I think some of the answer options are incorrect.
Let a+ib=√i, then a2+2aib-b2=i, where a and b are both real.
So, equating real and imaginary parts, 2ab=1 and a2-b2=0, b=±a.
±2a2=1. Therefore, since a is real, a=1/√2 which is the same as √2/2.
If a=1/√2, then b=1/√2 because 2ab>0 (product of a and b must be positive).
Hence √i (implying the positive square root) is (1+i)/√2.
However, if the square root is taken to be positive or negative then:
Square root of i=±(1+i)/√2, or ±(1+i)√2/2, which is not covered by any of the answer options.
CHECK FOR VALIDITY
[±(1+i)/√2]2=(1+2i-1)/2=i✔️