How do you use DeMoivre's theorem to simplify $(sqrt3+i)^8$ ?

Question

10314 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-21T09:07:30

$(sqrt3+i)^8=-128-128sqrt3*i$

DeMoivre's theorem states that given a complex number in its trigonometric form $r(costheta+isintheta)$ ,

$(r(costheta+isintheta))^n=r^n(cosntheta+isinntheta)$

Now let $sqrt3+i=rcostheta+irsintheta$ , which means

$rcostheta=sqrt3$ and $rsintheta=1$ and hence

$r=sqrt((sqrt3)^2+1^2)=sqrt(3+1)=sqrt4=2$

as such $costheta=sqrt3/2$ and $sintheta=1/2$ i.e. $theta=pi/6$

Hence, $(sqrt3+i)^8=(2(cos(pi/6)+isin(pi/6)))^8$

and using DEMoivre's Theorem, this is equal to

$2^8(cos((8pi)/6)+isin((8pi)/6))$

= $256(-1/2-i*sqrt3/2)$

= $-128-128sqrt3*i$

How do you use DeMoivre's theorem to simplify (sqrt3+i)^8(sqrt3+i)^8?