How do you use De Moivre's theorem to express (1 + i)^8?

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How do you use De Moivre's theorem to express (1 + i)^8?

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Answer 1 · 2018-03-03T09:08:31

$z = 16$ $z=16$

Explanation:

De Moivre's theorem states if $z = r (cos (θ) + i sin (θ))$ $color(orange)(z=r(cos(theta)+isin(theta)))$ ,

then $z^{n} = r^{n} (cos (n θ) + i sin (n θ))$ $color(orange)(z^n=r^n(cos(ntheta)+isin(ntheta)))$

First step convert from complex form to trig form

$a + b i \to r (cos (θ) + i sin (θ))$ $a+birarrr(cos(theta)+isin(theta))$

By using

$r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$ and $θ = arctan (\frac{b}{a})$ $theta=arctan(b/a)$

We have the number $z = 1 + i$ $z=1+i$ , thus

$r = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$ $r=sqrt(1^2+1^2)=sqrt(2)$

$θ = arctan (\frac{1}{1}) = \frac{π}{4}$ $theta=arctan(1/1)=pi/4$

$z = \sqrt{2} (cos (\frac{π}{4}) + i sin (\frac{π}{4})) \leftarrow Trig form$ $z=sqrt(2)(cos(pi/4)+isin(pi/4))larr"Trig form"$

Second step apply De Moivre's Theorem

$z^{8} = {(\sqrt{2} (cos (\frac{π}{4}) + i sin (\frac{π}{4})))}^{8}$ $z^8=(sqrt(2)(cos(pi/4)+isin(pi/4)))^8$

$= {\sqrt{2}}^{8} (cos (8 \frac{π}{4}) + i sin (8 \frac{π}{4}))$ $=sqrt(2)^8(cos(8pi/4)+isin(8pi/4))$

$= 16 (cos (2 π) + i sin (2 π)$ $=16(cos(2pi)+isin(2pi)$

$= 16$ $=16$

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How do you use De Moivre's theorem to express (1 + i)^8?

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