Question #07a2a

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Question #07a2a

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Answer 1 · 2017-04-13T16:44:57

$cos (14^{\circ}) + i sin (14^{\circ})$ $cos (14^@) + i sin (14^@)$

Explanation:

De Moivre's theorem
$Z^{n} = r^{n} (cos (n θ) + i sin (n θ))$ $Z^n = r^n(cos (ntheta) + i sin(ntheta))$

$Z^{\frac{1}{3}} = r^{\frac{1}{3}} (cos (\frac{1}{3} θ) + i sin (\frac{1}{3} θ))$ $Z^(1/3) = r^(1/3)(cos (1/3theta) + i sin(1/3theta))$

$Z = cos (42^{\circ}) + i sin (42^{\circ})$ $Z = cos(42^@) + i sin(42^@)$ where $r = 1$ $r =1$

$\sqrt[3]{Z} = Z^{\frac{1}{3}} = cos (\frac{1}{3}) (42^{\circ}) + i sin (\frac{1}{3}) (42^{\circ})$ $root(3)Z = Z^(1/3) = cos (1/3)(42^@) + i sin (1/3)(42^@)$

$Z^{\frac{1}{3}} = cos (14^{\circ}) + i sin (14^{\circ})$ $Z^(1/3) = cos (14^@) + i sin (14^@)$

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Question #07a2a

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