Question

How do you find the fourth root of `16(cos((4pi)/3)+isin((4pi)/3))`?

Answer 1

`root(4)(16(cos((4pi)/3)+i sin((4pi)/3))) = 2(cos(-pi/6)+i sin(-pi/6)) = sqrt(3)-i`

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 `2(cos(pi/3) + i sin(pi/3)) = 1 + sqrt(3)i` is a fourth root of `16(cos((4pi)/3)+i sin((4pi)/3))`

Is it the fourth root?

No.

Note that `(4pi)/3` is in Q3. So if we are using the normal range for `Arg(z) in (-pi, pi]` then the standard form of `(4pi)/3` is `(-2pi)/3` and the principal fourth root is:

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

which is in Q4.

Here are the four fourth roots of `16(cos((4pi)/3)+isin((4pi)/3))` plotted in the Complex plane...

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]}