How do you use DeMoivre's theorem to simplify ${(- \sqrt{3} + i)}^{5}$ $(-sqrt3+i)^5$ ?

Question

12996 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-23T14:29:51

${(- \sqrt{3} + i)}^{5} = 16 (\sqrt{3} + i)$ $(-sqrt3+i)^5=16(sqrt3+i)$ .

To use DeMoivre's Theorem, we need to convert

$z = - \sqrt{3} + i = x + i y$ $z=-sqrt3+i=x+iy$ into Polar Form, i.e.,

$r (cos θ + i sin θ), w h e r e, r > 0, &, θ \in (- π, π]$ $r(costheta+isintheta), where, r>0, &, theta in (-pi,pi]$ .

$z = x + i y = r (cos θ + i sin θ) = r c i s θ$ $z=x+iy=r(costheta+isintheta)=rcistheta$

$\Rightarrow x = r cos θ, y = r sin θ, r = \sqrt{x^{2} + y^{2}}$ $rArr x=rcostheta, y=rsintheta, r=sqrt(x^2+y^2)$

$:. r=sqrt{(-sqrt3)^2+1^2}=2$

Hence, from $x=rcostheta, costheta=-sqrt3/2, "&, similarly,"$ $sintheta=1/2$

$"clearly", theta=(5pi)/6$

Thus, $-sqrt3+i=2cis((5pi)/6)$ .

Now, by DeMoivre's Theorem, $(rcistheta)^n=r^ncis(ntheta)$

In our case, $n=5, r=2,theta=(5pi)/6$

$:.(-sqrt3+i)^5=(2cis(5pi)/6)^5$

$=(2^5)(cis(5*(5pi)/6))$

$=32(cis(25pi)/6)$

$=32(cis(4pi+pi/6))$

$=32(cos(4pi+pi/6)+isin(4pi+pi/6))$

$=32(cos(pi/6)+isin(pi/6))$

$=32(sqrt3/2+1/2i)$

$=16(sqrt3+i)$ .

Enjoy Maths.!

How do you use DeMoivre's theorem to simplify (-sqrt3+i)^5(−√3+i)5(-sqrt3+i)^5?