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.
