How do I use DeMoivre's theorem to find ${(2 + 2 i)}^{6}$ $(2+2i)^6$ ?

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Answer 1 · 2018-05-10T13:34:36

The answer is $= - 512 i$ $=-512i$

Let $z = 2 + 2 i = 2 (1 + i)$ $z=2+2i=2(1+i)$

Transform $z = a + i b$ $z=a+ib$ from algebraic form into polar form

$z = | z | (\frac{a}{| a |} + i \frac{b}{| z |})$ $z=|z|(a/(|a|)+ib/(|z|))$

where,

${(costheta=a/(|z|)),(sintheta=b/(|z|)):}$

Here,

$|z|=sqrt(1^2+1^2)=sqrt2$

$z=2sqrt2(1/sqrt2+i/sqrt2)$

${(costheta=1/sqrt2),(sintheta=1/sqrt2)):}$

$=>$ , $theta=1/4pi$ , $[2pi]$

The polar form is

$z=2sqrt2(cos(pi/4)+isin(pi/4))$

Demoivre's theorem is

$(costheta+isintheta)^n=cos(ntheta)+isin(ntheta)$

Therefore,

$(2+2i)^6=(2sqrt2(cos(pi/4)+isin(pi/4)))^6$

$=(2sqrt2)^6(cos(pi/4*6)+isin(pi/4*6))$

$=(2sqrt2)^6(cos(3/2pi)+isin(3/2pi))$

$=512(0-i)$

$=-512i$

How do I use DeMoivre's theorem to find (2+2i)^6(2+2i)6(2+2i)^6?