How do you use DeMoivre's Theorem to simplify $(sqrt5-4i)^3$ ?

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Answer 1 · 2016-11-15T16:26:55

$(sqrt5-4i)^3=-43sqrt5-4i$

According to DeMoivre's theorem

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

Now let $sqrt5-4i=r(costheta+isintheta)$

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

hence squaring and adding $r^2=((sqrt5)^2+(-4)^2)=5+16=21$

and $r=sqrt21$ , $costheta=sqrt(5/21)$ and $sintheta=-4/sqrt21$

Therefore using DeMoivre's theorem

$(sqrt5-4i)^3=(sqrt21)^3(cos3theta+isin3theta)$

As $cos3theta=4cos^3theta-3costheta=4(sqrt(5/21))^3-3sqrt(5/21)$

= $20/21sqrt(5/21)-3sqrt(5/21)=-43/21sqrt(5/21)$

and $sin3theta=3sintheta-4sin^3theta=3(-4/sqrt21)-4(-4/sqrt21)^3$

= $-12/sqrt21+256/(21sqrt21)=(-252+256)/(21sqrt21)=-4/(21sqrt21)$

Hence, $(sqrt5-4i)^3=(sqrt21)^3(-43/21sqrt(5/21)-i4/(21sqrt21))$

= $-43sqrt5-4i$

How do you use DeMoivre's Theorem to simplify (sqrt5-4i)^3(sqrt5-4i)^3?