Question

How do you use demoivre's theorem to simplify `4(1-sqrt3i)^3`?

Answer 1

`-32`

Explanation:

DeMoivre's theorem for exponents says that any complex number `z` can be written as `r(costheta+isintheta)`, or `rcolor(white)"." "cis"(theta)` for short.

It continues to say that when raising an imaginary number to a certain power, the result is:

`z^n = r^n color(white)".""cis"(ntheta)`

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To simplify this, let's first calculate `(1-isqrt3)^3`.

We find `r` by using Pythagoras' theorem:

`r = sqrt(1^2+sqrt(3)^2) = sqrt4 = 2`

We find `theta` by taking the inverse tangent of `(Im(z))/(Re(z)`.

`theta = tan^-1((-sqrt3)/(1))=-pi/3`

So `(1-isqrt3) = 2"cis"(-pi/3)`. Therefore:

`(1-isqrt3)^3 = 2^3"cis"(3(-pi/3)) = 8"cis"(pi)`

Finally, we expand `"cis"(pi)`.

`8"cis"(pi)=8cospi+8isinpi = 8(-1)+8i(0) = -8`

And don't forget to multiply by 4!

`-8*4 = -32`

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So `4(1-isqrt3)^3 = -32`.

Final Answer