How do you write #(4sqrt(3)-4i)^22# in the form of a+bi?
2 Answers
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 = 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#
Here's one way that doesn't use the Binomial Theorem.
Explanation:
Observe that
This will let us keep the coefficients down somewhat.
We will find the expansion of
Multiply by
The final answer is