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

1 Answer
George C.
Nov 20, 2016

#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]}

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