$z^2 = i$

By Euler's formula , we have $e^{i\theta} = \cos \theta + i \sin \theta$ . Then,

$\begin{aligned} z^2 & = i & \small \blue{\text{By Euler's theorem}} \\ & = e^{\left(2n + \frac 12\right)\pi i} & \small \blue{\text{where }n \in \mathbb Z} \end{aligned}$

$\implies z = \begin{cases} e^{\frac \pi 4 i} = \cos \frac \pi 4 + i \sin \frac \pi 4 = \dfrac {1+i}{\sqrt 2} & \text{if }n \text{ is even.} \\ e^{\frac {5\pi} 4 i} = \cos \frac {5\pi} 4 + i \sin \frac {5\pi 4} = \dfrac {-1-i}{\sqrt 2} & \text{if }n \text{ is odd.} \end{cases}$

Therefore, $A+B+C+E+D+F = 1 + 1 + 2 + 1 + 1 + 2 = \boxed 8$ .

$z^2=i=e^{\frac{iπ}{2}}\implies z=\pm \dfrac {1+i}{\sqrt 2 }$

So $A=B=D=E=1,C=F=2$ and the required answer is $1+1+2+1+1+2=\boxed 8$ .