Question

How do you write `(4sqrt(3)-4i)^22` in the form of a+bi?

Answer 1

`(4sqrt(3)-4i)^22 = 2^65+2^65sqrt(3) i`

`color(white)((4sqrt(3)-4i)^22) = 36893488147419103232+36893488147419103232sqrt(3)i`

Explanation:

Given:

`(4sqrt(3)-4i)^22`

Note that:

`abs(4sqrt(3)-4i) = sqrt((4sqrt(3))^2+4^2) = sqrt(48+16) = sqrt(64) = 8`

So `4sqrt(3)-4i` can be expressed in the form `8(cos theta + i sin theta)` for some suitable `theta`.

`4sqrt(3)-4i = 8(sqrt(3)/2-1/2i) = 8(cos(-pi/6) + i sin(-pi/6))`

So:

`(4sqrt(3)-4i)^22 = (8(cos(-pi/6)+isin(-pi/6)))^22`

`color(white)((4sqrt(3)-4i)^22) = 8^22(cos(-(22pi)/6)+isin(-(22pi)/6))`

`color(white)((4sqrt(3)-4i)^22) = 8^22(cos(pi/3)+isin(pi/3))`

`color(white)((4sqrt(3)-4i)^22) = 8^22(1/2+sqrt(3)/2 i)`

`color(white)((4sqrt(3)-4i)^22) = 2^65+2^65sqrt(3) i`

`color(white)((4sqrt(3)-4i)^22) = 36893488147419103232+36893488147419103232sqrt(3)i`

Answer 2

Here's one way that doesn't use the Binomial Theorem.

Explanation:

Observe that `(4sqrt3 - 4i)^22 = (4(sqrt3 - i))^22 = 4^22(sqrt3-i)^22`.
This will let us keep the coefficients down somewhat.
We will find the expansion of `(sqrt3-i)^22` and will multiply by `4^22 = 2^44` at the end.

`(sqrt3-i)^2 = (sqrt3-i)(sqrt3-i) = 3 -1 -2isqrt3 = 2-2isqrt3`
`(sqrt3-i)^3 = (2-2isqrt3)(sqrt3-i) = 2sqrt3 - 2i -6i - 2sqrt3 = -8i`
`(sqrt3-i)^21 = ((sqrt3-i)^3)^7 = (-8i)^7 = 2^21i`
`= (-8^7)(i^7) = (-2^21)(-i) = 2^21i`
`(sqrt3-i)^22 = (2^21i)(sqrt3 - i) = 2^21(1 + isqrt3)`
Multiply by `4^22 = 2^44`:

The final answer is
`=2^65(1+ isqrt3)`