Is the square root of "i", i.e.(-1)^(1/4), a number that exist?

i = cos(pi/2) + isin(pi/2)

sqrt(i) = cos(pi/4) + i sin(pi/4) = sqrt(2)/2 + i sqrt(2)/2

if z = r(cos(t) + i sin(t), to take the square root, you take the square root of r (in your case 1) and divide t by 2.

sqrt(i) = .707 + .707i

I agree with rt11guru

"Imaginary" numbers are a misnomer and the term causes much confusion. The complex numbers exist as an "extension" of the reals.

You need to define what a number is. Does the number "1" exist? There is after all no existence of a "1" on the earth - so it too is, in a sense, imaginary (only a construct in our heads)

What I think your question might be is "Is there a number in the complex numbers which is equal to the square root of i"? The answer then yes e.g. solve
(a+bi)^2=i and you'll find one possible solution is 2^(-1/2) + 2^(-1/2)i, just like the first answer from rt11guru.

Imaginary numbers are those which doe not exist. So this one can also be treated as an imaginary number

i = root (-1) itself is an imaginary number...it does not exist. so root(i) definitely doesnot exist.

If you want numbers that are not in the Argand Diagram, you might consider quaternions and Cayley numbers.