Question

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

Answer 1

`(1+isqrt(3))^3 = -8`

Explanation:

Let `omega=1+isqrt(3)`

First let us plot the point `omega` on the Argand diagram:

And we will put the complex number into polar form:

`|omega| = sqrt(1+3) = 2`
`arg(omega) = tan^(-1)sqrt(3) = pi/3`

So then in polar form we have:

`omega = 2(cos((pi)/3) + isin((pi)/3))`

By De Moivre's Theorem we have:

`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`