $128(-1+i)$ ?

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How do you find the fifth roots of $128 (- 1 + i)$ $128(-1+i)$ ?

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Answer 1 · 2018-06-14T05:55:55

$\Rightarrow (\sqrt{8} \cdot (cos 27 + i sin 27)$ $color(purple)(=> (sqrt8 * (cos 27 + i sin 27)$

Explanation:

$128 (- 1 + i) = 128 \sqrt{2} (- \frac{1}{\sqrt{2}} + i (\frac{1}{\sqrt{2}}))$ $128(-1+i) = 128sqrt2(-1/sqrt2 + i (1/sqrt2))$

$\Rightarrow 128 \sqrt{2} (- cos (\frac{π}{4}) + i sin (\frac{π}{4}))$ $=> 128sqrt2 (-cos (pi/4) + i sin (pi/4))$

$\Rightarrow 128 \sqrt{2} (cos (135) + i sin (135))$ $=> 128 sqrt2 (cos (135) + i sin (135))$

$\sqrt[5]{128 \sqrt{2} (cos 135 + i sin 135)}$ $root(5) (128sqrt2 (cos 135 + i sin 135)$

$\Rightarrow {(128 \sqrt{2})}^{\frac{1}{5}} (cos (\frac{135}{5}) + i sin (\frac{135}{5}))$ $=> (128sqrt2)^(1/5) (cos (135/5) + i sin (135/5))$

$\Rightarrow (\sqrt{8} \cdot (cos 27 + i sin 27)$ $=> (sqrt8 * (cos 27 + i sin 27)$ as

${(128 \sqrt{2})}^{\frac{1}{5}} = {(2^{7} \cdot \sqrt{2})}^{\frac{1}{5}} = {(\sqrt{2})}^{\frac{15}{5}} = \sqrt{8}$ $(128 sqrt2)^(1/5) = (2^7 * sqrt 2)^(1/5) = (sqrt2)^(15/5) = sqrt8$

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How do you find the fifth roots of $128 (- 1 + i)$ $128(-1+i)$ ?

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