How do you use DeMoivre's theorem to simplify ${(1 + i \sqrt{3})}^{3}$ $(1+isqrt3)^3$ ?

Question

How do you use DeMoivre's theorem to simplify ${(1 + i \sqrt{3})}^{3}$ $(1+isqrt3)^3$ ?

1 Answer

Impact of this question

6919 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-28T01:19:08

${(1 + i \sqrt{3})}^{3} = - 8$ $(1+isqrt(3))^3 = -8$

Explanation:

Let $ω = 1 + i \sqrt{3}$ $omega=1+isqrt(3)$

First let us plot the point $ω$ $omega$ on the Argand diagram:

And we will put the complex number into polar form:

$| ω | = \sqrt{1 + 3} = 2$ $|omega| = sqrt(1+3) = 2$
$a r g (ω) = {tan}^{- 1} \sqrt{3} = \frac{π}{3}$ $arg(omega) = tan^(-1)sqrt(3) = pi/3$

So then in polar form we have:

$ω = 2 (cos (\frac{π}{3}) + i sin (\frac{π}{3}))$ $omega = 2(cos((pi)/3) + isin((pi)/3))$

By De Moivre's Theorem we have:

$ω^{3} = {2 (cos (\frac{π}{3}) + i sin (\frac{π}{3}))}^{3}$ $omega^3 = {2(cos (pi/3) + isin(pi/3))}^3$
$\ \ \ = {2}^3{(cos(pi/3) + isin(pi/3))}^3$
$\ \ \ = 8(cos(3pi/3) + isin(3pi/3))$
$\ \ \ = 8(cos pi + isin pi)$
$\ \ \ = 8(-1 + 0)$
$\ \ \ = -8$

Science

Math

Humanities

... and beyond

How do you use DeMoivre's theorem to simplify ${(1 + i \sqrt{3})}^{3}$ $(1+isqrt3)^3$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you use DeMoivre's theorem to simplify (1+isqrt3)^3(1+i√3)3(1+isqrt3)^3?

1 Answer

Explanation:

Related questions

Impact of this question

How do you use DeMoivre's theorem to simplify ${(1 + i \sqrt{3})}^{3}$ $(1+isqrt3)^3$ ?