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

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Answer 1 · 2018-03-03T09:08:31

$z=16$

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

then $color(orange)(z^n=r^n(cos(ntheta)+isin(ntheta)))$

First step convert from complex form to trig form

$a+birarrr(cos(theta)+isin(theta))$

By using

$r=sqrt(a^2+b^2)$ and $theta=arctan(b/a)$

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

$r=sqrt(1^2+1^2)=sqrt(2)$

$theta=arctan(1/1)=pi/4$

$z=sqrt(2)(cos(pi/4)+isin(pi/4))larr"Trig form"$

Second step apply De Moivre's Theorem

$z^8=(sqrt(2)(cos(pi/4)+isin(pi/4)))^8$

$=sqrt(2)^8(cos(8pi/4)+isin(8pi/4))$

$=16(cos(2pi)+isin(2pi)$

$=16$