Root of iota plus root of minus iota

Answer 1

I assume by iota you mean i, which is √-1

√i = (1+i)/√2
√-i = (1-i)/√2

add them to get 2/√2 = √2

Answer 2

Let root iota &root -iota = y
Now square on both side we get i - i+ 2 i^1/2 -i^1/2
= 2 [-i^2]^1/2
= 2
As y^2 = 2
Then y 2^1/2