How do you use demoivre's theorem to simplify ${[2 (cos (\frac{π}{2}) + i sin (\frac{π}{2}))]}^{8}$ $[2(cos((pi)/2)+isin((pi)/2))]^8$ ?

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How do you use demoivre's theorem to simplify ${[2 (cos (\frac{π}{2}) + i sin (\frac{π}{2}))]}^{8}$ $[2(cos((pi)/2)+isin((pi)/2))]^8$ ?

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Answer 1 · 2016-10-15T16:55:23

${[2 (cos (\frac{π}{2}) + i sin (\frac{π}{2}))]}^{8} = 256$ $[2(cos(pi/2)+isin(pi/2))]^8=256$

Explanation:

According to de Moivre's Theorem we can calculate any integer power of a complex number given in trigonometric form.

The theorem says that:

If a complex number $z$ $z$ is given in a form:

$z = | z | (cos φ + i sin φ)$ $z=|z|(cosvarphi+isinvarphi)$

Then the $n - t h$ $n-th$ power of $z$ $z$ is:

$z^{n} = {| z |}^{n} (cos n φ + i sin n φ)$ $z^n=|z|^n(cosnvarphi+isinnvarphi)$

Using this formula we get:

${[2 (cos (\frac{π}{2}) + i sin (\frac{π}{2}))]}^{8} = 2^{8} [cos (\frac{8 π}{2}) + i sin (\frac{8 π}{2})] =$ $[2(cos(pi/2)+isin(pi/2))]^8=2^8[cos((8pi)/2)+isin((8pi)/2)]=$

$= 256 [cos 4 π + i sin 4 π] = 256$ $=256[cos4pi+isin4pi]=256$

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How do you use demoivre's theorem to simplify ${[2 (cos (\frac{π}{2}) + i sin (\frac{π}{2}))]}^{8}$ $[2(cos((pi)/2)+isin((pi)/2))]^8$ ?

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How do you use demoivre's theorem to simplify ${[2 (cos (\frac{π}{2}) + i sin (\frac{π}{2}))]}^{8}$ $[2(cos((pi)/2)+isin((pi)/2))]^8$ ?