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Use De Moivre's theorem to write the complex number in trigonometric form. (cos(4pi/7)+isin(4pi/7))^5 a. 5(cos(20pi/7)+isin(20pi/7)) b. 5(cos(4pi/7)+isin(4pi/7)) c. (cos(4pi/35)+isin(4pi/35)) d. (cos(20pi/7)+isin(20pi/7))

Respuesta :

[tex]\bf \qquad \textit{power of two complex numbers} \\\\\ [\quad r[cos(\theta)+isin(\theta)]\quad ]^{{ n}}\implies r^{{ n}}[cos({{ n}}\cdot \theta)+isin({{ n}}\cdot \theta)]\\\\ -------------------------------\\\\[/tex]

[tex]\bf z^5=\left[ 1\left[ cos\left( \frac{4\pi }{7} \right)+i\ sin\left( \frac{4\pi }{7} \right) \right] \right]^5 \\\\\\ z^5=1^5\left[ cos\left( 5\cdot \frac{4\pi }{7} \right)+i\ sin\left( 5\cdot \frac{4\pi }{7} \right) \right] \\\\\\ z^5=1\left[ cos\left( \frac{20\pi }{7} \right)+i\ sin\left( \frac{20\pi }{7} \right) \right]\implies z^5= cos\left( \frac{20\pi }{7} \right)+i\ sin\left( \frac{20\pi }{7} \right)[/tex]
pamelamorganbeeslyha

Answer: (cos(20pi/7)+isin(20pi/7))

Step-by-step explanation:

got it right on a p e x :)