How do you find all the cube roots of z, where #z = (sqrt3)/2 + 1/2i#?

1 Answer
Jan 6, 2016

#r_1 = cos(pi/18) + i sin(pi/18)#

#r_2 = cos((13pi)/18) + i sin((13pi)/18)#

#r_3 = cos((25pi)/18) + i sin((25pi)/18)#

Explanation:

This is an example of "Casus Irreducibilis" - a Complex cube root that is not expressible in the form #a+bi# where #a# and #b# are expressed in terms of Real #n#th roots.

You can express it in terms of trigonometric functions.

#z# itself lies on the unit circle:

#z = cos(pi/6) + i sin(pi/6)#

Use De Moivre's formula:

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

to deduce:

The cube roots also lie on the unit circle at intervals of #(2pi)/3#:

#r_1 = cos(pi/18) + i sin(pi/18)#

#r_2 = cos((13pi)/18) + i sin((13pi)/18)#

#r_3 = cos((25pi)/18) + i sin((25pi)/18)#