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..n-1#
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=n-1=2-1=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.
