What are the cube roots of #4+4sqrt(3)i# ?

1 Answer
George C.
Nov 10, 2017

The three cube roots of #4+4sqrt(3)i# can be written:

#2(cos(pi/9)+isin(pi/9))#

#2(cos((7pi)/9)+isin((7pi)/9))#

#2(cos((13pi)/9)+isin((13pi)/9))#

Explanation:

Note that:

#4+4sqrt(3)i = 8(1/2+sqrt(3)/2i) = 8(cos(pi/3)+i sin(pi/3))#

By de Moivre's formula:

#(cos theta + i sin theta)^n = cos n theta + i sin n theta#

Hence we find:

#(2(cos(pi/9) + i sin(pi/9)))^3 = 8(cos(pi/3)+i sin(pi/3))#

So one of the cube roots is:

#2(cos(pi/9) + i sin (pi/9))#

We can find the other two by multiplying by:

#omega = cos((2pi)/3) + i sin((2pi)/3)#

the primitive complex cube root of #1#

So the other two cube roots are:

#2(cos(pi/9) + i sin (pi/9))(cos((2pi)/3) + i sin((2pi)/3))#

#= 2(cos((7pi)/9)+isin((7pi)/9))#

#2(cos(pi/9) + i sin (pi/9))(cos((4pi)/3) + i sin((4pi)/3))#

#= 2(cos((13pi)/9)+isin((13pi)/9))#

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