How do you find the fourth root of #16(cos((4pi)/3)+isin((4pi)/3))#?
1 Answer
Explanation:
De Moivre's formula tells us that:
#(cos theta + i sin theta)^n = cos n theta + i sin n theta#
So we find:
#(2(cos(pi/3)+i sin(pi/3)))^4 = 2^4(cos((4pi)/3) + i sin((4pi)/3))#
#color(white)((2(cos(pi/3)+i sin(pi/3)))^4) = 16(cos((4pi)/3) + i sin((4pi)/3))#
So
Is it the fourth root?
No.
Note that
#2(cos(-pi/6)+i sin(-pi/6)) = sqrt(3)-i#
which is in Q4.
Here are the four fourth roots of
graph{((x-1)^2+(y-sqrt(3))^2-0.008)((x-sqrt(3))^2+(y+1)^2-0.008)((x+1)^2+(y+sqrt(3))^2-0.008)((x+sqrt(3))^2+(y-1)^2-0.008) = 0 [-5, 5, -2.5, 2.5]}