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

2 Answers
Feb 6, 2018

(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

Feb 6, 2018

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)