Let the answer be a complex number 
a+bi with a,b real. Then

(a+bi)� = 2i

a�+2abi+b�i� = 2i

a�+2abi+b�(-1) = 2i

a�+2abi-b�=2i

Equate real parts and equate imaginary
parts:

a�-b� = 0   2abi = 2i
a� = b�       ab = 1
              
So the only possible real solutions
are (a,b) = (1,1) and (a,b) = (-1,-1)

So the two square roots of 2i are

1+i and -1-i    <--answers 

Checking:

(1+i)� = 1+2i+i� = 1+2i+(-1) = 2i

Edwin