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
You can express it in terms of trigonometric functions.
#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
#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)#