How do you use DeMoivre's theorem to simplify ${(- 1 + i)}^{4}$ $(-1+i)^4$ ?

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Answer 1 · 2017-08-21T05:46:34

The answer is $= - 4$ $=-4$

We start by writing $z = - 1 + i$ $z=-1+i$ , in trigonometric form

$| z | = \sqrt{{(- 1)}^{2} + 1^{2}} = \sqrt{2}$ $|z|=sqrt((-1)^2+1^2)=sqrt2$

$z = \sqrt{2} (- \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i)$ $z=sqrt2(-1/sqrt2+1/sqrt2i)$

$cos θ = - \frac{1}{\sqrt{2}}$ $costheta=-1/sqrt2$ , $\Rightarrow$ $=>$ , $θ = \frac{3}{4} π$ $theta=3/4pi$

$sin θ = \frac{1}{\sqrt{2}}$ $sintheta=1/sqrt2$ , $\Rightarrow$ $=>$ , $θ = \frac{3}{4} π$ $theta=3/4pi$

The trigonometric form is

$z = \sqrt{2} (cos (\frac{3}{4} π) + i sin (\frac{3}{4} π))$ $z=sqrt2(cos(3/4pi)+isin(3/4pi))$

Applying Demoivre's theorem

${(cos θ + i sin θ)}^{n} = cos (n θ) + i sin (n θ)$ $(costheta+isintheta)^n=cos(ntheta)+isin(ntheta)$

$z^{4} = (\sqrt{2} {(cos (\frac{3}{4} π) + i sin (\frac{3}{4} π))}^{4}$ $z^4=(sqrt2(cos(3/4pi)+isin(3/4pi))^4$

$= {(\sqrt{2})}^{4} (cos (\frac{3}{4} π \cdot 4) + i sin (\frac{3}{4} π \cdot 4))$ $=(sqrt2)^4(cos(3/4pi*4)+isin(3/4pi*4))$

$= 4 (cos (3 π) + i sin (3 π))$ $=4(cos(3pi)+isin(3pi))$

$= 4 (- 1 + i \cdot 0)$ $=4(-1+i*0)$

$= - 4$ $=-4$

How do you use DeMoivre's theorem to simplify (-1+i)^4(−1+i)4(-1+i)^4?