Roots of Complex Numbers
Key Questions

To evaluate the
#nth# root of a complex number I would first convert it into trigonometric form:
#z=r[cos(theta)+isin(theta)]#
and then use the fact that:
#z^n=r^n[cos(n*theta)+isin(n*theta)]#
and:
#nsqrt(z)=z^(1/n)=r^(1/n)*[cos((theta+2kpi)/n)+isin((theta+2kpi)/n)]#
Where#k=0..n1# For example: consider
#z=2+3.46i# and let us try#sqrt(z)# ;
#z# can be written as:
#z=4[cos(pi/3)+isin(pi/3)]#
So:
#k=0#
#sqrt(z)=z^(1/2)=4^(1/2)[cos((pi/3+0)/2)+isin((pi/3+0)/2)]=#
#=2[cos(pi/6)+isin(pi/6))]#
And:
#k=n1=21=1#
#sqrt(z)=z^(1/2)=4^(1/2)[cos((pi/3+2pi)/2)+isin((pi/3+2pi)/2)]=#
#=2[cos(7pi/6)+isin(7pi/6))]#
Which gives, in total, two solutions. 
Answer:
If you express your complex number in polar form as
#r(cos theta + i sin theta)# , then it has fourth roots:#alpha = root(4)(r)(cos (theta/4) + i sin (theta/4))# ,#i alpha# ,#alpha# and# i alpha# Explanation:
Given
#a+ib# , let#r = sqrt(a^2+b^2)# ,#theta = "atan2"(b, a)# Then
#a + ib = r (cos theta + i sin theta)# This has one
#4th# root#alpha = root(4)(r)(cos (theta/4) + i sin (theta/4))# There are three other
#4th# roots:#i alpha# ,#alpha# and#i alpha# 
A root of unity is a complex number that when raised to some positive integer will return 1.
It is any complex number
#z# which satisfies the following equation:#z^n = 1# where
#n in NN# , which is to say that n is a natural number. A natural number is any positive integer: (n = 1, 2, 3, ...). This is sometimes referred to as a counting number and the notation for it is#NN# .For any
#n# , there may be multiple#z# values that satisfy that equation, and those values comprise the roots of unity for that n.When
#n = 1#
Roots of unity:#1# When
#n = 2#
Roots of unity:#1, 1# When
#n = 3#
Roots of unity =#1, (1 + sqrt(3)i)/2, (1  sqrt(3)i)/2# When
#n = 4#
Roots of unity =#1, i, 1, i# 
To evaluate the square root (and in general any root) of a complex number I would first convert it into trigonometric form:
#z=r[cos(theta)+isin(theta)]#
and then use the fact that:
#z^n=r^n[cos(n*theta)+isin(n*theta)]# Where, in our case,
#n=1/2# (remembering that#sqrt(x)=x^(1/2)# ).
To evaluate the#nth# root of a complex number I would write:#nsqrt(z)=z^(1/n)=r^(1/n)*[cos((theta+2kpi)/n)+isin((theta+2kpi)/n)]#
Where#k=0..n1# For example: consider
#z=2+3.46i# and let us try#sqrt(z)# ;
#z# can be written as:
#z=4[cos(pi/3)+isin(pi/3)]#
So:
#k=0#
#sqrt(z)=z^(1/2)=4^(1/2)[cos((pi/3+0)/2)+isin((pi/3+0)/2)]=#
#=2[cos(pi/6)+isin(pi/6))]#
And:
#k=n1=21=1#
#sqrt(z)=z^(1/2)=4^(1/2)[cos((pi/3+2pi)/2)+isin((pi/3+2pi)/2)]=#
#=2[cos(7pi/6)+isin(7pi/6))]#
Which gives, in total, two solutions.